Statement I : At STP, 22.4 liters of He will have the same volume as one mole of $$\displaystyle { H }_{ 2 }$$ (assume ideal gases). Statement II : One mole or 22.4 liters of any gas at STP will have the same mass.

true, true, correct explanation

true, true, not correct explanation

MCQ Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 13

A sample of a gas at STP contains

3.01 \times 10^{23}

$3.01 \times 10^{23}$ molecules and has a mass of 22.0 grams.
This gas is most likely :

0%
$CO_2$
0%
$O_2$
0%
$N_2$
0%
CO
0%
NO

Explanation

3.01 \times 10^{23}

$3.01 \times 10^{23}$ molecules

= \frac{3.01 \times 10^{23}}{6.022 \times 10^{23}} = \frac{1}{2} = 0.5 m o l

$=\dfrac {3.01 \times 10^{23}}{6.022 \times 10^{23}}= \dfrac {1}{2} = 0.5 mol$

6.022 \times 10^{23} p a r t i c l e s = 1 m o l

$6.022 \times 10^{23} particles = 1 mol$ .

so if half mol has mass = 22 g, then 1 mol has mass

= 2 \times 22 = 44 g

$= 2\times 22 = 44 g$ .

As 44 g is the molar mass of

C O_{2}

$CO_2$ .

Hence the correct option is A.

0.16 g

$0.16\ g$ of an organic compound containing sulphur produces

0.233 g

$0.233\ g$ of

B a S O_{4}

$BaSO_{4}$ . Percentage of sulphur in the compound is :

0%
$20$
0%
$80$
0%
$50$
0%
$10$

Explanation

The molar masses of barium sulphate and sulphur are

233 g / m o l

$233\ g/mol$ and

32.1 g / m o l

$32.1\ g/mol$ respectively.

Number of moles of

B a S O_{4} = \frac{0.233}{233} = 0.001 m o l

$\displaystyle BaSO_4 = \dfrac {0.233}{233} = 0.001\ mol$

= number of moles of S.

$\displaystyle = \text{number of moles of S.}$

Mass of

S = 0.001 \times 32.1 = 0.0321 g

$\displaystyle S = 0.001 \times 32.1 = 0.0321 \: g$

Percentage of

S = \frac{100 \times 0.0321}{0.16} = 20

$\displaystyle S = \dfrac {100 \times 0.0321}{0.16} = 20$ %

Hence, option A is correct.

Maximum percentage of chlorine is in :

0%
Chloral
0%
Pyrene
0%
Ethylidene chloride
0%
PVC

Explanation

Percentage of chlorine = \frac{Mass of chlorine in the compound}{Mass of compound} \times 100

$\text{Percentage of chlorine} = \dfrac{\text{Mass of chlorine in the compound}}{\text{Mass of compound}}\times 100$

Chloral is

C l_{3} C C H O

$Cl_3CCHO$ .

Percentage

= \frac{106.5}{147.5} \times 100 = 72.2

$=\dfrac{106.5}{147.5}\times100=72.2$

Pyrene is

C C l_{4}

$CCl_4$ .

Percentage

= \frac{142}{154} \times 100 = 92.20

$=\dfrac{142}{154}\times 100=92.20$

Ethylidene chloride is

C_{2} H_{2} C l_{2}

$C_2H_2Cl_2$ .

Percentage

= \frac{71}{97} \times 100 = 73.19

$=\dfrac{71}{97}\times100=73.19$

PVC is

[C H_{2} C H C l]_{n}

$[CH_2CHCl]_n$ .

Percentage

= \frac{35.5}{62.5} \times 100 = 56.8

$=\dfrac{35.5}{62.5}\times100=56.8$

Therefore, the maximum percentage of chlorine is present in pyrene.

Statement I : At STP, 22.4 liters of He will have the same volume as one mole of

H_{2}

$\displaystyle { H }_{ 2 }$ (assume ideal gases).
Statement II : One mole or 22.4 liters of any gas at STP will have the same mass.

0%
true, false
0%
false, true
0%
true, true, correct explanation
0%
true, true, not correct explanation

The system that contains the maximum number of atoms is :

0%
$4.25\ g$ of $N{H}_{3}$
0%
$8\ g$ of ${O}_{2}$
0%
$2\ g$ of ${H}_{2}$
0%
$4\ g$ of $He$

Explanation

Number of atoms

= \frac{m a s s}{m o l a r m a s s} \times N_{A} \times

$=\dfrac { mass }{ molar\ mass } \times { N }_{ A }\times$ Number of atoms in 1 mole

∴

$\therefore$ Number of atoms in

4.25 g N H_{3} = \frac{4.25}{17} \times N_{A} \times 4

$4.25\ g\ N{ H }_{ 3 } =\dfrac { 4.25 }{ 17 } \times { N }_{ A }\times 4$

= N_{A}

$={ N }_{ A }$
Number of atoms in

8 g

$8\ g$

O_{2} = \frac{8}{32} \times N_{A} \times 2 = \frac{N_{A}}{2}

${ O }_{ 2 }=\dfrac { 8 }{ 32 } \times { N }_{ A }\times 2=\dfrac { { N }_{ A } }{ 2 }$
Number of atoms in

2 g

$2\ g$

H_{2} = \frac{2}{2} \times N_{A} \times 2 = 2 N_{A}

${ H }_{ 2 }=\dfrac { 2 }{ 2 } \times { N }_{ A }\times 2=2{ N }_{ A }$
Number of atoms in

4 g

$4\ g$

H e = \frac{4}{4} \times N_{A} \times 1 = N_{A}

$He=\dfrac { 4 }{ 4 } \times { N }_{ A }\times 1={ N }_{ A }$
Thus,

2 g H_{2}

$2\ g\ { H }_{ 2 }$ contains the maximum number of atoms among the given.

What is the percentage of calcium in calcium carbonate

(C a C O_{3})

$(CaCO_3)$ ?

0%
30%
0%
40%
0%
50%
0%
60%

Explanation

Molecular weight of

{C a C O}_{3} = 40 + 12 + 3 \times 16

${ CaCO }_{ 3 }=40+12+3\times 16$

= 52 + 48 = 100 g

$=52+48=100g$

100 g m

$100gm$ of

{C a C O}_{3}

${ CaCO }_{ 3 }$ contains

40 g m

$40gm$ of

C a

$Ca$

∴

$\therefore$ % of

C a

$Ca$ in

{C a C O}_{3}

${ CaCO }_{ 3 }$

= \frac{40}{100} \times 100 = 40

$=\dfrac { 40 }{ 100 } \times 100=40$ %

What mass of

A g I

$AgI$ will dissolve in

1.0 L

$1.0L$ of

1.0 M

$1.0M$

{N H}_{3}

${NH}_{3}$ ? neglect change in conc. of

{N H}_{3}

${NH}_{3}$
(Given :

K_{s p} = 1.5 \times 10^{- 16}; K_{f} [A g {({N H}_{3})}_{2}^{+}]

${K}_{sp}=1.5\times {10}^{-16};{K}_{f}[Ag{({NH}_{3})}_{2}^{+}]$ ; (At.mass

A g = 108, I = 127

$Ag=108,I=127$ )

0%
$4.9\times {10}^{-5}g$
0%
$0.0056g$
0%
$0.035g$
0%
$0.011g$

According to Avogadro's law the correct statement are
a) Volume of gas is directly proportional to the no. of moles at constant T and P.
b) The pressure of gas is directly proportional to temp. of the gas under all conditions.
c) Equal volumes of different gases under similar conditions consist of equal no. of molecules.
d) Equal volumes of different gas under same conditions have equal no. of atoms.

