MCQ Questions for Class 12 Engineering Chemistry Solutions Quiz 6

$3$ moles of

$P$ and

$2$ moles of

$Q$ are mixed, what will be their total vapour pressure in the solution if their partial vapour pressures are

$80$ and

$60$ torr respectively?

0%
$80$ torr
0%
$140$ torr
0%
$72$ torr
0%
$70$ torr

Explanation

We know that :

$P_{total}=p_P+p_Q=p^0_Px_P + p^0_Qx_Q$

where

$p^0_P$ and

$p^0_Q$ are partial pressures of pure substances.

$x_P$ and

$x_Q$ are mole fractions.

$x_P=\dfrac{3}{5}=0.6$

$x_Q = \dfrac{2}{5}=0.4$

hence

$P_{total}=p^0_Px_P + p^0_Qx_Q= 80\times 0.6 + 60\times 0.4 =72torr$

Van't-Hoff factor(i) for this reaction is:

0%
$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

In the given reaction, Benzene (a non-ionizing compound) is converted into nitrobenzene(which has

$NO_2^+, C_6H_6$ ).

Therefore, the Van't Hoff factor is

$2$ .

What is the value of Van't-Hoff factor

$i$ for

$0.1\,m$ ideal solution?

0%
$1$
0%
$0$
0%
$0.1$
0%
$0.01$

On dissolving sugar in water at room temperature solution feels cool to touch. Under which of the following cases dissolution of sugar will be most rapid?

0%
Sugar crystals in cold water
0%
Sugar crystals in hot water
0%
Powdered sugar in cold water
0%
Powdered sugar in hot water

The value of van't Hoff factors for

$KCl, NaCl$ and

${K}_{2}{SO}_{4}$ , respectively, are :

0%
$2, \ 2$ and $2$
0%
$2, \ 2$ , and $3$
0%
$1, \ 1$ and $2$
0%
$1, \ 1$ and $1$

Explanation

$i$

i) For

$KCl$

$2$

$KCl\longrightarrow \underbrace { { K }^{ + }+{ Cl }^{ - } }$ {since

$i=n$ for strong electrolytes}

ii)

$NaCl$

$NaCl\longrightarrow \underbrace { { Na }^{ + }+{ Cl }^{ - } }$

$n=2$

iii)

${ K }_{ 2 }{ SO }_{ 4 }$

${ K }_{ 2 }{ SO }_{ 4 }\longrightarrow \underbrace { { 2K }^{ + }+{ SO }_{ 4 }^{ 2- } }$

$n=3$

$i=3$

What will be the abnormal molecular mass of

${{\text{K}}_3}\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_6}} \right]$ . If in a aqueous solution it is 40% dissociated? (Given, normal molecular weight of

${{\text{K}}_3}\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_6}} \right]$ is 329)

0%
$149.54$
0%
$723.8$
0%
$82.25$
0%
$329.0$

Explanation

$\mathop {\mathop {{{\text{K}}_3}\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_6}} \right]}\limits_1 }\limits_{1 - 0.4} \to \mathop {\mathop {3{{\text{K}}^ + }}\limits_{} }\limits_{3 \times 0.4} + \mathop {{{\mathop {\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_6}} \right]}\limits_{} }^{3 - }}}\limits_{0.4}$
Van't Hoff factor,

${\text{i}} = 1 - 0.4 + \left( {3 \times 0.4} \right) + 0.4$
= 1 + 1.2
= 2.2

${\text{i}} = \dfrac{{{\text{Normal molar mass}}}}{{{\text{Abnormal molar mass}}}}$

Abnormal molar mass =

$\dfrac{{{\text{Normal molar mass}}}}{{\text{i}}}$

$= \dfrac{{329}}{{2.2}}$
= 149.54
Thus, correct option is (A).

We have three aqueous solutions of

$NaCl$ labelled as

$'A','B'$ and

$'C$ with concentrations

$0.1M$ ,

$0.01M$ and

$0.001M$ , respectively. The value of van't Hoff factor for these solutions will be in the order.

0%
${i}_{A}< {i}_{B}< {i}_{C}$
0%
${i}_{A}> {i}_{B}> {i}_{C}$
0%
${i}_{A}= {i}_{B}= {i}_{C}$
0%
${i}_{A}< {i}_{B}> {i}_{C}$

The boiling point of liquid A, B, C and D are

$80{\,^o}C,\,\,60{\,^o}C,\,90{\,^o}C$ and

$100{\,^o}C$ respectively. Which will show highest vapour pressure at room temperature?

0%
A
0%
B
0%
C
0%
D

Explanation

Vapour pressure or equilibrium vapour pressure is defined as the pressure exerted by a vapour in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system.

