$3g$ of a metal

$M$ combines with

$2g$ of oxygen to form its metal oxide.

$0.03$ equivalent of this metal oxide is dissolved in

$500ml$ of water to form

$0.01$ molar aqueous solution. Molar mass of metal oxide is (assuming no change in volume of solution)

0%
$180g$
0%
$160g$
0%
$140g$
0%
$120g$

There are two common oxides of Sulphur, one of which contains

$50$ %

${O}_{2}$ by weight, the other almost exactly

$60$ %. The weights of sulphur which combine with

$1g$ of

${O}_{2}$ (Fixed) are in the ratio of:

0%
$1:1$
0%
$2:1$
0%
$2:3$
0%
$3:2$

Explanation

There are two common oxides of sulphur,one of which contains 50%

$O_2$ by weight the other almost exactly 60%.

% of sulphur % of oxygen

First oxide 50% 50%

Second oxide 40% 60%

In first oxide 1 part of oxygen combines with sulphur =

$\dfrac{50}{50} = 1$

Im second oxide 1 part of oxygen will combine with sulphur =

$\dfrac {40}{60} = 0.67$

Therefore weight of sulphur which combines with 1 g of

$O_2$ are in the ratio :

$\dfrac{1}{0.67}= 1:0.67 = 3:2$

Hence option D is correct answer.

Iron forms two oxides, in first oxide

$56$ grams of Iron is found to be combined with

$16$ gram oxygen and in second oxide

$112$ gram of Iron is found to be combined with

$48$ gram oxygen. This data satisfy the law of

0%
Conservation of mass
0%
Reciprocal proportion
0%
Multiple proportions
0%
Combining volume

What is the correct order of occurance (% by weight) in air of Ne, Ar and Kr ?

0%
Ne > Ar > Kr
0%
Ar > Ne > Kr
0%
Ar > Kr > Ne
0%
Ne > Kr > Ar.

$1$ mole of oxygen gas weighs ______________

0%
$3g$
0%
$32g$
0%
$1g$
0%
$6.022 \times {10^{22}}$

Mole fraction of glycerine

$(C_{ 3 }H_{ 5 }(OH)_{ 3 })$ in a solution of 36 g of water and 46 g of glycerine is :

0%
$0.46$
0%
$0.36$
0%
$0.20$
0%
$0.40$

The density of gas is 0.0002136 g/mL. Its vapour density is :-

0%
12
0%
24
0%
48
0%
32

${\text{2}}{{\text{5}}^{\text{o}}}{\text{C}}$ the density of

${\text{15}}\;{\text{M}}\;{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\;{\text{is}}\;{\text{1}}{\text{.8}}\;{\text{g}}\;{\text{c}}{{\text{m}}^{ - 3}}.$ Thus, mass percentage of

${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is aqueous solution is

0%
2%
0%
81.6%
0%
18%
0%
1.8%

All homologues of which of the following hydrocarbons have the same percentage composition

0%
Alkanes
0%
Alkynes
0%
Alkenes
0%
Alkadienes

The fractional abundance of

$Cl^{35}$ in a sample of chlorine containing only

$Cl^{35}$ (atomic weight

$=34.9$ ) and

$Cl^{37}$ (atomic weight

$=36.9$ ) isotopes, is

$0.6$ . The average mass number of chlorine is _______________.

0%
$35.7$
0%
$35.8$
0%
$18.8$
0%
$35.77$

Explanation

$\dfrac {C1^{35}}{C1^{37}}\Rightarrow \dfrac {0.6}{(1-0.6)}=\dfrac {0.6}{0.4}$

Average atomic mass

$=\dfrac {34.9\times 0.6+36.9\times 0.4}{1}$

Average Atomic mass

$=35.7$

Option

$A$ is the answer.

In which of the following sets all the compounds are ionic?

0%
$BCl_3, BeCl_2, MgO$
0%
$LiF, MgCl_2, Na_2O$
0%
$NH_4Cl, BF_3, CO_2$
0%
$AgCl, AlCl_3, PCl_5$

An element

$X$ have three isotopes

$X^{20}, X^{21}$ and

$X^{22}$ . The percentage abundance of

$X^{20}$ is

$90\%$ and the average atomic mass of element is

$20.18$ . The percentage abundance of

$X^{21}$ should be ______________.