0%
b,c
0%
a,c
0%
d,b
0%
c,d

Natural abundances of

^{12} C

$^{12}C$ and

^{13} C

$^{13}C$ isotopes of carbon are

99 %

$99\%$ and

1 %

$1\%$ , respectively. Assuming they only contribute to the mol. wt. of

C_{2} F_{4}

$C_2F_4$ , the percentage of

C_{2} F_{4}

$C_2F_4$ having a molecular mass of

101

$101$ is:

0%
$1.00$
0%
$98$
0%
$0.198$
0%
$99$

Explanation

The molecular mass of

C_{2} F_{4}

$\displaystyle C_2F_4$ containing C-12 is

2 (12) + 4 (19) = 100

$\displaystyle 2(12)+4(19)=100$ g/mol
The molecular mass of

C_{2} F_{4}

$\displaystyle C_2F_4$ containing one C-12 and one C-13 is

12 + 13 + 4 (19) = 101

$\displaystyle 12+13+4(19)=101$ g/mol
Natural abundances of C-12 and C-13 isotopes of carbon are 99% and 1%, respectively.
100

C_{2} F_{4}

$\displaystyle C_2F_4$ molecules will have 200 C atoms out of whic 1% or 2 C atoms will be C-13. Assuming that these two C-13 atoms will be in different

C_{2} F_{4}

$\displaystyle C_2F_4$ molecules, the percentage of

C_{2} F_{4}

$\displaystyle C_2F_4$ having a molecular mass of 101 is

\frac{2}{200} \times 100 = 1

$\displaystyle \dfrac{2}{200} \times 100 = 1$ %

1 g of silver gets distributed between

10 c m^{3}

$10\ cm^3$ of molten zinc and

100 c m^{3}

$100\ cm^3$ of molten lead at

800^{o} C

$800^o C$ . The percentage of silver in the zinc layer in approximately:

0%
89%
0%
91%
0%
97%
0%
94%

Explanation

Concentration of

A g

$Ag$ is zinc

= \frac{x}{10}

$=\dfrac{x}{10}$

∴

$\therefore$ concentration of

A g

$Ag$ in

P b

$Pb$

= \frac{1 - x}{100}

$=\dfrac{1-x}{100}$

\frac{\frac{x}{10}}{\frac{1 - x}{100}} = 300

$\dfrac{\dfrac{x}{10}}{\dfrac{1-x}{100}}=300$
On solving, we get

x = 97 %

$x=97\%$

How do you determine the mass percent of a solute in a solution?

0%
$\displaystyle \text {Mass percent } = \dfrac { \text { mass of solute} }{ \text { mass of solution }} \times 100$ %
0%
$\displaystyle \text {Mass percent } = \dfrac { \text { mass of solution} }{ \text { mass of solute }} \times 100$ %
0%
$\displaystyle \text {Mass percent } = \dfrac { \text { moles of solute} }{ \text { mass of solution }} \times 100$ %
0%
$\displaystyle \text {Mass percent } = \dfrac { \text { mass of solute} }{ \text { moles of solution }} \times 100$ %
0%
$\displaystyle \text {Mass percent } = \dfrac { \text { moles of solute} }{ \text { moles of solution }} \times 100$ %

Explanation

Mass percent = \frac{mass of solute}{mass of solution} \times 100

$\displaystyle \text {Mass percent } = \dfrac { \text { mass of solute} }{ \text { mass of solution }} \times 100$ %
Mass percentage is the grams of solute present in 100 gram of the solution.

What is the percentage of sulphur in sulphuric acid

(H_{2} S O_{4})

$(H_2SO_4)$ ?

0%
32%
0%
32.45%
0%
32.65%
0%
32.55%

Explanation

Molecular weight of

H_{2} {S O}_{4} = 2 + 32 + 64 = 98 g

${ H }_{ 2 }{ SO }_{ 4 }=2+32+64=98g$

98 g

$98g$ contains

32 g m

$32gm$ of

S

$S$

∴

$\therefore$ % of

S

$S$ in

H_{2} {S O}_{4}

${ H }_{ 2 }{ SO }_{ 4 }$

= \frac{32}{98} \times 100 = 32.65

$=\dfrac { 32 }{ 98 } \times 100=32.65$ %

The density of a liquid is 1.2 g/mL. That are 35 drops in 2 mL. The number of molecules in 1 drop is (molecular weight of liquid = 70):

0%
$\frac{1.2}{35} N_A$
0%
$\left (\frac{1}{35} N_A \right )^2$
0%
$\frac{1.2}{(35)^2} N_A$
0%
$1.2 N_A$

Explanation

Volume of one drop

= \frac{2}{35} m L

$=\frac{2}{35}\, mL$

Moles in one drop

= \frac{2 \times 1.2}{35 \times 70} = \frac{1.2}{(35)^{2}} m o l

$=\frac{2 \times 1.2}{35 \times 70}=\frac{1.2}{(35)^2}\ mol$

Number of molecules in one drop

= \frac{1.2}{(35)^{2}} \times N_{A}

$=\frac{1.2}{(35)^2}\times N_A$

Avogadro's law shows the relationship between which two variables?

0%
Volume and number of moles
0%
Pressure and number of moles
0%
Volume and pressure
0%
Temperature and pressure
0%
Temperature and number of moles

Percentage purity of a sample of gold is

85 %

$85\%$ . How many atoms of gold are present in its

1

$1$ gram sample? (Atomic mass of gold

= 197

$=197$ u).

0%
$3.0\times 10^{21}$
0%
$2.6\times 10^{23}$
0%
$2.6\times 10^{21}$
0%
$4.5\times 10^{26}$

Explanation

Given,

85

$85$ % pure gold sample

That means in

100 g m

$100gm$ sample

85 g m

$85gm$ gold is present

We are given with

1 g m

$1gm$ sample

∴

$\therefore$ Mass of gold=

85 \times 10^{- 2}

$85\times 10^{-2}$

We know,

1

$1$ mole of gold=

197 g m

$197gm$

197 g m

$197gm$ gold=

6.023 \times 10^{23}

$6.023\times 10^{23}$ atoms of gold

∴ 85 \times 10^{- 2}

$\therefore 85\times 10^{-2}$ gm gold contains

\frac{85 \times 10^{- 2} \times 6.023 \times 10^{23}}{197}

$\cfrac {85\times 10^{-2}\times 6.023\times 10^{23}}{197}$ atoms

= 2.598 \times 10^{21}

$=2.598\times 10^{21}$ atoms

= 2.6 \times 10^{21}

$=2.6\times 10^{21}$ atoms

Calculate the mole percentage of

C H_{3} O H

$CH_{3}OH$ and

H_{2} O

$H_{2}O$ respectively in

60

$60$ % (by mass) aqueous solution of

C H_{3} O H

$CH_{3}OH$ .