The equilibrium vapour pressure is an indication of a liquid's evaporation rate. Thus it is related to the ease of vapour formation.

Higher is the boiling point difficult is the vapourization thus lower is the vapour pressure or vice-versa.

Among A, B, C, D lowest boiling point is for B i.e

$60^{\circ}C$ thus will have the highest vapour pressure at room temperature.

Therefore, option B is correct.

Let

$P_s$ and

$P$ be the saturated partial pressure and partial pressure of water respectively, then the relative humidity is given by:

0%
$\dfrac{P_s+P}{P_s}\times 100$
0%
$\dfrac{P}{P_s}\times 100$
0%
$\dfrac{P_s}{P}\times 100$
0%
$(P+P_s)\times 100$

Explanation

Relative humidity

$\phi=\cfrac {\text{partial pressure of water vapour}}{\text{(Saturated pressure of pure water vapour at same temperature)}}\times 100$

$\Rightarrow \phi= \cfrac {P}{P_s}\times 100$

The vapour pressure of a solvent at

$293$ K is

$100$ mm Hg. Then the vapour pressure of a solution containing

$1$ mole of a strong electrolyte

$(AB_2)$ in

$99$ moles of the solvent at

$293$ K is:

(Assume complete dissociation of solute)

0%
$103$ mm Hg
0%
$99$ mm Hg
0%
$97$ mm Hg
0%
$101$ mm Hg

Explanation

Let us consider the problem:

$\frac{{({P_A}^0 - {\rm{ }}{P^0})}}{{{P_A}^0}} = i\frac{{({n_B})}}{{({n_A} + {n_B})}}$

Since,

$\frac{{\left( {100 - {P_A}} \right)}}{{100}} = 3{\rm{ }}\left[ {\frac{1}{{\left( {99 + 1} \right)}}} \right]$

Since,

$100 - {P_A} = 3$

Hence the solution is

${P_A} = 97 mm$ of

$Hg$ .

Which of the following has maximum vapour pressure at a given temperature?

Explanation

Vapour pressure increases with decreasing net molecular weight of a compound. Now among

$A,B,C$ and

$D$ in and

$D$ intermolecular

$H-$ bonding is present as a result their net molecular weight increases but in

$A$ and

$C$ due to presence of intermolecular

$H-$ bonding their net molecular weight is approximately equals to their absolute molecular weight. Now intermolecular

$H-$ bonding is more feasible in

$C$ due to presence of negative charge on

${ NO }_{ 3 }^{ \left( - \right) }$ ion.

But absolute molecular weight of

$A$ is less than

$C$ . So vapour pressure of

$A$ will be highest.

1101541_1028448_ans_471a893c9e2142179d5645ff29b506a0.png

Calculate Vant Hoff factor (i) of the compound acetic acid in benzene if their $\alpha=50$ %.

0%
$0.25$
0%
$1.5$
0%
$0.75$
0%
$2$

Explanation

We have,

$\alpha = \dfrac{1-i}{1/n-1}$

since, dimerization of acetic acid in benzene takes place

so,

$n=2$

in acetic acid dimerisation takes place

$\alpha= 50$ %

$\alpha= 50/100=0.5$

$0.5 = \dfrac{1-i}{1/2 -1}$

$0.5 =\dfrac{1-i}{0.5}$

$i = 0.75$

Two liquids X and Y form an ideal solution. At

$300$ K, vapour pressure of the solution containing

$1$ mol of X and

$3$ mol of Y is

$550$ mm Hg. At the same temperature, if

$1$ mol of Y is further added to this solution, vapour pressure of the solution increases by

$10$ mmHg. Vapour pressure (in mm Hg) of X and Y in their pure states will be ________ respectively.

0%
$200$ and $300$
0%
$300$ and $400$
0%
$400$ and $600$
0%
$500$ and $600$

Explanation

Let

$P=x_xP^0_x+x_yP^0_y$

$(\dfrac{1}{1+3})P_x^0+(\dfrac{3}{1+3})P_y^0=550\;mm\;of\;Hg$

$(\dfrac{1}{1+4})P_x^0+(\dfrac{4}{1+4})P_y^0=560\;mm\;of\;Hg$

on solving for

$P_y^0$ and

$P_x^0$

We get,

$P_x^0=400\;mm\;of\;Hg$

$P_y^0=600\;mm\;of\;Hg$

A solution at

$20^o$ C is composed of

$1.5$ mol of benzene and

$3.5$ mol of toluene. If the vapour pressure of pure benzene of pure benzene and pure toluene at this temperature are

$74.7$ torr and

$22.3$ torr. respectively, then the total vapour pressure of the solution and the benzene mole fraction in equilibrium with it will be, respectively.

