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CBSE Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 14 - MCQExams.com

$60$ g of solution containing

$40\%$ by mass of NaCl are mixed with

$100$ g of a solution containing

$15\%$ by mass NaCl. Determine the mass percent of sodium chloride in the final solution.

0%
$24.4\%$
0%
$78\%$
0%
$48.8\%$
0%
$19.68\%$

Explanation

Mass of NaCl in 60 g of solution which contains 40% by mass of NaCl =

$60 \times \dfrac{40}{100} = 24 g$

Mass of NaCl in 100 g of solution which contains 15% by mass of NaCl =

$100 \times \dfrac{15}{100} = 15 g$

Total mass of NaCl contained in both solutions = 24 + 15 = 39 g

Total mass of both solutions = 60 + 100 = 160 g

Mass percent of NaCl in the final solution =

$\dfrac{39}{160} \times 100$ = 24.375 % = 24.4 % (approximately)

An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. What will be the masses of the carbon dioxide and water produces when 0.20g of this substance is subjected to the complete combustion?

0%
0.69 g and 0.048g
0%
0.506 g and 0.086g
0%
0.345 g and 0.0024 g
0%
0.91 g and 0.72 g

Explanation

Given :

% of

$C = 69\%$ , % of

$H = 4.8 \%$

$44g$ of

$CO_2$ has

$12g$ of C

$1 g$ of

$CO_2 = \dfrac{12}{44} g$ of C

$\% C = \dfrac{12}{44} \times \dfrac{w_c}{0.2} \times 100$

$\dfrac{44 \times 69 \times 0.2}{12 \times 100} = w_c$

$w_c = 0.506$

$18 g$ of

$H_2 O$ has

$2g$ of H

$1g$ of

$H_2 O = \dfrac{2}{18} gH$

$w_{H_2} O$ of

$H_2O$ has

$\dfrac{2}{18} \times w_{H_2O}$

$\%$ of

$H = \dfrac{2}{18} \times \dfrac{w_{H_2O}}{0.2} \times 100$

$4.8 = \dfrac{2}{18} \times \dfrac{w_{H_2O}}{0.2} \times 100$

$w_{H_2O} = 0.0864 g$

Weight of Carbon and Hydrogen in the organic compound are

$0.506 g$ and

$0.0864 g$ , respectively.

If a solution contains 34.2 g of sugar in 100 g of water. Find the concentration inn terms of mass by mass percentage of the solution?

0%
20 %
0%
25.48 %
0%
50 %
0%
10 %

Explanation

Mass of sugar

$=34.2\,gm$

Mass of the solution

$34.2+100=134.2\,gm$

$\therefore$ Concentration

$=\cfrac{34.2}{134.2}\times 100=25.48\%$

A mixture of

$NaHCO_3$ and

$Na_2CO_3$ weighs 950 gm. When the mixture is heated strongly in an open vessel, till constant weight, the mass of sample reduced by 155 gm. The correct information(s) is/ are ?

0%
The mass percent of $NaHCO_3$ in initial mixture = 50%
0%
The mole percent of $Na_2CO_3$ in initial mixture = 50%
0%
Final sample contains only $Na_2O$
0%
The loss in mass is only due to $CO_2$ gas evolved

Explanation

${ Na }_{ 2 }{ CO }_{ 3 }$ is a strong base and does not dissociate on normal heating.

$2NaH{ CO }_{ 3 }\longrightarrow { Na }_{ 2 }{ CO }_{ 3 }+{ CO }_{ 2 }+{ H }_{ 2 }O$

No. of moles

${ CO }_{ 2 }=\dfrac { 0.950 }{ 22400 } =4.24\times { 10 }^{ -5 }$

No. of moles of

${ NaHCO }_{ 3 }=4.24\times { 10 }^{ -5 }\times 22400=0.084\times { 10 }^{ +3 }=84g$

Initial

${ NaHCO }_{ 3 }$ mass

$=155g$

$\Rightarrow$ Mass percent

$=\dfrac { 84 }{ 155 } \times 100$ %

$\approx 50$ %

P and Q are two elements which form

${ P }_{ 2 }{ Q }_{ 3 }$ and

${ PQ }_{ 2 }$ . If 0.15 mole of

${ P }_{ 2 }{ Q }_{ 3 }$ weight 15.9 g and 0.15mole of

${ PQ }_{ 2 }$ weight 9.3 g. what are atomic weights of P and Q respectively?