0%
$45.8, 54.2$
0%
$54.2, 45.8$
0%
$50, 50$
0%
$60, 40$

Explanation

Given,

60

$60$ % by mass solution of

C H_{3} O H

$CH_3OH$ . That means in

100 g m

$100gm$ solution

60 g m

$60gm$

C H_{3} O H

$CH_3OH$ is dissolved in

40 g m

$40gm$

H_{2} O

$H_2O$

∴

$\therefore$ Molar mass of

C H_{3} O H = 32 g / m o l

$CH_3OH=32g/mol$

Molar mass of

H_{2} O = 18 g / m o l

$H_2O=18g/mol$

\Rightarrow

$\Rightarrow$ No. of moles of

C H_{3} O H = \frac{60 g m}{32 g / m o l} = 1.875 m o l

$CH_3OH=\cfrac {60gm}{32g/mol}=1.875\ mol$

. . . . . . . . . . (1)

$..........(1)$

No. of moles of

H_{2} O = \frac{40 g m}{18 g m / m o l} = 2.222 m o l

$H_2O=\cfrac {40gm}{18gm/mol}=2.222\ mol$

. . . . . . . . . . (2)

$..........(2)$

Total no. of moles=

1.875 + 2.222 = 4.097

$1.875+2.222=4.097$ moles

. . . . . . . . . . (3)

$..........(3)$

\Rightarrow

$\Rightarrow$ Mole percentage=

\frac{M o l e s o f c o m p o u n d}{T o t a l n o . o f m o l e s} \times 100

$\cfrac {Moles\ of\ compound}{Total\ no.\ of\ moles}\times 100$

Mole % of

C H_{3} O H = \frac{1.875}{4.097} \times 100

$CH_3OH=\cfrac {1.875}{4.097}\times 100$

= 45.8

$=45.8$ % (From 1 & 3)

Mole % of

H_{2} O = \frac{2.222}{4.097} \times 100

$H_2O=\cfrac {2.222}{4.097}\times 100$

= 54.23

$=54.23$ % (from 2,3)

Hence, the correct option is

A

$A$

The vapour pressure of a solution of a non -volatile electrolyte (A) in a solvent (B) is

95 %

$95\%$ of the vapour pressure of the solvent at the same temperature. If

M_{b} = 0.3 M_{A}

$M_b=0.3 M_A$ , where

M_{B}

$M_B$ and

M_{A}

$M_A$ are molecular weights of B and A respectively, the weight ratio of the solvent and solute are:

0%
0.15
0%
5.7
0%
0.2
0%
4.0

Explanation

Given,

vapour pressure of solution having non volatile solute & solvent =

35

$35$ % of vapour pressure of pure solvent.

At constant temperature

Let,

P^{o}

$P^o$ be vapour pressure of pure solvent

Let

P_{s}

$P_s$ be vapour pressure of solution

\Rightarrow

$\Rightarrow$ If

P^{o} = 100

$P^o=100$ then

P_{s} = 95

$P_s=95$

[∵ g i v e n]

$[\because given]$

95

$95$ % of

100 = 95

$100=95$

From Relative lowering of vapour pressure , we know

\frac{P^{o} - P_{s}}{P^{o}} = x_{s o l u t e}

$\cfrac {P^o-P_s}{P^o}=x_{solute}$

[x_{s o l u t e}

$[ x_{solute}$ is mole fraction of solute

]

$]$

\Rightarrow \frac{5}{100} = x_{s o l u t e}

$\Rightarrow \cfrac {5}{100}=x_{solute}$

\Rightarrow x_{s o l u t e} = 0.05 ⟶ (1)

$\Rightarrow x_{solute}=0.05 \longrightarrow (1)$

Now, we know

x_{s} o l u t e = \frac{N o . o f m o l e s o f s o l u t e}{N o . o f m o l e s o f s o l v e n t}

$x_solute=\cfrac {No.\quad of\quad moles\quad of\quad solute}{No.\quad of\quad moles\quad of\quad solvent}$

\Rightarrow X_{s o l u t e} = \frac{n_{A}}{n_{B}}

$\Rightarrow X_{solute}=\cfrac {n_A}{n_B}$

[∵ n_{A} <<< n B]

$[\because n_A <<< nB]$

\Rightarrow X_{s o l u t e} = \frac{W_{A}}{M_{A}} \div \frac{W_{B}}{M_{B}}

$\Rightarrow X_{solute}= \cfrac {W_A}{M_A}\div \cfrac {W_B}{M_B}$

\Rightarrow X_{s o l u t e} = \frac{W_{A}}{W_{B}} \times \frac{M_{B}}{M_{A}}

$\Rightarrow X_{solute}=\cfrac {W_A}{W_B}\times \cfrac {M_B}{M_A}$

\Rightarrow 0.05 = \frac{W_{A}}{W_{B}} \times 0.3

$\Rightarrow 0.05=\cfrac {W_A}{W_B}\times 0.3$

[∵ M_{B} = 0.3 M_{A}

$[\because M_B=0.3 M_A$ given

]

$]$

\Rightarrow \frac{W_{A}}{W_{B}} = \frac{0.05}{0.3}

$\Rightarrow \cfrac {W_A}{W_B}=\cfrac {0.05}{0.3}$

= 0.166 \approx 0.15

$=0.166\approx 0.15$

Two vessels of volumes

16.4 L

$16.4\ L$ and

5 L

$5\ L$ contains two ideal gases of molecular existence at the respective temperature of

27^{\circ} C

$27^{\circ}C$ and

227^{\circ} C

$227^{\circ}C$ and exert

1.5

$1.5$ and

4.1

$4.1$ atm, respectively. The ratio of the number of molecules of the former to that of the later is:

0%
$2$
0%
$1$
0%
$\dfrac {1}{2}$
0%
$\dfrac {1}{3}$
0%
$3$

Explanation

From the ideal gas equation we have,

\frac{p_{1} V_{1}}{n_{1} T_{1}} = \frac{p_{2} V_{2}}{n_{2} T_{2}} = R

$\dfrac {p_{1}V_{1}}{n_{1}T_{1}} = \dfrac {p_{2}V_{2}}{n_{2}T_{2}}=R$

\frac{n_{1}}{n_{2}} = \frac{p_{1} V_{1} T_{2}}{p_{2} V_{2} T_{1}} = \frac{1.5 \times 16.4 \times 500}{4.1 \times 5 \times 300} = 2

$\dfrac {n_{1}}{n_{2}} = \dfrac {p_{1}V_{1}T_{2}}{p_{2}V_{2}T_1} = \dfrac {1.5\times 16.4\times 500}{4.1\times 5\times 300} = 2$

∴ n_{1} : n_{2} = 2 : 1

$\therefore n_{1} : n_{2} = 2 : 1$ .