0%
$38.0$ torr and $0.589$
0%
$30.5$ torr and $0.389$
0%
$35.8$ torr and $0.280$
0%
$35.0$ torr and $0.480$

Explanation

Let us the problem.

According to the raoult's law is used which states that for a solution of volatile liquids, the partial vapour pressure of each liquid in the solution is directly proportional to its mole fraction.

${X_{benzene}} = \frac{{1.5}}{5} = 0.3$

${X_{toluence}} = \frac{{3.5}}{5} = 0.7$

${P_T} = 74.7 \times 0.3 + 0.7 \times 22.3$

$= 22.41 + 15.61$

$= 38.02\,torr$

${Y_{benzene}} = \frac{{22.41}}{{38.02}} = 0.589$

At what relative humidity will

$Na_2SO_4$ be deliquescent (absorb moisture) when exposed to the air at

$0^o$ C?

Given:

$Na_2SO_4\cdot 10H_2O(s)\rightleftharpoons Na_2SO_4(s)+10H_2O(g); K_p=4.08\times 10^{-25}$ and vapour pressure of water

$0^o$ C

$=4.58$ Torr.

0%
Below $50.5\%$ but Above $30.5\%$
0%
Above $50.5\%$
0%
Above $30.5\%$
0%
Below $30.5\%$

An ideal solution has equal mol-fractions of two volatile components A and B in the vapour above the solution, the mol-fractions of A and B.

0%
are both $0.50$
0%
are equal but necessarily $0.50$
0%
are not very likely to be equal
0%
are $1.00$ and $0.00$ respectively

What is the density of wet air with

$75\%$ relative humidity at

$1$ atm and

$300$ K?

[Given: vapour pressure of

$H_2O$ is

$30$ torr and average molar mass of air is

$29$ g

$mol^{-1}$ ]

0%
$1.614$ g/L
0%
$0.96$ g/L
0%
$1.06$ g/L
0%
$1.164$ g/L

Explanation

Let us consider the equation

${P_{wet\,air}} = \dfrac{{{P_d}}}{{{R_d}T}} + \dfrac{{{P_v}}}{{{R_v}T}} = \dfrac{{{P_d}{M_d} + {P_v}{M_v}}}{{RT}}$

Given that :

$T=$ temperature

$=300\;K$

$M_d=$ molar mass of dry air

$=29\;gmol^{-1}$

$M_v=$ molar mass of water vapourr

$=18\;gmol^{-1}$

$R=0.821\;L\;atm\;mol^{-1}K^{-1}$

$P_v=30\;torr=0.0394\;atm$

$P_d=P-P_v=1-0.0394=0.9606\;atm$

where

$P_d$ partial pressure of dry air

$R_d=$ specific gas constant for dry air

$R_v=$ Specific gas constant for water vapour

${P_{wet\,air}} = \dfrac{{0.9606 \times 29 + 0.0394 \times 18}}{{24.63}}$

$=1.164\;gL^{-1}$

Hence, the correct option is

$D$

Phenol-water system is a/an __________.

0%
element
0%
compound
0%
homogeneous system
0%
heterogeneous system

A binary liquid solution of n-heptane and ethyl alcohol is prepared. Which of the following statements correctly represents the behaviour of this liquid solution?

0%
The solution formed is an ideal solution
0%
The solution formed is a non-ideal solution with positive deviations from Raoult's law
0%
The solution formed is non-ideal solution with negative deviations from Raoult's law
0%
Normal-heptane exhibits positive deviations, whereas ethyl alcohol exhibits negative deviations from Raoult's law

The vapour pressure of two pure liquids

$(A)$ and

$(B)$ are

$100$ and

$80\ torr$ respectively. The total vapour pressure of solution obtained by mixing

$2$ moles of

$(A)$ and

$3$ moles of

$(B)$ would be :-

0%
$120\ torr$
0%
$36\ torr$
0%
$88\ torr$
0%
$180\ torr$

Explanation

Total vapour pressure

$P={ P }_{ A }+{ P }_{ B }$

${ P }_{ A }=100\times \dfrac { 2 }{ 2+3 } =100\times \dfrac { 2 }{ 5 } =40$ torr

${ P }_{ B }=80\times \dfrac { 3 }{ 2+3 } =80\times \dfrac { 3 }{ 5 } =48$ torr

${ P }_{ Total }={ P }_{ A }+{ P }_{ B }=40+48=88$ torr.