0%
18 and 26
0%
26 and 26
0%
26 and 18
0%
18 and 18

Explanation

Using- no. of mole

$=\cfrac{\text {weight}}{\text{molecular weight}}$

$P_2Q_3 \Rightarrow$

$2P + 3Q =15.9/0.15= 106$ ....(1)

$PQ_2 \Rightarrow$

$P + 2Q = 9.3/0.15=62$ ........(2)

On solving eqn. (1) & (2) we get-

$P =26, Q = 18$

What percent of a sample of nitrogen must be allowed to escape if its temperature, pressure, and volume are to be changed from

$220^oC$ ,

$3$ atm and

$1.65$ litre to

$110^oC$ ,

$0.7$ atm and

$1.00$ litre respectively?

0%
$81.8$ %
0%
$71.8$ %
0%
$76.8$ %
0%
$86.8$ %

Explanation

Let

$n_{i}$ and

$n_{f}$ be initial & final moles of nitrogen respectively.

$\because PV=nRT$

$\therefore 3\times 1.65=n_{i}\times R\times 493$

$n_{i}=\cfrac{4.95}{493 R}$

Similarly,

$0.7\times 1=n_{f}\times R\times 383$

$\therefore n_{f}=\cfrac{0.7}{383\,R}$

$\therefore \%$ nitrogen escaped

$=\cfrac{(n_{i}-n_{f})}{n_{i}}\times 100$

$=\cfrac{\left (\cfrac{4.95}{493R}-\cfrac{0.7}{383R} \right)}{\cfrac{4.95}{493R}}\times 100$

$=81.8\%$

$\because R= 0.082057 L.atm.K^{-1}.mol^{-1}$

Hence, the correct option is

$A$

$3 g$ of activated charcoal was added to

$50 mL$ of acetic acid solution (

$0.06 N$ ) in a flask. After an hour it was filtered and the strength of the filtrate was found to be

$0.042 N$ . The amount of acetic acid absorbed (per gram of charcoal) is ?

0%
$18 mg$
0%
$36 mg$
0%
$42 mg$
0%
$54 mg$

Explanation

Given:

The initial mass of charcoal $= 3g$

Vol. of acetic acid = $50mL$ Normality $= 0.06N$

Strength of filtrate after reaction $= 0.042N$

We know,

Molarity of acetic acid solⁿ $= \cfrac{{Wt/M}}{{Vol\ of\ so{l^n}}}$

$\Rightarrow Wt = 0.06 \times 60 \times 0.05 = 0.180g$

Also, normality after reaction, $0.042 = \cfrac{W}{{60 \times 0.05}}$

$W = 0.126g$

Change in mass $= 0.180 - 0.126 = 0.054g = 54mg$

Amount of charcoal available $= 3g$

Amount of acetic acid absorbed $= \cfrac{{54mg}}{{3.0g}} \times 1.0g = 18g$

Correct option is A.

A substance contains 3.4%

$S$ . if it contain two atom of sulphur per molecule then minimum molecular weight of substance will be:

0%
188.2
0%
94.1
0%
1882
0%
282.3

Explanation

According to question-

$3.4g S=100g$ insulin

$\therefore 32g\ S= \dfrac{100\times 32}{3.4}$

$\Rightarrow 941.176$
For two atoms of

$S$

$\Rightarrow 941.176\times 2 = 1882.4$

Here correct unit is

$\dfrac{18882.4}{10}=188.2$

Insulin must contain at least Two atom of S in its one molecule.

Hence

$(A)$ is the correct answer.

Find out the weight of solute (M. wt. 60) that is required to dissolve in

$180g$ water to reduce the vapour pressure to

$\dfrac{4}{5}th$ of pure water:

0%
$120g$
0%
$150g$
0%
$300g$
0%
$75g$

Explanation

Relative lowering in vapour pressure

$=\cfrac{P_{0}-P_{S}}{P_{0}}=\cfrac{W_{1}\times M_{2}}{M_{1}\times W_{2}}$

where 1 & 2 denote solute and solvent respectively.