The mass of one twelfith of the mass of one carbon 12 atoms is

0%
$1.68 \times 10^{-24}g$
0%
1.008 g
0%
-12 g
0%
$3.66 \times 10^{24}g$

How many hydroxide ions are present in 2.5 mol of

C a (O H)_{2}

$Ca(OH)_2$ ?

(N_{A} = 6 \times 10^{23})

$(N_A=6\times10^{23})$

0%
$3\times10^{23}$
0%
$6\times10^{23}$
0%
$1.5\times10^{24}$
0%
$3\times10^{24}$

A 1.85 g sample of an arsenic-containing pesticide was chemically converted to

A s O_{4}^{3 -}

$AsO_4^{3-}$ (atomic mass of As = 74.9) and titrated with

P b^{2 +}

$Pb^{2+}$ to form

P b_{3} (A s O_{4})_{2}

$Pb_3(AsO_4)_2$ . If 20 mL of 0.1 M

P B^{2 +}

$PB^{2+}$ is required to reach the equivalence point, the mass percentages of arsenic in the pesticide sample is closest to

0%
8.1
0%
2.3
0%
5.4
0%
3.6

Explanation

Given, Mass of sample=

1.85 g

$1.85g$

The reaction involved is

3 P b^{2 +} + 2 A s O_{4}^{3 -} ⟶ P b_{3} (A s O_{4})_{2}

$3Pb^{2+}+2AsO_4^{3-}\longrightarrow Pb_3(AsO_4)_2$

Now, Volume of

P b^{2 +}

$Pb^{2+}$ consumed for neutralization=

20 m l

$20ml$

Molarity of

P b^{2 +} = 0.1 M

$Pb^{2+}=0.1M$

∴

$\therefore$ No. of moles of

P b^{2 +} = 0.1 \times \frac{20}{1000}

$Pb^{2+}=0.1\times \cfrac {20}{1000}$

= 2 \times 10^{- 3}

$=2\times 10^{-3}$ moles

Now, from reaction it is clear that

3

$3$ moles

P b^{2 +}

$Pb^{2+}$ produce

1

$1$ mole

P b_{3} (A s O_{4})_{2}

$Pb_3(AsO_4)_2$

We know

1

$1$ mole

P b_{3} (A s O_{4})_{2}

$Pb_3(AsO_4)_2$ has

1

$1$ mole

A s

$As$ in it.

\Rightarrow 3

$\Rightarrow 3$ moles

P b^{2 +}

$Pb^{2+}$ produce

1

$1$ mole

A s

$As$

∴ 2 \times 10^{- 3}

$\therefore 2\times 10^{-3}$ moles

P b^{2 +}

$Pb^{2+}$ will give

\frac{2 \times 10^{- 3}}{3}

$\cfrac {2\times 10^{-3}}{3}$ moles of

A s

$As$

∴

$\therefore$ No. of moles of

A s

$As$ in the compound=

0.66 \times 10^{- 3}

$0.66\times 10^{-3}$ moles

Now,

We know,

1

$1$ mole

A s = 74 g

$As=74g$

\Rightarrow 0.66 \times 10^{- 3}

$\Rightarrow 0.66\times 10^{-3}$ moles

A s = 0.66 \times 10^{- 3} \times 74 g

$As=0.66\times 10^{-3}\times 74g$

= 48.84 \times 10^{- 3} g

$=48.84\times 10^{-3}g$

= 0.0488 g

$=0.0488g$

Now % calculation,

Mass% of

A s = \frac{M a s s o f A s}{M a s s o f S a m p l e} \times 100

$As=\cfrac {Mass\quad of\quad As}{Mass\quad of\quad Sample}\times 100$

= \frac{0.0488}{1.85} \times 100

$=\cfrac {0.0488}{1.85}\times 100$

= 0.0263 \times 100

$=0.0263\times 100$

= 2.6

$=2.6$ %

\approx 2.3

$\approx 2.3$ %

20 ml of

H_{2} O_{2}

$H_2O_2$ after acidification with dil

H_{2} S O_{4}

$H_2SO_4$ required 30 ml of N/2

K M n O_{4}

$KMnO_4$ for complete oxidation .Calculate the

%

$\%$ of

H_{2} O_{2}

$H_2O_2$ in g/lit.

0%
10.75 g/lit
0%
11.75 g/lit
0%
12.75 g/lit
0%
13.75 g/lit

Explanation

Given,

Volume of

H_{2} O_{2} = 20 m l

$H_2O_2=20ml$

Volume of

K M n O_{4} = 30 m l

$KMnO_4=30ml$

Normality of

K M n O_{4} = \frac{1}{2} N

$KMnO_4=\cfrac {1}{2}N$

Now,

We know at equivalence

Gram equivalent Weight of

H_{2} O_{2}

$H_2O_2$ = Gram equivalent weight of

K M n O_{4}

$KMnO_4$

i.e.

\frac{W t_{H_{2} O_{2}}}{E q . W t ._{H_{2} O_{2}}} = N_{K M n O_{4}} \times V_{K M n O_{4}}

$\cfrac {Wt_{H_2O_2}}{Eq.Wt._{H_2O_2}}=N_{KMnO_4}\times V_{KMnO_4}$

= \frac{1}{2} N \times 0.03 L

$=\cfrac {1}{2}N\times 0.03L$

= 0.015

$=0.015$

\Rightarrow

$\Rightarrow$ Weight of

H_{2} O_{2} = 0.015 \times E q . W t o f H_{2} O_{2}

$H_2O_2=0.015\times Eq.Wt of H_2O_2$

= 0.015 \times \frac{34}{2}

$=0.015\times \cfrac {34}{2}$

= 0.255

$=0.255$

Now, Weight/Volume % of

H_{2} O_{2} = \frac{W e i g h t o f H_{2} O_{2}}{V o l u m e o f H_{2} O_{2}} \times 100

$H_2O_2=\cfrac {Weight\quad of\quad H_2O_2}{Volume\quad of\quad H_2O_2}\times 100$

= \frac{0.255}{20 \times 10^{- 3}} \times 100

$=\cfrac {0.255}{20\times 10^{-3}}\times 100$

= 12.75

$=12.75$ %

If the molar concentration of

P b l_{2}

$Pbl_2$ is then the

1.5 \times 10^{- 3} m o l L^{- 1}

$1.5 \times 10^{-3}\,\, mol \, L^{-1}$ concentration of iodide ions in g ion

L^{- 1}

$L^{-1}$ is:

0%
$3.0 \times 10^{-3}$
0%
$6.0 \times 10^{-3}$
0%
$0.3 \times 10^{-3}$
0%
$0.6 \times 10^{-3}$

Explanation

Given,

[P b I_{2}] = 1.5 \times 10^{- 3}

$[PbI_2]=1.5\times 10^{-3}$

m o l L^{- 1}

$mol L^{-1}$

The reaction involved here is

P b I_{2} (s) ⇄ P b^{2 +} (a q) + 2 I^{-} (a q)

$PbI_2(s) \rightleftarrows Pb^{2+} (aq)+2I^{-}(aq)$

t = 0

$t=0$

1.5 \times 10^{- 3}

$1.5\times 10^{-3}$

0

$0$

0

$0$

At equilibrium

1.5 \times 10^{- 3} - x

$1.5\times 10^{-3}-x$

x

$x$

2 x

$2x$

∴

$\therefore$ According to law of conservation of mass

1.5 \times 10^{- 3} - x = x + 2 x

$1.5\times 10^{-3}-x=x+2x$

1.5 \times 10^{- 3} = 4 x

$1.5\times 10^{-3}=4x$

∴ x = 0.375 \times 10^{- 3}

$\therefore x=0.375\times 10 ^{-3}$

∴ x =

$\therefore x=$ Concentration of

P b^{2 +}

$Pb^{2+}$ &

2 x =

$2x=$ Concentration of

I^{-}

$I^-$

∴ 2 x = 0.6 \times 10^{- 3}

$\therefore 2x=0.6\times 10^{-3}$

∴

$\therefore$ Concentration of Iodide ion=

0.6 \times 10^{- 3} g L^{- 1}

$0.6\times 10^{-3}g L^{-1}$

Volume of

O_{2}

$O_{2}$ remained after the first combustion?