The vapour pressure of two liquids

$P$ and

$Q$ are

$80$ torr and

$60$ torr respectively. The total vapour pressure obtained by mixing

$3$ mole of P and

$2$ mole of

$Q$ would be:

0%
$68$ torr
0%
$20$ torr
0%
$140$ torr
0%
$72$ torr

Explanation

Mole fraction of

$P(X_P)=\cfrac {3}{3+2}={3}{5}$

Mole fraction of

$Q(X_Q)=\cfrac {2}{3+2}=\cfrac {2}{5}$

Now, Total vapour pressure

$(P_T)=X_PP_P+X_QP_Q$

$=\cfrac {3}{5}\times 80+\cfrac {2}{5}\times 60$

$\therefore$ Total vapour pressure

$(P_T)=72 torr$

The relative lowering of the vapour pressure of an aqueous solution containing a non volatile solute is

$0.0125$ . The molality of the solution is:

0%
$0.70$
0%
$0.50$
0%
$0.80$
0%
$0.40$

Explanation

Relative lowering of vapour pressure

$=$ mole fraction of solute (Raoult's law)

$\frac{{p - {p_s}}}{p} = {\chi _2}$

Implies that

$\frac{{p - {p_s}}}{p} = \frac{{wM}}{{mW}}$

Implies that

$0.0125 = \frac{{wM}}{{mW}}$

$\frac{w}{{mW}} = 0.012518 = 0.00070$

Hence,

Molality

$= \frac{{wm}}{W} \times 1000$

$= 0.0007 \times 1000 = 0.70$ .

A substance is completely trimerized on dissolution in a solvent. The Van't Hoff factor (

$i$ ) for such change is:

0%
$1$
0%
$2$
0%
$3$
0%
$1/3$

Explanation

Given that,

$\alpha =1$ .

Now,

$3A\longrightarrow A_3$

1 0

$1-\alpha$

$\dfrac{\alpha}{3}$

Van't Hoff Factor

$(i)=1-\alpha+ \dfrac{\alpha}{3}=1-1+\dfrac{1}{3}=\dfrac{1}{3}$

What will b the result if $100ml{\text{ of }}0.06{\text{ M }}Mg{\left( {N{O_3}} \right)_2}$ is added to $50ml{\text{ of 0}}{\text{.06M }}N{a_2}{C_2}{O_4}?$ $\left[ {{K_{sp}}{\text{ of }}Mg{C_2}{O_4} = 8.6{\text{ x }}{{10}^{ - 5}}} \right]$

0%
A precipitate will not be formed
0%
A precipitate will form and an excess of $Mg{^{2 + }}$ ions will remain in the solution
0%
A precipitate will form and an excess of ${C_2}{O_4}^{2 - }$ ions will remain in the solution
0%
A precipitate is form but neither ion is present in excess

What would be the vapour pressure of 0.5 molal solution of non volatile solute in benzene at

$30^oC$ ? The vapour pressure of pure benzene at

$30^oC$ is 119.6 torr.

0%
119.6 torr
0%
121.23 torr
0%
118.52 torr
0%
None of these

Explanation

$\dfrac{\triangle p}{p^0}=x_2$

$\dfrac{\triangle p}{p^0}=\dfrac{n_2}{n_1}$

$=\dfrac{m_2\times M_1}{M_2\times m_1}$

$=\dfrac{m\times m_1}{1000}$

$\dfrac{\triangle p}{p^0}=\dfrac{0.5\times 18}{1000}$

$=0.009$

$\dfrac{p^0-p}{p^0}=0.009$

$[m=molarity=\dfrac{m_2}{M_2}\times \dfrac{1000}{m_1}$

$\dfrac{119.6-p}{119.6}=009$

$19.6-p=1.0764$

$\boxed{p=118.52}torr$

If 2 moles of A and 3 moles of B are mixed to form an ideal solution vapour pressure of A and B are 120 and 180 mm of Hg respectively. then the composition of A and B in the Vapour phase when the first traces of vapour are formed in the above case is:

0%
$X^l A = 0.407$
0%
$X^l A = 0.8$
0%
$X^l A = 0.109$
0%
$X^l A = 0.307$

Explanation

$X_A=\dfrac{2}{2+3}=\dfrac{2}{5}$

$X_B=\dfrac{3}{2+3}=\dfrac{3}{5}$

$X^1_A=\dfrac{P^0_A\times A}{P_{total}}$

$=\dfrac{120\times \dfrac{2}{5}}{120\times \dfrac{2}{5}+180\times \dfrac{3}{5}}$

$=\dfrac{48}{48+108}$

$X^1_A=0.307$

The vapour pressure of a pure liquid A is

$70$ torr at

$27^o$ C. It forms an ideal solution with another liquid B. The mole fraction of B is

$0.2$ and total vapour pressure of the solution is

$84$ torr at

$27^o$ C. The vapour pressure of pure liquid B at

$27^o$ C.