Given,

$P_{S}=\cfrac{4}{5}P_{0}$

$\Rightarrow \cfrac{1}{5}=\cfrac{W\times 18}{60\times 180}$

$\therefore W=120\,g$

A sample of diesel has the same knocking characteristics as a 60 ml mixture of cetane and

$a$ - methyl naphthalene mixed in 2 :1 ratio .The cetane number of the diesel sample is :

0%
60
0%
40
0%
66.6
0%
33.3

4.24 g of an organic compound on combustion produces 8.45 g of

${ CO }_{ 2 }$ and 3.46 g of water. The mass percentage of

$C$ and

$H$ in the compound respectively are:

0%
$27.2, 18.2$
0%
$54.4,9.1$
0%
$9.1,54.4$
0%
$18.2,27.2$

Explanation

% of carbon

$=\dfrac { 12 }{ 44 } \times \dfrac { Weight\ of\ { CO }_{ 2 } }{ Weight\ of\ organic\ compound } \times 100$

$=\dfrac { 12 }{ 44 } \times \dfrac { 3.45 }{ 4.24 } \times 100=54.35\approx 54.4$

% of Hydrogen

$=\dfrac { 2 }{ 18 } \times \dfrac { weight\quad of\quad { H }_{ 2 }O }{ weight\quad of\quad organic\quad compound } \times 100$

$=\dfrac { 2 }{ 18 } \times \dfrac { 3.46 }{ 4.24 } \times 100=9.06=9.1$

Ans :-

$C=54.4$ %,

$H=9.1$ %

Hence, option

$B$ is correct.

Air has three major components : nitrogen,oxygen and argon. Given that one mole of air at sea level is made up of 78% nitrogen, 21% oxygen, and 1% argon, by volume. What is the density of air? Assume that gases behave in an ideal manner. (Atomic mass of argon is 40 amu)

0%
14.62 g/L
0%
1.3 g/L
0%
29 g/L
0%
0.65 g/L

The volume of

$0.1\ M\ H_2SO_4$ required to neutralize 30 mL of

$2.0\ M\ NaOH$ is:

0%
100 mL
0%
300 mL
0%
400 mL
0%
200 mL

Explanation

Solution:- (B)

$300 \; mL$

Let

$V$ be the volume of

${H}_{2}S{O}_{4}$ required.

Given:-

${M}_{{H}_{2}S{O}_{4}} = 0.1 M$

${M}_{NaOH} = 2 M$

${V}_{NaOH} = 30 \; mL$

Now, as we know that

${N}_{1} {V}_{1} = {N}_{2} {V}_{2}$

$\because N = n \times M$

Whereas

$n$ is the valence factor

Therefore,

${n}_{1} {M}_{1} {V}_{1} = {n}_{2} {M}_{2} {V}_{2}$

For

${H}_{2}S{O}_{4}; \; n = 2$

For

$NaOH; \; n = 1$

Therefore,

$2 \times 0.1 \times V = 1 \times 2 \times 30$

$\Rightarrow V = \cfrac{60}{0.2}$

$\Rightarrow V = 300 \; mL$

Hence

$300 \; mL$ of

${H}_{2}S{O}_{4}$ is required to neutralize

$30 \; mL$ of

$2 M \; NaOH$ .

When 15gm of a mixture of

$NaCl$ and

$Na_2CO_3$ is heated with dilute hydrochloric acid, 2.5gm of

$CO_2$ is evolved at NTP. Calculate percentage composition of

$Na_2CO_3$ in a mixture.

0%
$59.85\%$
0%
$40.15\%$
0%
$30\%$
0%
$70\%$

A mixture of

${ C }_{ 2 }{ H }_{ 2 }$ AND

${ C }_{ 3 }{ H }_{ 8 }$ occupied a certain volume at 80 mm Hg . The mixture wsa completely burnt to

${ CO }_{ 2 }$ and

${ H }_{ 2 }O$ (l). When the pressure of

${ CO }_{ 2 }$ was found to be 230 mm Hg at the same temperature and volume, the mole fraction of

${ C }_{ 3 }{ H }_{ 8 }$ in the mixture is

0%
0.125
0%
0.875
0%
0.6
0%
0.8

$\cfrac{N_{A}}{4}$ atoms of an element ionise by absorbing

$X\ kJ$ of energy, then the ionisation energy of the elements is: (

$N_{A}$ = Avogadro number)

0%
$2X$
0%
$4X$
0%
$\cfrac{4}{X}$
0%
$\cfrac{1}{4X}$

Two oxide of metal contain 27.6% and 30% oxygen respectively. If the formula of first oxide is

$M_3O_4$ then formula of second oxide is -

0%
$MO$
0%
$M_2O$
0%
$M_2O_3$
0%
$MO_2$

$81gm$ mixture of

$MgC{O_3}\left( s \right)$

$N{H_2}COON{H_4}\left( s \right)$ in equimolar ratio is heated in open to constant mass. If vapour density of gaseous mixture evolved was found to be

$\cfrac{61}{4}$ , then find the mole % of

$MgCO_3$ in original sample.