0%
$4\ ml$
0%
$2\ ml$
0%
$0$
0%
$8\ ml$

The percentage of

S e

$Se$ in peroxidase anhydrous enzyme is

0.5

$0.5$ % by weight (atomic weight

= 78.4

$= 78.4$ ). Then minimum molecular weight of peroxidase anhydrous enzyme is:

0%
$1.568\times 10^{4}$
0%
$1.568\times 10^{3}$
0%
$15.68$
0%
$3.136\times 10^{4}$

Explanation

Correct Option: A

Hint: $Concentration_{ByMass}$ $Percentage = \dfrac{Mass_{solute}}{Mass_{solvent}}\times 100$

Explanation

Step 1: Preparing data

Let mass of $Peroxidase$ $anhydrous$ $enzyme = x$ $gram$ ...(1)
Mass percentage of $Se = 0.5 \%$ ...(2)
Atomic mass of $Se = 78.4g$ ...(3)
Therefore, the minimum amount of an element which can be present in a substance is $1$ atom ...(4)
Thus, minimum mass that can be present is atomic mass.
Therefore, minimum molecular weight of peroxidase anhydrous enzyme is that which contains mass of $Se = Atomic$ $Mass$

Step 2: Solving using data

As $Concentration_{ByMass}$ $Percentage = \dfrac{Mass_{solute}}{Mass_{solvent}}\times 100$ ...(5)
Putting value from (1), (2), (3) and (4) in (5), we have
$0.5 = \dfrac{78.4}{x}\times 100$
$x = \dfrac{784}{5} \times 100 = 1.568 \times 10^4$

Hence, minimum molecular mass of $Se$ present in $peroxidase$ $anhydrous$ $enzyme$ is $1.568 \times 10^4$

The mass of a mixture containing HCl and

H_{2} S O_{4}

$H_2SO_4$ is 0.1 g. On treatment with an excess of an

A g N O_{3}

$AgNO_3$ solution, reacted with this acid mixture gives 0.1435 g of AgCl. Mass % of the

H_{2} S O_{4}

$H_2SO_4$ mixture is :

0%
30
0%
60
0%
59
0%
None of these

During the preparation of

H_{2} S_{2} O_{8}

$H_{2}S_{2}O_{8}$ (per disulphuric acid)

O_{2}

$O_{2}$ gas also releases at anode as by product. When

9.72 L

$9.72\ L$ of

H_{2}

$H_{2}$ releases at cathode and

2.35 L O_{2}

$2.35\ L\ O_{2}$ at anode at STP, the weight of

H_{2} S_{2} O_{8}

$H_{2}S_{2}O_{8}$ produced in gram is:

0%
$87.12$
0%
$43.56$
0%
$83.42$
0%
$51.74$

The amonia evolved from 2g of a compound in Kjeldahl's estimation of nitrogen neutralizes 10 mL of 2 M ${ H }_{ 2 }S{ O }_{ 4 }$ solution. The weight percentage of nitrogen in the compound is

0%
$28$
0%
$14$
0%
$56$
0%
$7$

The common isotopes of carbon are

^{12} C

$^{ 12 }C$ and

^{13} C

$^{ 13 }C$ . The average mass of carbon is

12.01115

$12.01115$ amu. What is the abundance of

^{13} C

$^{ 13 }C$ isotope ?.

0%
$1.115\%$
0%
$98.885\%$
0%
$0.480\%$
0%
$99.52\%$

A gas mixture contains

50 %

$50\%$ helium and

50 %

$50\%$ methane by volume. What is the percentage by mass of methane in the mixture?

0%
$20\%$
0%
$30\%$
0%
$50\%$
0%
$80\%$

224 m l

$224ml$ of triatomic gas has a mass of

1 g

$1g$ at

273 K

$273K$ and

1 a t m

$1atm$ pressure, then the mass of one atom is-

0%
$8.30\times {10}^{-23}gm$
0%
$2.08\times {10}^{-23}gm$
0%
$5.53\times {10}^{-23}gm$
0%
$6.24\times {10}^{-23}gm$

Explanation

1

$1$ mole gas occupies

22.4

$22.4$ litres

= 22, 400

$= 22,400$ ml at STP condition.

224

$224$ ml will then be occupied by

0.01

$0.01$ mol

The weight of 0.01 mol

= 1

$= 1$ g

1 mole of the given gas weighs

= 100

$= 100$ g

6.023 \times 10^{23} \times 3

$6.023 \times 10^{23} \times 3$ (since its trimolecular) atoms has weight of 100 g.

Mass of one atom

= \frac{100 g}{6.023 \times 10^{23} \times 3}

$= \dfrac {100g}{6.023 \times 10^{23} \times 3}$

= 0.0553 \times 100 g = 5.53 \times 10^{- 23} g

$= 0.0553 \times 100g = 5.53 \times 10^{-23}g$

Hence, option C is correct.

A 5 mL (specific gravity 1.02) of chlorine water is treated with an excess of KI.The liberate iodine required 26 mL of

0.15 M - N a_{2} S O_{4}

$0.15 \,M-Na_2SO_4$ solution.Calculate the percentage of

C l_{2}

$Cl_2$ (by weight) the chloride water.Chloride water is a solution of free chlorine in water

0%
$0.388 \%$
0%
$0.301 \%$
0%
$0.307 \%$
0%
$3.02 \%$

0.3 g

$0.3\ g$ of an oxalate salt was dissolved in

100 m L

$100\ mL$ solution. The solution required

90 m L

$90\ mL$ of

N / 20 K M n O_{4}

$N/20\ KMnO_4$ for complete oxidation.The % of oxalate ion in salt is:

0%
$33\%$
0%
$66\%$
0%
$70\%$
0%
$40\%$

Explanation

Reaction taking between potassium permanganate and axalate-

M n O_{4}^{-} + C_{2} O_{4}^{2 -} + H^{+} \to M n^{2 +} + C O_{2} + H_{2} O

$MnO_4^- +C_2O_4^{2-} + H^+ \to Mn^{2+}+CO_2 +H_2O$

Hence equivalent of

K M n O_{4}

$KMnO_4$ oxidizes

2.5

$2.5$ equivalences of axalate

Moles of

K M n O_{4} = 0.05 \times \frac{0.09}{5} = 9 \times 10^{- 4} . . . . . (1)