0%
$14$
0%
$56$
0%
$140$
0%
$70$

Explanation

Here we will use the simple total vapour pressure equation:

$P_T=x_AP_A^o+x_BP_B^o$

$\Rightarrow 84\ torr= 0.8 \times 70\ torr+ 0.2 \times P_B^o$

$\Rightarrow P_B^o=140\ torr$ .

The vapour pressure of pure water is 760 mm at

$25^{ 0 }C$ .The vapour pressure of solution containing 1(m) solution of glucose will be:

0%
761.0 mm
0%
746.5 mm
0%
734.5 atm
0%
746.5 Pa

Explanation

Vapor pressure

$\dfrac{P^{o}- P}{P^{o}}= \dfrac{W_{2}}{M_{2}} \times \dfrac{M_{1}}{W_{1}}$

$P^{o}= 760\ mm$

$M_{1} = 18$

Molality

$(m)= 1m= \dfrac{mole\ of\ solute}{(weight)\ kg\ of\ solvent }= \dfrac{W_{2}/M_{2}}{W_{i} m \ gm}$

$\dfrac{760-P}{760} = 1 \times 10^{-3} \times 18$

$P= 746.5\ mm$

Hence, the correct option is

$B$

The van't Hoff factor is 3 for:

0%
$Na_2SO_4$ with 50% dissociation
0%
$Al_2(SO_4)_3$
0%
$CaCl_2$ with 80% dissociation
0%
$K_4[Fe(CN)_6]$ with 50 % dissociation

Explanation

(A)

$Na_2SO_4 \rightleftharpoons 2Na^++SO_4^{2-}$

$1-0.5$

$2 \times 0.5$

$0.5$ (

$50$ % dissociation)

$\therefore$ Vant Hoff factor,

$i= 1-0.5+2\times 0.5+0.5=2$

(B)

$Al_2(SO_4)_3\rightleftharpoons 2Al^{3+}+3SO_4^{2-}$

$i=5$

(C)

$CaCl_2\rightleftharpoons Ca^{2+}+2Cl^-$

$1-0.8$

$0.8$

$2\times 0.8$ (

$80$ % dissociation)

$i=1-0.8+0.8+2\times 0.8=2.6$

(D)

$K_4[Fe(CN)_6]\rightleftharpoons 4K^++[Fe(CN)_6]^{4-}$

$1-0.5$

$4 \times 0.5$

$0.5$ (

$50$ % dissociation)

$\therefore i=1-0.5+4 \times 0.5+0.5=3$ .

A Solution of non-volatile solute in water freezes at

$-0.{ 30 }^{ }C.$ The vapour-pressure of pure water at

$298K$ is

$23.51mm Hg$ and

${ K }_{ 1 }$ for water is

$1.86 K. kg { mol }^{ -1 }$ Calculate the vapour-pressure of this solution at

$298K.$

0%
$23.4mm$
0%
$24.8mm$
0%
$34.8mm$
0%
$40mm$

Explanation

We know that,

$\triangle { T }_{ f }={ K }_{ f }m$

$\therefore \quad m=\cfrac { \triangle { T }_{ f } }{ { K }_{ f } } =\cfrac { 0.30 }{ 1.86 } =0.161$

According to Raooh's law,

${ P }^{ \circ }-P/{ P }^{ \circ }=$ No. of moles of Solute/No of moles of solvent

$\therefore \quad 23.51-P/23.51=\cfrac { 0.161 }{ 1000/18 } =0.0020898$

$\therefore \quad P=23.51-23.51\times 0.0020898$

$\therefore \quad P=23.44mmHg$

The composition of vapour when first bubble formed is:

0%
$y_A = 0.6; y_B = 0.4$
0%
$y_A = 0.48; y_B = 0.52$
0%
$y_A = 0.52; y_B = 0.48$
0%
$y_A = 0.5; y_B = 0.5$

The vapour pressure of water at

$20^{o}C$ is

$17.54mm Hg$ . Then the vapour pressure of the water in the apparatus is lowered, decreasing the volume of the gas above the liquid to one half of its initial volume (temp. constant) is:

0%
$5.77\ mmHg$
0%
$16\ mmHg$
0%
$35.08\ mmHg$
0%
between $8.77$ and $15.54\ mmHg$

Explanation

We know that,

\dfrac {\text{ Moles of sucrose} }{\text{ Moles of sucrose}+\text{ Moles of water} }