0%
$2\%$
0%
$8\%$
0%
$10\%$
0%
$0\%$

Explanation

Average atomic mass

$=\dfrac {\displaystyle \sum \%\ abundance\ \times \ Atomic \ mass }{100}$

$20.18=\dfrac {90\times 20+x\times21+(10-x)22}{100}$

$2018=1800+21x+220-22x$

$x=2020-2018$

$x=2\%$

Option

$(A).$

$1.0\ g$ of

$Mg$ is burnt with

$0.28\ g$ of

$O_{2}$ in a closed vessel. Which reactant is left in excess and how much?

0%
$Mg, 5.8\ g$
0%
$Mg, 0.58\ g$
0%
$O_{2}, 0.24\ g$
0%
$O_{2}, 2.4\ g$

Explanation

$2Mg + O_{2}\rightarrow 2MgO$

$\therefore 1\ mole$ of

$O_{2}$ reacts with

$2$ moles of

$Mg$

$\because$ Moles of

$O_{2} = \dfrac {0.28}{32} = 0.00875$

$0.00875\ mole$ of

$O_{2}$ reacts with

$\dfrac {2}{1} \times 0.00875\ moles$ of

$Mg$

$= 0.0715\ Moles$ of

$Mg$

Hence, mass of magnesium that reacts

$= moles\times molar\ mass$

$= 0.0715\times 24 = 0.42\ g$

That means, out of the

$1\ g$ of

$Mg$ , only

$0.42\ g$ is used.

Therefore magnesium is in excess by

$(1 - 0.42) = 0.58\ g$ .

Hence, option

$B$ is correct.

Percentage of oxygen in urea is about :

0%
$20.24$ %
0%
$26.64$ %
0%
$46.19$ %
0%
$7.86$ %

A sample of clay contains

$50\%$ silica and

$10\%$ water. The sample is partially dried by which it loses

$8\ g$ water. If the percentage of silica in the partially dried clay is

$52$ , what is the percentage of water in the partially dried clay?

0%
$2.0\%$
0%
$6.4\%$
0%
$10.4\%$
0%
$2.4\%$

Explanation

Let weight of clay is

$x\ gm$

$\therefore \$ Weight of silica

$=\dfrac {50}{100}\times x$

$=0.5\ x\ gm$

$\therefore \$ Weight of water

$=\dfrac {10}{100}\times x=0.1\ x\ gm$
Now,

$8\ gm$ water is reduced

$\therefore \$ weight of clay

$=(x-8)\ gm$

Weight of silica

$=\dfrac {52}{100}\times (x-8)\ gm$

$=(0.52 \ x-4.16)\ gm$

Now, Weight of silica in both case are equal.

$\therefore \ 0.5\ x=0.52\ x-4.16$

$0.02\ x=4.16$

$\therefore \ x=208\ gm\ =$ weight of clay initially.

$\therefore \$ weight of water

$=0.1\ x=20.8\ gm$

Now,

$8\ g$ water is reduced.

$\therefore \$ Weight of water

$=20.8-8\ x =12.8\ gm$

Weight of clay

$=208-8\ gm=200\ gm$

$\therefore \ \%$ of water

$=\dfrac {12.8}{200}\times 100=6.4\%$

Compound Q contains

$40\%$ carbon by mass.
What could Q be?

$1$ . glucose,

$C_6H_{12}O_6$

$2$ . starch,

$(C_6H_{10}O_5)_n$

$3$ . sucrose,

$C_{12}H_{22}O_{11}$ .

0%
$1, 2$ and $3$ are correct
0%
$1$ and $2$ only are correct
0%
$2$ and $3$ only are correct
0%
$1$ only is correct

Explanation

$\text{percentage mass of C}=\dfrac{\text{mass of C X 100%}}{\text{total mass}}$

$\text{1. percentage of C in glucose=40%}$

$\text{2. percentage of C in starch=44%}$

$\text{3. percentage of C in sucrose=42%}$

Solid ammonium dichromate decomposes as under:

$( NH_4)_2Cr_2O_7 \rightarrow N_2 + Cr_2O_3 +4H_2O$ if 63 g of ammonium dichromate decomposes, calculate the volume of

$N_2$ evolved at STP.