$KMnO_4 =0.05\times \dfrac {0.09}{5}=9\times 10^{-4} .....(1)$

Moles of oxalate that

M K n O_{4}

$MKnO_4$ axidizes

= 9 \times 10^{- 4} \times 2.5

$=9\times 10^{-4} \times 2.5$

= 22.5 \times 10^{- 4}

$=22.5\times 10^{-4}$ (from

(1)

$(1)$ )

. . . . . . (2)

$......(2)$

Molecolar weight of oxalate

= 2 \times 12 + 4 \times 16 = 88

$=2\times 12+4\times 16=88$

(

∵ C_{2} O_{4}

$\because C_2 O_4$ , where

C = 12

$C=12$ &

O = 16

$O=16$ )

. . . . . . (3)

$......(3)$

Oxalate ion amount

= 22.5 \times 10^{- 4} \times 88 = 0.198

$=22.5\times 10^{-4}\times 88 =0.198$ from

(2)

$(2)$ &

(3)

$(3)$

Percentage of oxalate ion

= \frac{0.198 \times 100}{0.3} = 66 %

$=\dfrac {0.198\times 100}{0.3}=66\%$

Ans

B) 66 %

$B)\ 66\%$

5 g of a sample of bleaching powder is treated with excess acetic acid and KI solution. The liberated

I_{2}

$I_2$ required 50 ml of N/10 hypo solution. The % available chlorine in sample is

0%
3.55
0%
7.0
0%
35.5
0%
28.2

1.85 g

$1.85\ g$ sample of an arsenic-containing pesticide was chemically converted to

A s O_{4}^{3 -}

$AsO_{4}^{3-}$ (atomic mass of

A s = 74.9)

$As = 74.9)$ and titrated with

P b^{2 +}

$Pb^{2+}$ to form

P b_{3} (A s O_{4})_{2}

$Pb_{3}(AsO_{4})_{2}$ . If

20 m L

$20\ mL$ of

0.1 M P b^{2 +}

$0.1\ M\ Pb^{2+}$ is required to reach the equivalence point, the mass percentage of arsenic in the pesticide sample is closest to:

0%
$8.1$
0%
$2.3$
0%
$5.4$
0%
$3.6$

Choose the correct answers from the alternatives given :
Age of fossil may be found out by determining the ratio of two isotopes of carbon. The isotopes are ____________________.

0%
C-12 and C-13
0%
C-13 and C-14
0%
C-12 and C-14
0%
C-12 and carbon black

Which one of the following is not required for the formation of photochemical smog?

0%
Nitrogen oxide
0%
Methane
0%
Volatile organic compounds
0%
Ozone

4.0 g of a mixture of

N a C l

$NaCl$ and an unknown metal iodide

M I_{2}

$MI_{2}$ was dissolved in a water to form its aqueous solution. To this aqueous solution, The aqueous solution

A g N O_{3}

$AgNO_{3}$ was added gradually so that silver halides are precipitated. The precipitates were weighed at regular interval and the given curve for the mass of precipitate versus volume of

A g N O_{3}

$AgNO_{3}$ added was obtained. With the knowledge of the fact that halides are precipitated successively i.e. when one halide is precipitating , the other halide remains in the solution, answer the following questions:
(Molar mass of

A g = 108, I = 127, N a = 23

$Ag = 108,\ I = 127,\ Na = 23$ ).
Points

P

$P$ and

Q

$Q$ are:

P (20, 4.8), Q (37, 7.25)

$P(20, 4.8),\ Q(37, 7.25)$
What is the approximate mass percentage of

M I_{2}

$MI_{2}$ ?

0%
25
0%
40
0%
60
0%
75

Explanation

Mass of

A g C l

$\mathrm{AgCl}$ (formed in

2^{nd}

$2^{\text {nd }}$ step

)

$)$

= 7.25 - 4.80 = 2.45 g

$=7.25-4.80=2.45 \mathrm{~g}$

∵ 143.5 g A g C l

$\because 143.5 \mathrm{~g} \mathrm{AgCl}$ form

58.5 g

$58.5 \mathrm{~g}$ Nacl

2.45 g

A g C l

$\mathrm{AgCl}$ form

\frac{58.5}{143.5} \times 2.45

$\frac{58.5}{143.5}\times 2.45$

= 1.0 g N a C l

$=1.0 \mathrm{~g} \mathrm{NaCl}$

Mass of

M I_{2} = 4 - 1 = 3 g

$M I_{2}=4-1=3 g$

\begin{matrix} % of M I_{2} & = \frac{3}{4} \times 200 = 75 % \end{matrix}

$\begin{aligned}\% \text { of } M I_{2} &=\frac{3}{4} \times 200 \\&=75 \%\end{aligned}$

∴

$\therefore$ (D) is correct coption.

0.5g sample of a sulphite salt was dissolved in 20ml solution and 20ml of this solution required 10ml of 0.002M acidified permanganate solution. Hence, the

%

$\%$ by mass of sulphite in the sulphite salt is:

0%
$2\%$
0%
$4\%$
0%
$6\%$
0%
$8\%$

2

$2$ g of brass containing

C u

$Cu$ and

Z n

$Zn$ only reacts with

3

$3$ M

H N O_{3}

$HNO_3$ solution. Following are the reactions taking place.

C u (s) + H N O_{3} (a q) \to C u^{2 +} (a q) + N O_{2} (g) + H_{2} O (l)

$Cu(s)+HNO_3(aq)\rightarrow Cu^{2+}(aq)+NO_2(g)+H_2O(l)$

Z n (s) + H^{+} (a q) + {N O}_{3}^{-} (a q) \to N H_{4}^{+} (a q) + Z n^{2 +} (a q) + H_{2} O (l)

$Zn(s)+H^+(aq)+{NO}^-_3(aq)\rightarrow NH^+_4(aq)+Zn^{2+}(aq)+H_2O(l)$

The liberated

N O_{2} (g)

$NO_2(g)$ was found to be

1.04

$1.04$ L at

25^{o}

$25^o$ C and

1

$1$ atm

[C u = 63.5, Z n = 65.4]

$[Cu =63.5, Zn=65.4]$ .

How many grams of

N H_{4} N O_{3}

$NH_4NO_3$ will be obtained in the reaction?