So,

$X_{ sucrose }=\dfrac { \dfrac { 200g }{ 342.13gmol^{ -1 } } }{ \dfrac { 200g }{ 342.13gmol^{ -1 } } +\dfrac { 112.3g }{ 18.01gmol^{ -1 } } } =0.0857$

Mole fraction of water:

$X_{ water }=\dfrac { \dfrac { 112.3g }{ 18.01gmol^{ -1 } } }{ \dfrac { 112.3g }{ 18.01gmol^{ -1 } } +\dfrac { 200g }{ 342.13gmol^{ -1 } } } =0.914$

Note that by definition,

$x_{ sucrose }+x_{ water }=1$

Because sucrose is involatile, the vapour pressure of the solution is proportional to the mole fraction of water:

$P_{ solution }=X_{ water }+17.54\ mm$ of

$Hg$

$=0.914\times 17.54\ mm$ of

$Hg=16\ mm$ of

$Hg$
Conclusion: hence the correct answer is not mentioned in the above options.

A liquid A and B form an ideal solution. If vapour pressure of pure A and B are

$500N{ m }^{ -2 }$ and

$200N{ m }^{ -2 }$ respectively, the vapour pressure of a solution of A and B containing 0.2 mole fraction of A would be:

0%
$700N{ m }^{ -2 }$
0%
$300N{ m }^{ -2 }$
0%
$260N{ m }^{ -2 }$
0%
$140N{ m }^{ -2 }$

Explanation

Vapour pressure of a solution containing A & B liquid is

$P_T=P_A+P_B$

$P_A= X_A-P^o_A$

$X_A$ = mole fraction of A;

$P^o_A$ = Pure vapour pressure of A.

$\Rightarrow P_A= 0.2 \times 500 Nm^{-2}$ &

$P_B= (1-0.2)\times 200 Nm^{-2}$

Thus

$P_T= 0.2 \times 500 Nm^{-2} + 0.8 \times 200 Nm^{-2}$

$\Rightarrow P_T= 260 Nm^{-2}$

Thus total vapour pressure of solution is

$260 Nm^{-2}$ .

A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of

$290\ mm$ at

$300\ K$ . The vapour pressure of propyl alcohol is

$200\ mm$ . If the mole fraction of ethyl alcohol is

$0.6$ , its vapour pressure (in

$mm$ ) at the same temperature will be:

0%
$300$
0%
$700$
0%
$360$
0%
$350$

Explanation

By Roult's law,

${ P }_{ T }={ P }_{ A }^{ \circ }{ x }_{ A }+{ P }_{ B }^{ \circ }{ x }_{ B }$

$\therefore \quad 290mm={ P }_{ A }^{ \circ }\times 0.6+200\times \left( 1-0.6 \right)$

$={ P }_{ A }^{ \circ }\times 0.6+80$

$\therefore \quad { P }_{ A }^{ \circ }=350\ mm$

Hence, the correct option is

$D$

$3.00\ L$ oxygen gas is collected over water at

${27}^{o}C$ when the barometric pressure is

$787\ Torr$ . If vapour pressure of water is

$27\ Torr$ at

${27}^{o}C$ , moles of

$O_{2}$ gas will be?

0%
$0.122$
0%
$0.126$
0%
$0.004$
0%
$0.152$

Explanation

Given Volume (V)

$=3.00$ L

Temperature(T)

$=27^{0}C=300 K$

Barometric pressure

$=787$ torr

Vapour pressure of water

$=27$ torr

total pressure

$=787-27=760\quad{torr}=1\quad{atm}$

Applying in ideal gas equation

PV=nRT

$n=\dfrac{PV}{RT}$

$n=\dfrac{{1}\times{3}}{{0.0821}\times{300}}=0.122$

Which of the following condition is not followed for an ideal solution?

0%
$\Delta { H }_{ mixing }$ =0
0%
$\Delta { S }_{ mixing }$ =0
0%
$\Delta { v }_{ mixing }$ =0
0%
All A, B, and C are followed

Explanation

An ideal solution satisfies the following conditions.

1)There will be no change in volume on mixing two components.

$\triangle V_{mix}=0$

2)There will be no charge in enthalpy on mixing two components.

$\triangle H_{mix}=0$

So, the

$\Delta S_{mixing}=0$ is not followed by the ideal solution.

Which of the following pair will form an ideal solution?

0%
Chlorobenzen , chloro ethane
0%
Benzene, Toluene
0%
Acetone, chloroform
0%
Water, HCl

Explanation

The substances that have similar structures and polarities form nearly ideal solution benzene and toulene will form an ideal solution.