Given that:

$MgC{O_3}\left( s \right)\xrightarrow{\Delta }MgO\left( s \right) + C{O_2}\left( g \right)$

${H_2}COON{H_4}\left( s \right)\xrightarrow{\Delta }2N{H_3}\left( g \right) + C{O_2}\left( g \right)N$

0%
$50\%$
0%
$51.9\%$
0%
$70.9\%$
0%
$59.1\%$

In the mixture of

$(NaHCO_3 + Na_2CO_3$ ) in 1:1 molar ratio, the volume of

$HCl$ required is

$'x'$ ml with methyl orange and

$'y'$ ml with phenolphthalein indicator for 2 different titrations from start for equal volumes of original solution. The relation between

$x$ and

$y$ is:

0%
$x = 3y$
0%
$y = 3x$
0%
$x =6y$
0%
$y = 6x$

Substance	Boiling point ${ ( }^{ 0 }C)$	Melting point ${ ( }^{ 0 }C)$
A	-183	-219
B	445	119
C	78	-15

Point out the physical states of A, B, and C at room temperature

$({ 30 }^{ 0 }C)$ .

0%
Gas, solid, liquid
0%
Gas, liquid, solid
0%
Liquid, solid, gas
0%
Solid, liquid, gas

Explanation

$(i)$ If Boiling point

$>$ Room Temperature

$>$ Melting Point

Then the physical state is liquid.

$(ii)$ If Room temperature

$>$ Boiling point

Then the physical state is gas.

$(iii)$ If Room temperature

$<$ Melting point

Then the physical state is solid.

Therefore,

for

$A\Rightarrow$ Physical state is gas.

For

$B\Rightarrow$ Physical state is solid.

For

$C\Rightarrow$ Physical state is liquid.

100 gm impure

$CaCO_3$ on heating gives 5.6 lit.

$CO_2$ gas at NTP. Find the percentage of calcium in

$CaCO_3$ sample.

0%
1
0%
40
0%
10
0%
30

If a sample of

$MgBr_2$ contains

$4\times 10^{20} Br^-$ ions, then the number of

$Mg^{2+}$ ions present in it is:

0%
$4 \times 10^{20}$
0%
$8 \times 10^{20}$
0%
$2 \times 10^{20}$
0%
$1 \times 10^{20}$

Explanation

correct answer: option 'C'

$MgBr_2\rightarrow Mg^{2+} +2Br^-$

Explanation: because the number of

$Mg^{2+}$ ions are half of

$Br^-$ ions.

Therefore option 'C' is the correct answer

Vapour density of the equilibrium mixture of

$NO_2$ and

$N_2O_4$ is found to be

$40$ for the equilibrium. Calculate percentage of

$NO_2$ in the mixture:

$N_2O_4 \leftrightharpoons 2 NO_2$

0%
$10$ %
0%
$5$ %
0%
$26.08$ %
0%
None of these

0.1 g of a solution containing

$Na_2CO_3$ and

$NaHCO_3$ requires 10 mL of 0.01 N HCl for neutralization using phenolphthalein as an indicator. Wt % of

$Na_2CO_3$ in the mixture is

0%
25
0%
32
0%
50
0%
10.6

$12\ g$ of impure cyanogen undergoes hydrolysis by two different pathways :
i)

$(CN)_{2}+4H_{2}O \rightarrow (NH_{4})_{2} C_{2}O_{4}$
ii)

$(CN)_{2}+2H_{2}O \rightarrow NH_{2}CONH_{2}$
The same amount of urea was obtained when

$11.52\ g$ of pure ammonium carbonate was heated. If

$20\ mL$ of

$1.6\ M$ acidic

$KMnO_{4}$ was required to completely oxidise

$(NH_{4})_{2}C_{2}O_{4}$ then which of the following statement is incorrect ?