0%
$0.405$ g
0%
$0.0428$ g
0%
$0.2018$ g
0%
$0.358$ g

Explanation

Balance equation

$\text { Balance equation }$

Cu + 4 H N O_{3} ⟶ C u {(N O_{3})}_{2} + 2 N O_{2} + 2 H_{2} O

$\text { Cu }+ 4 \mathrm{H} \mathrm{N} \mathrm{O}_{3} \longrightarrow \mathrm{Cu}\left(\mathrm{N} \mathrm{O}_{3}\right)_{2}+2 \mathrm{~N} \mathrm{O}_{2}+2 \mathrm{H}_{2} \mathrm{O}$

P V = n R T

$P V= n R T$

\begin{matrix} V = 1.04 L P = a t m T = 25^{\circ} C = 290 K R = 0.0821 . atm / {m o l}^{- 1} k^{- 1} \end{matrix}

$\begin{array}{l} V=1.04 L \\ P=atm \\ T=25^{\circ} C=290 K \\ R=0.0821 \text { . atm } / \mathrm{mol}^{-1} \mathrm{k}^{-1} \end{array}$

\begin{matrix} moles of {N O}_{2} & = \frac{1 \times 1.04}{0.0821 \times 298} = 0.0425 \end{matrix}

$\begin{aligned} \text { moles of } \mathrm{NO}_{2} &=\dfrac{1 \times 1.04}{0.0821 \times 298} \\ &=0.0425 \end{aligned}$

\begin{matrix} moles of C u = \frac{1}{2} \times & mole of N O_{2} = & \frac{1}{2} \times 0.0425 = 0.02125 \end{matrix}

$\begin{aligned} \text { moles of } C u=\dfrac{1}{2} \times & \text { mole of } NO_2 \\ =& \dfrac{1}{2} \times 0.0425 \\ &=0.02125 \end{aligned}$

$\begin{aligned} \text { weight of } Cu &=\operatorname{mole} x \text { molecular weight } \\ &=0.02125 \times 63.5 . \\ &=1.3496 g \end{aligned}$

$\begin{aligned} \text { weight of } Zn=& \text { weight of brass - weight of Cu} \\ =& 2-1.3496 \\ &=0.6504 \end{aligned}$

$\text { more of } Z n=\dfrac{0.6504}{65.4}$

$\text { moles of } \mathrm{N} \mathrm{H}_4\mathrm{~N}O_3 \mathrm{~L}=\dfrac{2}{1} \dfrac{1}{4} \times \text { mole of } Z n$

$=\dfrac{1}{2} \times \dfrac{0.6504}{65.4}=2.6 \times 10^{-3}$

$\begin{aligned} \text { weight of } NH_4NO_3 &=\text { mole } \times \text { molecular weight } NH_4NO_3 \\ &=2.6 \times 10^{-3} \times 80 \\ &=0.2018 \mathrm{~g} \end{aligned}$

When a mixture of

$NaBr$ and

$NaCl$ is repeatedly digested with sulphuric acid, all the halogens are expelled and

$Na_{2}SO_{4}$ is formed quantitatively. With a particular mixture, it was found that the weight of

$Na_{2}SO_{4}$ obtained was precisely the same as the weight of

$NaBr - NaCl$ mixture taken. Calculate the ratio of the weights of

$NaCl$ and

$NaBr$ in the mixture.

0%
$1:1.454$
0%
$2:1.31$
0%
$1.454 : 1$ .
0%
$1.31:2$

Explanation

$\begin{array}{l} \text { NaBr + } \quad \mathrm{H}_{2} \mathrm{SO}_{4} \quad \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4} \\ \text { a } \mathrm{g} \\ \mathrm{NaCl +} \quad \mathrm{H}_{2} \mathrm{SO}_{4} \quad \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4} \\ \text { b } \mathrm{g} \end{array}$

$\begin{aligned} \text { Meq of } \mathrm{NaBr}+\text { meq of } \mathrm{NaCl} &=\text { meg of } \mathrm{Na}_{2} \mathrm{SO}_{4} \\ \text { meq of } \mathrm{Na}_{2} \mathrm{SO}_{4} \text { formed }=&(9 / 103) \times 1000+(\mathrm{b} / \mathrm{} 8.5) \times 1000 \\ =& 9.71 \mathrm{a}+17.09 \mathrm{~b} \\ & \Rightarrow \frac{\omega}{142 / 2} \times 1000=9.71 \mathrm{a}+17.09 \mathrm{~b} \\ & \Rightarrow \quad \omega=\frac{142}{2000}(9.71 \mathrm{a}+17.09 \mathrm{~b}) \end{aligned}$

$\begin{array}{l} \text { Also given, } \\ \text { mass of } \mathrm{NaCl}+\text { mass of } \mathrm{NaBr}=\text { mass of } \mathrm{Na}_{2} \mathrm{SO}_{4} \text { formed. } \\ \qquad \begin{aligned} a+b &=\frac{142}{2000}(9.7 a+17.09 \mathrm{~b}) \\ \frac{b}{a}=\frac{1.455}{1} &=1.455: 1 \end{aligned} \\ (\mathrm{C}) \text { is correct option } \end{array}$

Calculate the mass percent (w/w) of sulphuric acid in a solution prepared by dissolving

$4$ g of sulphur trioxide in a

$100$ ml sulphuric acid solution containing

$80$ mass percent (w/w) of

$H_2SO_4$ and having a density of

$1.96\ g/ml$ (molecular weight of

$H_2SO_4=98$ ) Take reaction

$SO_3+H_2O\rightarrow H_2SO_4$ .

0%
$80.8\%$
0%
$84\%$
0%
$41.65\%$
0%
None of these

Explanation

$SO_3 + H_2O \rightarrow H_2SO_4$

100 ml sulphuric acid solution

density =

$\dfrac{Mass}{volume}$

$1.96 \, g/ml = \dfrac{Mass}{100}$

Mass

$= 1.96 g$

Amount of water

$= 196 \times 20$

$= 39.2 g$

Amount of

$H_2SO_4 = 156.8 g$

$SO_3 + H_2O \rightarrow H_2 SO_4$

Molar mass of

$SO_3 = 32 + 16 \times 3 = 80$

No. of mole =

$49$

$80 g = 0.05 mol$

0.05 mol of sulphuric acid formed

Now total amount of sulphuric acid

$= 0.05 \times 98 + 156.8$

$= 4.9 + 156.8$

$= 161.7$

total amount of water

$= 39.2 -$ total water consume

$= 39.2 - 0.05 \times 18$

$=39.2 - 0.9 = 38.3$

Mass percent (w/w)

$= \dfrac{161.7}{161.7 + 383} = \dfrac{161.7}{200} \times 100 \%$

$= 80.85\%$

$\approx 80.8\%$

Option

$A$ is correct.

Mole fraction of ethyl alcohol in aqueous ethyl alcohol

$(C_2H_5OH)$ solution is

$0.25$ . Hence percentage of ethyl alcohol by weight is:

0%
$54\%$
0%
$25\%$
0%
$75\%$
0%
$46\%$

Explanation

If ethanol is 0.25, then H

$_2$ O will be 0.75.

Mol wt of ethanol is 46.06 g/mol,

So, 0.25 mol=11.515g of ethanol

Mol wt of H

$_2$ O is 18.01528g/mol,

So, 0.75 mol=13.51g of H

$_2$ O

Total wt=25.02646 g

wt % of etanol is

$={\dfrac{ 11.515}{25.02646}} \times 100 = 46.011 \approx 46$ %

Hence, the correct option is

$D$

32 g of a sample of

$FeSO_4$ .