0%
11.2 L
0%
5.6 L
0%
22.4 L
0%
1 L
0%
44.8 L

A quantity of

$13.5\ g$ of aluminium when change to

$Al^{3+}$ ion in solution, will lose

$(Al=27)$ .

0%
$18.0\times 10^{23}$ electrons
0%
$6.02\times 10^{23}$ electrons
0%
$3.01\times 10^{23}$ electrons
0%
$9.0\times 10^{23}$ electrons

Explanation

Mole of

$Al=\dfrac {13.5}{27}=0.5\ mole$

$Al\to Al^{3+}$

As there is a loss of

$3e^-$

$\therefore \$ Total no. of

$e^-=0.5\times 6.02\times 10^{23}\times 3$

$=9.03\times 10^{23}\ e^-$

So, the correct option is

$D$

A certain mixture of

$MnO$ and

$MNO_2$ contains

$66.6\ mol$ per cent of

$MnO_2$ . What is the approximate mass percent of

$Mn$ in it?

$(Mn=55)$

0%
$66.67$
0%
$24.02$
0%
$72.05$
0%
$69.62$

Explanation

Man of

$MnO=66.67\times 70.93$

$=4728.9\ g$
man of

$MnO_2=33.33\times 86.93$

$=2897.37$
Total man

$=4728.9+2897.37$

$=7626.2\ g$
man of

$Mn=55\times 100=5500$

$\%Mn=\dfrac {5500}{7626.2}\times 100$

$=72.11\%$

The mass composition of universe may be given as

$90\%\ H_2$ and

$10\%\ He$ . The average molecular mass of universe should be ____________.

0%
$2.20$
0%
$2.10$
0%
$3.80$
0%
$3.64$

Explanation

Average M.M of universe

$=\dfrac{0.90 \times 2 + 0.10 \times4}{0.90+0.10}$

$=\dfrac{1.80+0.40}{1}=2.20$

Option

$B$ ic correct.

From

$\ 2 \ mg$ calcium

$1.2\times 10^{19}$ atoms are removed. The number of

$g$ - atoms of calcium left is

$(Ca=40)$ :

0%
$5\times 10^{-5}$
0%
$2\times 10^{-5}$
0%
$3\times 10^{-5}$
0%
$5\times 10^{-6}$

Explanation

$2\ mg$ Calcium moles

$=\dfrac{2\times 10^{-3}}{40}=5\times 10^{-5}$

$1$ mole

$=6.022\times 10^{23}$ atoms

$x=\dfrac{1.2\times 10^{19}}{6.022\times 10^{23}}$

$=1.99\times 10^{-5}$ moles removed

$\therefore$ Remaining moles

$=5\times 10^{-5}-1.99\times 10^{-5}$

$=3.00\times 10^{-5}$ moles

$1\ g$ atom

$=1$ mole

$\therefore$ Option

$C$ correct.

In the atomic weight determination. Dalton suggest the formula of water as

$HO$ and the composition of water as hydrogen

$=12.5\%$ and oxygen

$=87.5\%$ by weight. What should be the atomic weight of oxygen on

$H-$ scale, on the basis of this information?

0%
$16$
0%
$8$
0%
$14$
0%
$7$

Explanation

Let mass of oxygen be

$'x'$

$\%$ of oxygen in

$HO$

$87.5=\dfrac {mass\ of\ oxygen}{mass\ of\ compound}\times 100$

$87.5=\dfrac {x}{1+x}\times 100$

$87.5(1+x)=100x$

$87.5+87.5x=100x$

$100x-87.5x=87.5$

$12.5x=87.5x$

$x=\dfrac {87.5}{12.5}$

$\boxed {x=}$

$\therefore \$ Atomic weight of oxygen

$=7$

Option

$'d'$ correct

Dopamine is neurotransmitter, a molecular that serves to transmit message in the brain. The chemical formula of dopamine is

$C_{8}H_{11}O_{2}N$ . How many moles are there in

$1\ g$ of dopamine?