0%
$\%$ purity of cyanogen is $86.67\%.$
0%
$\%$ purity of cyanogen is $60.67\%.$
0%
$\%$ progress in case $(i)$ is $40\%.$
0%
$\%$ progress in case $(ii)$ is $60\%.$

0.1 g of a solution containing

${ Na }_{ 2 }$

${ Co }_{ 3 }$ and

${ NaHCO }_{ 3 }$ requires 10 mL of 0.01 N HCl for neutralization using phenolphthalein as an indicator.Wt.% of

${ Na }_{ 2 }$

${ Co }_{ 3 }$ in the mixture is

0%
25
0%
32
0%
50
0%
10.6

What weight of

$NaHSO_3$ is required to react with

$100mL$ of solution containing

$0.33g$ of

$NaIO_3$ according to the following reaction

${IO}_{3}^{\Theta}+HS{O}_{3}^{\Theta}\longrightarrow I^{\Theta}+{SO}_{4}^{2-}$

0%
$0.52g$
0%
$5.2g$
0%
$1.04g$
0%
$10.4g$

Avogadro number is the number of molecule present in

0%
1 lit of any gas at NTP
0%
10 lit of any gas at NTP
0%
11.2 lit of any gas at NTP
0%
22.4 lit of any gas at NTP

A gas cylinder of the capacity of 20

$dm^{3}$ is filled with gas

$X$ , the mass of which is

$10\ gm$ . When the same cylinder is filled with hydrogen gas at the same temperature and pressure the mass of the hydrogen is

$2 \ gm$ hence the relative molecular mass of the gas is:

0%
5
0%
10
0%
15
0%
20

Explanation

Correct option is (b)

$\text{Explanation:}$

$\text{Step 1: Finding the vapour density}$

The weight of the

$X$ gas is

$10g$

Since,

$Vapour\ density=\dfrac{Weight\ of\ n\ molecules\ of\ gas}{Weight\ of\ n\ molecules\ of\ hydrogen}$

Therefore,

$Vapour\ density=\dfrac{10}{2}$

$Vapour\ density=5$

$\text{Step 1: Calculation of relative molecular mass}$

Since,

$Relative\ density=2\times vapour\ density$

Therefore,

$Relative\ density=2\times 5$

$Relative\ density=10$

Hence, the relative molecular mass of gas

$X$ is

$10g$ .

The

$O^{18}/O^{16}$ ratio in some meteorites is greater than the used to calculate the average atomic mass of oxygen one earth. The average mass of an atom of oxygen in these meteorites is ______ that of terrestrial oxygen atom?

0%
equal to
0%
greater than
0%
less than
0%
none of these

Explanation

Given :

$\dfrac {O^{18}}{O^{16}}$ ratio is greater than that which is used for calculation on earth on numerically,
on earth,
Let

$O_{16}\%$ be

$x\%$

$\therefore O_{18}\%$ be

$(100-x)\%$

$\therefore \$ Avg at mass on earth

$=\dfrac {16\times x+(100-x)\times 18}{100}$

$=\dfrac {16x-14x+1800}{100}$

$=\dfrac {-2x+1800}{100}$
on meteorites as

$O^{18}/O^{16}$ is greater

$\therefore$ Let

$O^{18} (100-x+2)$ i.e.,

$100-cn-2\%$

$\therefore O^{16}\%$ be

$x-2$
Avg at mass

$=\dfrac {16(x-2)18(100-x+2)}{100}$

$=\dfrac {16x-32+1800-18x+36}{100}$

$=\dfrac {-2x+180}{100}$
compare both atomic mass

$\dfrac {-2x+1800}{100} < \dfrac {-2x+180^4}{100}$

$\therefore \$ Avg atomic mass of

$O$ on meteorites will be greater

$\to$ You can simply check if

$O^{18}$ is greater amount.

$\therefore \$ Avg atomic mass will be greater.

An oxide,

$X_{2}O_{3}$ is oxidized to

$X_{4}^{-}$ by

$Cr_{2}O_{7}^{2-}$ in acid medium. The number of moles of

$X_{2}O_{3}$ oxidized per mole of

$Cr_{2}O_{7}^{2-}$ is

0%
2
0%
3
0%
3/2
0%
2/3

Gold crystallizes with FCC lattice for which the side length of the unit cell is

$5.0\mathring { A }$ . If the density of gold is

$10.5g/{cm}^{3}$ , the value of Avagadro number is

$(Au=198)$

0%
$6.022\times { 10 }^{ 23 }$
0%
$6.034\times { 10 }^{ 23 }$
0%
$5.966\times { 10 }^{ 23 }$
0%
$6.022\times { 10 }^{ 22 }$

What percentage of hydrogen fluoride molecules is dimerized?