$7H_2O$ were dissolved in dilute sulphuric acid and water and its volume was made up to 1 litre. 25 mL of this solution required 20 mL of 0.02 M

$KMnO_4$ solution for complete oxidation. Calculate the mass % of

$FeSO_4$ .

$7H_2O$ in the sample.

0%
34.75
0%
69.5
0%
89.5
0%
None of these

Explanation

$Fe^{+2}\longrightarrow Fe^{+3}$

$n_{factor}=1$

Comparing equivalents of

$FeSO_4\cdot 7H_2O$ and

$KMnO_4$

Let molarity of

$FeSO_4\cdot 7H_2O=M$

$25\times M\times 1=20\times 0.02\times 5$

$M=0.08$

$Molar$

No. of moles of

$FeSO_4\times 7H_2O=0.08$

$moles$ (Volume

$=1$ Litre)

Mass of

$FeSO_4\cdot 7H_2O=0.08\times 278=22.24$

$gm$

$Mass\%=\cfrac{22.24}{32}\times 100=69.5\%$

What is the mass percentage of carbon tetrachloride if

$22g$ of benzene is dissolved in

$122g$ of carbon tetrachloride?

0%
$84.72$ %
0%
$15.28$ %
0%
$50$ %
0%
$44$ %

Explanation

Given:-

Mass of

$C_6H_6 = 22g$

Mass of carbon tetrachloride

$CCl_4 = 122g$

Formula:-

Mass % of benzene

$=\dfrac{Mass \ of \ solute }{Total\ mass\ of\ solution }\times 100$

Mass of solute

$=\dfrac{22}{22 + 122}\times 100= 15.28\%$

Mass percentage of

$CCl_4 = 100 -$ mass % of

$C_6H_6$

$= 100 - 15.28 =84.72\%$

What is the mass per cent of oxygen in ethanol?

0%
$52.14\%$
0%
$13.13\%$
0%
$16\%$
0%
$34.73\%$

Explanation

Ethanol is

$C_2H_6O$

molar mass of ethanol =46 g

Mass % of oxygen =

$100 \times \dfrac{(mass \ of\ oxygen\ in\ the\ molecule)}{(molar\ mass\ of\ molecule)}$

$\therefore$ m% of oxygen

$=100 \times (16/46)=34.78$ %

Hence, option

$D$ is correct.

0.2 g of an organic compound contains

$C,\ H,\ O$ on combustion, it yields 0.15 g of

$CO_2$ and 0.12 g

$H_2O$ . The percentage of

$C,\ H$ and

$O$ respectively is:

0%
$C=15$ %, $H= 20$ %, $O=65$ %
0%
$C=10$ %, $H=8.2$ %, $O=81.8$ %
0%
$C=12.2$ %. $H=8.8$ %, $O=79$ %
0%
$C=20$ %, $H=6.66$ %, $O=73.34$ %

Explanation

Weight of organic compound = 0.2g

Weight of

$C{O}_{2}$ = 0.15g

Weight of

${H}_{2}O$ = 0.12g

Molecular weight of

$C{O}_{2}$ = 44

Molecular weight of

${H}_{2}O$ = 18

Atomic weight of carbon = 12u

Atomic weight of

${H}_{2}$ = 2u

$\text{Percentage of carbon} = \cfrac{\text{Atomic weight of carbon}}{\text{Molecular weight of }C{O}_{2}} \times \cfrac{\text{Weight of }C{O}_{2}}{Weight of organic compound} \times 100$

$\text{Percentage of carbon} = \cfrac{12}{44} \times \cfrac{0.15}{0.2} \times 100 = 20.45 \% \approx 20 \%$

$\text{Percentage of Hydrogen} = \cfrac{\text{Atomic weight of Hydrogen}}{\text{Molecular weight of }{H}_{2}O} \times \cfrac{\text{Weight of }{H}_{2}O}{Weight of organic compound} \times 100$

$\text{Percentage of Hydrogen} = \cfrac{2}{18} \times \cfrac{0.12}{0.2} \times 100 = 6.66 \%$

Percentage of oxygen = 100 - (Percentage of carbon + Percentage of hydrogen) = 100 - (20 + 6.66) = 100 - 26.66 = 73.34

$\%$

Flask I contains a gaseous mixture of

$He(g)$ ,

$H_2(g)$ , in the molar ratio of

$1:\sqrt{2}$ . The gas mixture is allowed to effuse through a fine orifice in the flask l and is collected in flask ll. What is the average molar mass of gas mixture initially collected in flask ll?

0%
3.20
0%
2.33
0%
4.00
0%
None of these

Explanation

$(A)$ Given

$\dfrac{\ ^{n}H_{e}}{\ ^{n}H_{2}}=\dfrac{1}{\sqrt{2}}\Rightarrow \ ^{n}H_{e}=\dfrac{\ ^{n}H_{2}}{\sqrt{2}}$

Let

$\ ^{n}H_{2}=x$ then

$\ ^{n}H_{e}=\dfrac{x}{\sqrt{2}}$

mole fraction of

$H_{2}=\dfrac{x}{\dfrac{x}{\sqrt{2}}+x}=\dfrac{\sqrt{2}}{1+\sqrt{2}}=Y_{A}$

mole fraction of

$H_{e}=\dfrac{1}{1+\sqrt{2}}=Y_{B}$

arg molecular weight

$(\bar{M}w)=M_{H_{e}}Y_{B}+M_{H_{2}}Y_{A}$

32g of a sample of

${ FeSO }_{ 4 }.{ 7H }_{ 2 }O$ where the dissolved in dilute Sulphuric acid and water and it's volume was made up to 1 litre, 25 mL of this solution required 20 mL of 0.02

$M\ KMn{ O }_{ 4 }$ solution of complete oxidation. Calculate the weight % of

$FeS{ O }_{ 4 }.{ 7H }_{ 2 }O$ in the sample.

0%
34.75
0%
69.5
0%
89.5
0%
None of these

Explanation

$M_{eq}$ of

$FeSO_4.7H_2O$ in

$25mL$ =

$M_{eq}$ of

$KMnO_4$

$=20\times 0.2 \times 5=80$

So,

$M_{eq}$ of

$FeSO_4.7H_2O$ in 1000mL

$=20\times 0.2 \times 5\times\dfrac{1000}{25}=80$

$\implies \dfrac{W}{278}\times 1\times 1000=80$

$\implies W=22.24$

So,%

$feSO_4.7H_2O=\dfrac{22.4}{32}\times 100=69.5$

Natural water contains about

$0.02\%$

$D_2O$ (heavy water). When it is enriched to

$20\%$ (by volume), calculate the fraction of weight due to neutrons in 1 mole of sample.

0%
$0.355$
0%
$0.444$
0%
$0.455$
0%
$0.500$

Explanation

Assume total sample is of

$100gm$

Now,

$20gm = D_2O$ and

$80gm = H_2O$ ,

Mass of neutrons in

$80gm$ of

$H_2O = \dfrac{320}{9}$

mass of neutrons in

$20gm$ of

$D_2O = \dfrac{100}{11}$

Fraction of weight

$= \dfrac{\dfrac{320}{11}+\dfrac{100}{11}}{100} = .455$

CBSE Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 13 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 13 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.