0%
$0.00654$
0%
$153$
0%
$0.0654$
0%
$None\ of\ these$

Explanation

$C_{8}H_{11}O_{2}N$
Molar mass

$=8\times 12+11\times 1+2\times 16+1\times 14$

$=153\ g/mol$

$1\ mol=153\ g$

$x\ mol=1\ g$

$x=\dfrac{1}{153} mol$

$\boxed{x=0.00654}$
Option

$a$

A sample of impure cuprous oxide contains

$66.67\%$ copper, by ,mass. What is the percentage of pure

$Cu_2O$ in the sample?

$(Cu=63.5)$

0%
$66.67$
0%
$75$
0%
$70$
0%
$80$

Explanation

$100\ gm$ of sample contains

$=66.6\ gm$
mole of

$Cu=\dfrac {66.6}{63.6}=1.05\ mole$

$1$ mole of oxygen in

$2$ mole

$Cu$

$2$ mole of

$Cu$ is with

$1$ mole oxygen

$1.05$ mole contains

$\dfrac {1}{2}\times 1.05$
weight of oxygen

$=52\times 16=84$
pure

$Cu_2O=66.6+8.42$

$=75$

Molecular mass of dry air is ________________.

0%
less than moist air
0%
greater than moist air
0%
equal to moist air
0%
may be greater or less than moist air

Explanation

The molecular mass of dry ice is greater than moist air.

As dry air consists of nitrogen and oxygen while moist air contains water vapour which less mass than that of oxygen and nitrogen making the mass of moist air lower.

$M.M$ of dry air

$\simeq 28.9\ gm$

$M.M$ of moist air

$< 28.9\ gm$

When

$2.5\,g$ of a sample of Mohr's salt reacts completely with

$50\,mL$ of

$\dfrac{N}{10} KMnO_4$ solution. The % purity of the sample of Mohr's salt is:

0%
$78.4$
0%
$70$
0%
$37$
0%
$40$

Explanation

meq.

$FeSO_4 (NH_4)_2SO_4. 6H_2O$ = meq. of

$KMnO_4$

$(n = 1)$

$\dfrac{W}{392} \times 1 \times 1000 = 0.1 \times 50$ ;

$W = 1.96\,g$
Hence,

$\%$ purity of Mohr's salt

$= \dfrac{1.96}{2.5} \times 100 = 78.4 \%$

$Fe_{0.95} O_{1.00}.$ What is the percentage of Fe(III) in the sample?

0%
10.52%
0%
3.46%
0%
13.84%
0%
93.08%

Explanation

The composition of a sample of wustite is

$Fe_{0.95}O$ . Formula of ferric oxide is

$FeO$ . So,number of

$Fe^{+2}$ ion missing=0.05

Each

$Fe^{+2}$ ion contains +2 charge. So total charge missing =2

$\times$ 0.05=0.10.

To maintain electrical neutrality the 0.10 positive charge is compensated by

$Fe^{+3}$ ions.

Replacement of one ferrous ion by one ferric ion increases +1 charge. So no of ferric ions required to compensate the 0.10 charge is 0.10.

So 0.95

$Fe^{+2}$ ions require 0.1

$Fe^{+3}$ ions

So, in 100 Fe atoms, percentage of

$Fe^{+3}$ is =

$(.1/.95)\times 100$ =10.52

A vessel contains

$1$ mole of

$O_2$ gas at a temperature

$T$ . The pressure of the gas is

$P$ . An identical vessel containing

$1$ mole of

$He$ gas at a temperature

$2$ T has a pressure of?

0%
$P/8$
0%
$P$
0%
$2P$
0%
$8P$

Explanation

Moles of

$O_2=n_1=1$

$T_1=T$ K

$P_1=P$ atm
Moles of He

$=h_2=1$

$T_2=2T$ K
Let pressure be

$P_2$

$V_1=V_2=V$ (given)

$P_1V_1=n^2T_1$

$\therefore PV=(1)(R)(T)$ …….

$(1)$

$\therefore P_2V=(1)(R)(2T)$ …….

$(2)$
Equation

$(1)$ divided by equation

$(2)$ , we get

$\therefore \dfrac{P}{P_2}=\dfrac{1}{2}$

$\therefore P_2=2P$
Option C.