0%
$50$
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

$=4+10+3=17$

$(4)\ \%$ dimer

$=\dfrac{10}{13}\times 100=58.82\%$ Option

$(b)$

If the ratio of mole fraction of solute and solvent is unity, then the mass per cent of solute is (Molar masses of solute and solvent are

$X$ and

$Y$ respectively)

0%
$50\%$
0%
$\dfrac{X}{X+Y}\times 100\%$
0%
$\dfrac{X}{Y}\times$ mass per cent of solvent
0%
$\dfrac{Y}{X}\times$ mass per cent of solvent

Explanation

$X_{solute}+Y_{solvent}=1$

$Y_{solvent}=1-X_{solute}$
given,

$\dfrac{X_{solute}}{Y_{solvent}}=1$

$\therefore X_{solute}=1\times X_{solvent}$

$Y_{solute}=1(1-X_{solute})\Rightarrow 2X_{xolute} =1$

$X_{solute}=\dfrac{1}{2}=0.5$

$\therefore Y_{solvent }=1-0.5=0.5$
Mass

$\%$ of solute

$=\dfrac{Mass\ of\ solute}{Mass\ of\ soln}\times 100\%$

$=\dfrac{X_{solute}X\ X}{Y_{solute} X+X_{solvent}Y}\times 100\%$

$=\dfrac{0.5}{0.5 X+0.5 Y}\times 100=\dfrac{0.5\ X}{0.5(X+Y)}\times 100$

$=\dfrac{X}{X+Y}\times 100\%$ Option

$(b)$
Mass

$\%$ of solvent

$=\dfrac{X_{solvent}Y}{X_{solvent}Y+X_{solute} X}\times 100\%$

$=\dfrac{0.5\ Y}{0.5\ X+0.5 Y}\times 100\%$

$=\dfrac{Y}{X+Y}\times 100\%$

$\dfrac{X}{X+Y}\times 100\%=\dfrac{X}{Y}\times \dfrac{Y}{X+Y}\times 100\%$
Mass

$\%$ of solute

$=\dfrac{X}{Y}\times$ Mass

$\%$ of solvent Option

$(c)$
Option

$b,c$ are correct

If in esterification, initially equal moles of alcohol and acid are taken. At equilibrium, unreacted alcohol and acid, on the treatment of sodium, produced

$H_2$ which occupied 44.8 L at

$0^o C$ and 1 atm. The percentage esterification of acid and alcohol is:

0%
25 %
0%
33.3 %
0%
50 %
0%
66.7 %

A solution containing 4.2 g of KOH and

$Ca(OH)_2$ is neutralised by an acid. If it consumes 0.1 equivalent of acid, composition of sample in solution is which of the following?

0%
45%, 55%
0%
35%, 65%
0%
25%, 75%
0%
15%, 85%

Complete combustion of

$858.0 \,g$ of compound

$X$ gives

$63.2 \,g$ of

$CO_{2}$ and

$1.28 \,g$ of

$H_{2}O.$ The lowest molecular mass

$X$ can have:

0%
$43 \,g$ `
0%
$86 \,g$
0%
$129 \,g$
0%
$172 \,g$

The correct sequence in decreasing order of the percentage of nitrogen in the given compounds is

0%
Urea > Ammonium chloride > Ammonium nitrate > Ammonium nitrite
0%
Urea > Ammonium nitrate > Ammonium nitrite > Ammonium chloride
0%
Urea > Ammonium nitrite > Ammonium Nitrate > Ammonium chloride
0%
Urea > Ammonium nitrite > Ammonium chloride > Ammonium nitrate

Explanation

Correct sequence is:

Urea > Ammonium nitrite > Ammonium Nitrate > Ammonium chloride

$\underset{47\%}{NH_{2}CONH_{2}} > \underset{43.75\%}{NH_{4}NO_{2}} > \underset{35\%}{NH_{4}NO_{3}} > \underset{26.2\%}{NH_{4}Cl}$ .

Chemistry plays an important role in national economy.

0%
True
0%
False

Explanation

chemistry plays an important role in national economy. chemistry is used in all the research done around us, from Water we drink, food we eat, clothes we wear, and medicines we Take.

Chemical industries have been really contributing to GDP of country and generates large amount of employment around the globe.

correct option

$\rightarrow$ True

$(A)$

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CBSE Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 14 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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