Acetone, chloroform, chlorobenzene chloroethane ,water

$HCl$ will form non-ideal solutions due to the difference of polarites between the two components.

$3$ moles of

$P$ and

$2$ moles of

$Q$ are mixed, what will be their total vapour pressure in the solution if their partial vapour pressures are

$80$ and

$60$ respectively?

0%
$80$ torr
0%
$140$ torr
0%
$720$ torr
0%
$70$ torr

Explanation

Mole fraction of

$P=\dfrac { 3 }{ 3+2 } =\dfrac { 3 }{ 5 }$

Mole fraction of

$Q=\dfrac { 2 }{ 3+2 } =\dfrac { 2 }{ 5 }$

Hence,

Total vapour pressure =(Mole fraction of

$P$

$\times$ Vapour pressure of

$P$ ) + (mole fraction of

$Q$

$\times$ Vapour pressure of

$Q$ )

$=\left( \dfrac { 3 }{ 5 } ×80 \right) +\left( \dfrac { 2 }{ 5 } ×60 \right) =48+24=72\ torr$

For which of the following binary solution

${P}_{Total}={P}_{A}^{o}\times {X}_{A}+ {P}_{B}^{o}\times {X}_{B}$ is followed?

0%
Phenol+ Aniline
0%
Benzene + Toluene
0%
Acetone + Chloroform
0%
water + $HCl$

How many grams of sucrose must be added to

$320g$ of water to lower the vapour pressure by

$1.5mm$

$Hg$ at

${25}^{o}C$ ?
(Given: The vapour pressure of water at

${25}^{o}C$ is

$23.8mm$

$Hg$ and molar mass of sucrose is

$324.3g/mol$ )

0%
$21.5g$
0%
$140g$
0%
$363.36g$
0%
$160.12g$

Explanation

$w_{1}=$ Wsucrose = ?

$M_{1}=$ Molecular weight

$+32,4.3\ g/mol$

$w_{2}=320\ g$

$M_{2}=H_{2}O=18\ g/mol$

$p-ps=1.5\ mm$

$p=23.8\ mm$

$\dfrac{p-ps}{p}=\dfrac{W_{1}}{M_{1}}\times \dfrac{M_{2}}{W_{2}}$

$\dfrac{1.5}{23.8}=\dfrac{W_{1}}{324.3}\times \dfrac{18}{320}$

$W_{1}=363.36\ g$

If two liquids

$A({P}_{A}^{o}=100torr)$ and

$B({P}_{B}^{o}=200torr)$ are completely immiscible with each other, each one will behave independently of the other are present in closed vessel. The total pressure of the system will be:

0%
less than $100$ torr
0%
less than $200$ torr
0%
between $100$ and $200$ torr
0%
$300$ torr

Explanation

${ P }_{ TOTAL }={ P }^{ 0 }A\quad +{ P }^{ 0 }B\\ \quad \quad \quad \quad \quad =100+200\\ \quad \quad \quad \quad \quad =300\quad torr\\ \quad \quad \quad$

$P^{0}_{A}$ and

$P^{0}_{B}$ are

$108$ and

$36$ torr respectively. What will be the mole fraction of

$A$ is vapour phase if

$B$ has mole fraction in solution

$0.5$ :

0%
$0.25$
0%
$0.75$
0%
$0.60$
0%
$0.35$

Explanation

$P=X_AP^{0}_{A}+ X_BP^{0}_{B}$

$=0.5\times 108+0.5\times 36$

$=54+18=72$ torr

$Now,\quad { X }_{ a }=\dfrac { { P }_{ a } }{ P } =0.5\times \dfrac { 108 }{ 72 } =\dfrac { 54 }{ 72 } =0.75$

Hence, the correct option is

$A$

The vapour pressure of a liquid decreases by

$10 torr$ when a non-volatile solute is dissolved. The mole fraction of the solute in solution is

$0.1$ . What would be the mole fraction of the liquid if the decrease in vapour pressure is

$20torr$ , the same solute being dissolved?