A mixture of

$He(4)$ and

$Ne(20)$ in a

$5-$ litre flask at

$300\ K$ and

$1$ atm weighs

$4\ g$ . The mole

$\%$ of

$He$ is

0%
$2$
0%
$0.02$
0%
$20$
0%
$4$

Explanation

Using

$PV=nRT$

$n= \dfrac{PV}{RT} = \dfrac{1 \times 5}{0.082}{300} =0.203 \ moles$

Let the amount of

$He$ in the mixture be

$x$ , thus amount of

$Ne$ =

$4-x$ g

Total moles = 0.203

$\therefore \dfrac{x}{4} + \dfrac{4-x}{20} = 0.203$

$x= 0.012 \ g$

Thus moles of

$He$ in the mixture is

$\dfrac{0.012}{4} = 0.003$

which is 2% moles of the mixture.

What is involved in

$CO_2$ and

$CH_4$ inclusion?

0%
Diamond inclusion
0%
Fluid inclusion
0%
Natural inclusion
0%
Crystal inclusion

What percentage of hydrogen fluoride molecules is trimerized?

0%
$40$
0%
$58.8$
0%
$76.47$
0%
$17.65$

Explanation

Consider mole of

$HE=4$

$H_{3}F_{3}=1$ then

$HF$ in it

$=1$ molecules

$H_{2}F_{2}=y$ then

$HF$ in it

$=2$ molecules
Total mole

$=4+1+y=5+y$

$X_{HE}=\dfrac{4}{5+0}, X_{H_{2}F}=\dfrac{y}{5+y}, X_{H_{3}F_{3}}=\dfrac{1}{5+y}$

$20\left(\dfrac{y}{5+y}\right)+40\left(\dfrac{4}{5+y}\right)+60\left(\dfrac{1}{5+y}\right)=34$

$80+40y+60=170+94 y$

$40y-34y=170-140$

$6y=30$

$y=5$
Total molecular

$m$ monomer

$=4$
dimmer

$=5\times 2=10$
trimer

$=1\times 3=3$
Total molecules

$=4+10+3=17$

$(5)\ \%$ trimer

$=\dfrac{3}{17}\times 10=17.64\%$ Option

$(d)$

10 grams of a solute is dissolved in 90 grams of a solvent. Its mass percent in solution is

0%
0.01
0%
11.1
0%
10
0%
9

Explanation

$\%$ by wt.

$= \dfrac {wt. \,of \,the \,solute (g)}{wt. \,of \,the \,solution(g)} \times 100$

$\%$ by wt

$= \dfrac {10}{90 + 10} \times 100 = 10\%$

One of the statements of Dalton's atomic theory is given below:
Compounds are formed when atoms of different elements combine in a fixed ratio.
Which of the following laws is/are not related to this statement?

0%
Law of conservation of mass
0%
Law of definite proportions
0%
Law of multiple proportions
0%
Avogadro law

Which of the following reactions is not according to the law of conservation of mass?

0%
$2Mg(s)+O_2(g)\to 2MgO(s)$
0%
$C_3H_3(g)+O_2(g)\to CO_2(g)+H_2O(g)$
0%
$P_4(s)+5O_2(g)\to P_4O_{10}(s)$
0%
$CH_4(g)+2O_2(g)\to CO_2(g)+2H_2O(g)$

Mass is measure of the amount of matter.

0%
True
0%
False

The accepted unit of atomic and molecular mass is:

0%
kilogram
0%
gram
0%
pound
0%
atomic mass unit

Which of the given laws of chemical combination is satisfied by the figure?

0%
Law of multiple proportion
0%
Gay Lussac's law
0%
Avogadro law
0%
Law of definite proportion

"Equal volumes of all gases at the same temperature and pressure contain equal number of particles." This statement is a direct consequence of:

0%
Avogadro's law
0%
Charles law
0%
Ideal gas equation
0%
Law of partial pressure

$40\,mL$ gaseous mixture of

$CO , CH_4$ and Ne was exploded with

$10\,mL$ of oxygen. On cooling the gases occupied

$36.5\,mL$ . After treatment with KOH the volume reduced by

$9\,mL$ again on treatment with alkaline pyrogallol, the volume further reduced.
Percentage of

$CH_{4}$ in the original mixture is:

0%
$22.5$
0%
$77.5$
0%
$7.5$
0%
$15$

Explanation

Let the volume of

$CO , CH_4$ and

$Ne$ be

$x , y$ and

$z$ respectively

$CO + \dfrac{1}{2} O_2 \rightarrow CO_2$ ;

$x$

$x/2$

$x$

$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O (l)$

$y$

$2y$

$y$
remaining volume of

$O_2 = 10 - \dfrac{x}{2}- 2y$

Volume after reaction :

$x + y + 10 - \dfrac{x}{2} - 2y + z = 36.5$ ............(i)

$x + y = 9$ ..........(ii)

$x + y + z = 40$ ...............(iii)
by Eq. (i) , (ii) & (iii)
Volume of

$CH_4 = 6\,mL$

$\%$ of

$CH_4 = \dfrac{6}{40} \times 100 \Rightarrow 15$

How many molecules are there in one mole of a compound?

0%
$60.22\times 10^{-23}$
0%
$6.022\times 10^{-23}$
0%
$60.22\times 10^{23}$
0%
$6.022\times 10^{23}$

$4$ mole of a mixture of Mohr's salt and

$Fe_2(SO_4)_3$ requires

$500\,mL$ of

$1\,M \,K_2Cr_2O_7$ for complete oxidation in acidic medium. The mole % of the Mohr's salt in the mixture is:

0%
$25$
0%
$50$
0%
$60$
0%
$75$

Explanation

$Cr_2O_{7}^{2-} + 6Fe^{2+} + 14 H^+ \rightarrow 2Cr^{3+} + 6Fe{3+} + 7H_2O$

$( n = 1 )$ (Mohr's salt)
Equivalent of

$Fe^{2+}$ = moles of Mohr's salt
= equivalent of

$K_2Cr_2O_7$
=

$500 \times 10^{-3} \times 6 \times 1 = 3.0$
Hence. mole percent of Mohr's salt

$= \dfrac{3}{4} \times 100 = 75$

The weight percentage of NaCl solution isIf the weight of the solution is 150 gms, then calculate the weight of Na and water in gms.

0%
15,135
0%
25,125
0%
50,100
0%
75,75

Explanation

Weight percent

$=\cfrac{weight \,of\, solute}{weight\, of\,solution}\times 100\%$

Given, Weight percent

$=10$

Weight of solution

$=150\,gms$

$\Rightarrow\, 10=\cfrac{weight\,of\,solute}{150\,gms}\times100\%$

$\therefore$ Weight of solute

$=15\,gms$

$\therefore$ Weight of water

$=(150-15)=135\,gms$ .

$5$ g of crystalline salt, when rendered anhydrous, lost

$1.89$ g of water. The formula weight of anhydrous salts is

$160$ . The number of molecules of water of crystallisation in the salt is:

0%
$1$
0%
$2$
0%
$3$
0%
$5$

Explanation

Given,

$1.89$ g of water is present in

$5$ g of hydrated salt.

Therefore, weight of anhydrous salt is

$(5 -1.89)$ g

$= 3.11$ g.

No of moles of anhydrous salt

$= \dfrac{3.11}{160} = 0.02 moles$

No of moles of water

$= \dfrac{1.89}{18} = 0.1 moles$

Ratio of anhydrous salt to water,

Salt: Water

$= 0.02 : 0.1$

$1:5$

Therefore, no of molecules of water

$=5$ .

$Na_2SO_xH_2O$ has 50%

$H_2O$ by mass. Hence,

$x$ is:

0%
$4$
0%
$45$
0%
$6$
0%
$7$

Explanation

$Na_{2}SO_{3}.x H_{2}O$ has 50%

$H_{2}O$ .

$\therefore \dfrac{18x}{126+18x}=\dfrac{50}{100}$

$\therefore x=7$

Given that the abundances of isotopes

$^{54}Fe, ^{56}Fe$ and

$^{57}Fe$ are

$5\%, 90\%$ and

$5\%$ respectively, the atomic mass of

$Fe$ is:

0%
$55.85$
0%
$55.95$
0%
$55.75$
0%
$56.05$

Explanation

$\bar{A}=\dfrac{\sum A_{i}x_{i}}{\sum x_{i}}$

$\bar{A}=54\times 0.05+56\times 0.90+57\times 0.05$

$\bar{A}=55.95$

In a gaseous mixture, if an alkane

$(C_xH_{2x+ 2})$ and an alkene

$(C_yH_{2y})$ are taken in

$2 : 1$ mole ratio, the average molecular weight of the mixture is observed to be