0%
$0.2$
0%
$0.9$
0%
$0.8$
0%
$0.6$

Explanation

Answer is option

$C.$

According to Rault's law,

$\cfrac{P^0-P}{P}=X_a\longrightarrow (1)$ Where

$X_a=\text{Mole fraction of solute.}$

$\underline {\text{Case 1}:}$

$P^0-P=10$

$torr,$

$X_a=0.1$

$\therefore$

$\cfrac{10}{P}=0.1$ or

$P=\cfrac{10}{0.1}=100$

$torr$

$\underline {\text{Case 2}:}$

$P^0-P=20$

$torr,$

$\therefore (1)\longrightarrow \cfrac{20}{100}=X_a\Longrightarrow X_a=0.2$

We have

$X_a+X_b=1,$ where

$X_b=\text{Mole fraction of solvent.}$

$\therefore X_b=1-0.2=0.8$

A manometer attached to a flask containing ammonia gas have no difference in mercury level initially as shown in diagram. After sparking into the flask, ammonia is partially dissociated as

${ 2NH }_{ 3 }(g)\longrightarrow N_{ 2 }(g){ +3H }_{ 2 }(g)$ now it have difference of

$6\ cm$ in mercury level in two columns, what is partial pressure of

${ H }_{ 2 }(g)$ at equilibrium?

0%
$9\ cm\ Hg$
0%
$18\ cm\ Hg$
0%
$27\ cm\ Hg$
0%
$None\ of\ these$

If the relative lowering in vapour pressure for he ethanolic solution containing non volatile solute is

$0.02$ , then the molality of solution will be.(molecular mass of ethanol =

$46$ gm/mole)

0%
$0.22 m$
0%
$1.33 m$
0%
$1.09 m$
0%
$0.44 m$

Explanation

By Roulits law

Relative lowering vapour pressure= mole fraction of solute

$\dfrac { P-P(s) }{ P } ={ X }_{ 2 }$

$\dfrac { [P-P(S)] }{ P } =\dfrac { W\times M }{ W\times M } =0.02$

$\dfrac { W }{ m } \times W=\dfrac { 0.02 }{ 46 } =4.347\times { 10 }^{ -4 }$

$Molality\quad =4.347\times { 10 }^{ -4 }\times 1000=0.434M$

18 g glucose is added to 178.2 g of water the vapour pressure of water for this aqueous solution at

${ 100 }^{ 0 }C$ is:

0%
704 torr
0%
759 torr
0%
7.6 torr
0%
None of these

Explanation

Molar mass of water $=18$ $g$

For $178.2$ $g$ of $H_2$ , $n_A = 9.9$ $X_B= \cfrac {0.1}{0.1 + 9.9} =0.01$

Molar mass of Glucuse $=180$ $g$ $X_A = 0.99$

For $18$ $g$ , $n_B = 0.1$

For lowering of vapour pressure ,

$P = P^o \times A= P^oA(1-X_B) = 760(1-0.01)$

$P = 760-7.6$

$P= 752.4$ $torr$

The mass of glucose that should be dissolved in

$100g$ water in order to produce same lowering of vapour pressure as produced by dissolving

$1g$ urea in

$50g$ of water is?

0%
$1g$
0%
$2g$
0%
$6g$
0%
$12g$

Explanation

$\left( { H }_{ 2 }NCON{ H }_{ 2 } \right) \rightarrow M=60g/mol$

$m=$ molar mass

${ M }_{ { C }_{ 6 }{ H }_{ 12 }{ O }_{ 6 } }=12\times 6+12\times 1+6\times 6=72+12+96+180 g/mol$

Relative lowering in Vapour pressure

$=no.\quad of\quad moles\times \cfrac { molar\quad mass\quad of\quad solvent }{ mass\quad of\quad water } \\ =0.33\times \cfrac { 18 }{ 50 }$

$\cfrac { W }{ 180 } \times \cfrac { 18 }{ 100 } =\cfrac { 1 }{ 60 } \times \cfrac { 18 }{ 50 } \\ W=6g$

The vapour pressure of o-nitrophenol at any given temperature is predicted to be:

0%
Higher than that of p-nitrophenol
0%
Lower than that of p-nitrophenol
0%
Same as that of p-nitrophenol
0%
Higher or lower depending upon the size of the vessel

The vapour pressure is least for?

0%
pure water
0%
0.1m aqueous urea
0%
0.2m aqueous urea
0%
0.3m aqueous urea

Explanation

Vapour pressure decreases as content of any impurity increases in solution.

$\therefore V.P$ will be least for

$0.3m$ aq.urea

Relative lowering of vapour pressure is maximum for:

0%
0.1m urea
0%
0.1m NaCI
0%
$0.1m\quad Mg{ CI }_{ 2 }$
0%
$0.1m\quad AI_{ 2 }{ \left( { SO }_{ 4 } \right) }_{ 3 }$

Explanation

$RLVP=\dfrac{\Delta P}{P}=i_{OL_{solute}}$

$\therefore$ concentration is same we will look out for

$i$ (vont-haff factor)

$A\rightarrow i=2$

$B\rightarrow i=2$

$C\rightarrow i=3$

$D\rightarrow i=5$

$\therefore$ Max. for

$D$

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