$20$ . If the same alkane and alkene are taken in

$1 : 2$ mole ratio, the average molecular weight of the mixture is observed to be

$24$ . Then, the value of

$x$ and

$y$ are respectively :

0%
$2, 1$
0%
$1, 2$
0%
$2, 3$
0%
$3, 2$

Explanation

If the alkane,

$C_xH_{2x+2}$ and alkene,

$C_yH_{2y}$ is in the ratio of

$2:1$ , then

$M_{mix}=\dfrac{2.(14x + 2) + 1.(14y)}{3}=20$

$28x + 14y = 56$ ...( 1 )

If the alkane,

$C_xH_{2x+2}$ and alkene,

$C_yH_{2y}$ is in the ratio of

$1:2$ , then

$M_{mix}=\dfrac{1.(14x + 2) + 2.(14y)}{3}=24$

$14x + 28y = 70$ ...( 2 )

Solving equation (1) and (2), we get

$x = 1, y = 2$

In what ratio should a 15% solution of acetic acid be mixed with a 3% solution of the acid to prepare a 10% solution?

[All percentages are mass/mass percentages.]

0%
7:3
0%
5:7
0%
7:5
0%
7:10

$Cu_{2}S$ and

$M_{2}S$ are isomorphous in which percentage of sulphur is

$20.14$ % and

$12.94$ % respectively. The atomic weight of M is:

[Atomic wt. of Cu = 63.5]

0%
208
0%
108
0%
112
0%
106

Explanation

Assume atomic weight of

$S$ and

$M$ to be

$x$ and

$y$ respectively.

Percentage of sulfur is

$20.14$ % and

$12.94$ % in

$Cu_2S$ and

$M_2S$ respectively.

For

$Cu_2S$ , 2 moles Cu = 1 mole S.

1g of

$Cu_2S$ contains 0.7986g Cu and 0.2014g S respectively.

Hence,

$2\times \dfrac {0.7986}{63.5} =1 \times \dfrac {0.2014 }{x}$

1g of

$M_2S$ contains 0.8706g M and 0.1294g S respectively.

Hence,

$2 \times \dfrac {0.8706}{y} =1 \times \dfrac {0.1294}{x}$

Solving tow equation we get,

$y=107.70 \approx 108.$

A 1.24 M aqueous solution of KI has density of

$1.15 g / cm^{3}$ . What is the percentage composition of solute in the solution?

0%
$17.89$
0%
$27.89$
0%
$37.89$
0%
$47.89$

Explanation

1.24 M aqueous solution of KI means 1.24 moles of KI in 1 litre of solution.

Given,

Density of solution

$= 1.15\ \dfrac{g}{cm^3}$

Volume of solution

$= 1\ L\ or\ 1000\ cm^3$

$\therefore$ Mass of solution

$= Density\times volume$

$= 1.15\times 1000$

$= 1150\ g$

Mass of solute

$= 1.24\times 166 = 205.8\ g$

Percentage composition of solute

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

$= \dfrac{205.8}{1150}\times 100$

$= 17.89\%$

Hence, option A is correct.

$4.9$ g sample of

$KClO_{3}$ was heated under such conditions that a part of it decomposed according to the equation:

$2KClO_{3}\rightarrow 2KCl+3O_{2}$ and remaining underwent change according to the equation:

$4KClO_{3}\rightarrow 3KClO_{4}+KCl$ . If the amount of

$O_{2}$ evolved was

$672$ mL at

$1$ atm and

$273$ K, the percentage by weight of

$KClO_{4}$ in the residue is :

0%
$52.72 \%$
0%
$46.64\%$
0%
$54.86\%$
0%
$42.35\%$

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

$Fe_{0.95} O_{1.00}.$ What is the percentage of Fe(III) in the sample?

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

The composition of a sample of wustite is Fe0.95O1.00. Fe_{0.95} O_{1.00}. What is the percentage of Fe(III) in the sample?

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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$Fe_{0.95} O_{1.00}.$ What is the percentage of Fe(III) in the sample?