MCQ Questions for Class 12 Engineering Chemistry Solutions Quiz 7

Two liquids A and B form an ideal solution. The vapour pressure of pure A and pure B are 66mm of Hg and 88mm of Hg, respectively. Calculate the composition of vapour A in the solution which is equilibrium and whose molar volume is 36%.

0%
0.43
0%
0.70
0%
0.30
0%
0.50

Explanation

$P_t= P_A+P_B =66mm+88mm=154mm$ of

$Hg$

Now,

$P_A$

$=X_A \times P_T$

Therefore,

$X_A$ =

$\cfrac{P_A}{P_T}=\cfrac{66}{154}$

$=0.43$

Therefore, composition of Vapour A is

$0.43$ .

Vapour pressure of

${ CCI }_{ 4 }$ at

${ 24 }^{ o }C$ is

$143 mmHg 0.05$ g of the non-volatile solute (mol.wt.=65) is dissolved in

$100 ml { CCI }_{ 4 }$ . Find the vapour pressure of the solution (Density of

${ CCI }_{ 4 }$ =

$1.58 g/{ cm }^{ 2 }$ )

0%
$143.99 mm$
0%
$94.39 mm$
0%
$199.34 mm$
0%
$14.197 mm$

Explanation

The equation of relative lowering vapour pressure is given as-

$\cfrac{{p}^{0} - p}{{p}^{0}} = {X}_{2}$

Whereas,

${p}^{0} =$ Vapour prssure of solvent

$p =$ Vapour pressure of solution

${X}_{2} =$ Mole fraction of solute

Given that density of

$C{Cl}_{4}$ is

$1.58 \; {g}/{{cm}^{3}}$

As we know that,

$\text{density} = \cfrac{\text{mass}}{\text{volume}}$

$\Rightarrow 1.58 = \cfrac{\text{mass}}{100}$

$\Rightarrow \text{mass of } C{Cl}_{4} = 158 \; g$

Molecular weight of

$C{Cl}_{4} = 154 \; g$

$\therefore$ No. of moles of

$C{Cl}_{4} = \cfrac{158}{154} = 1.026 \text{ mol}$

Given weight of solute

$= 0.05 \; g$

Molecular weight of solute

$= 65 \; g$

$\therefore$ No. of moles of solute

$= \cfrac{0.05}{65} = 0.00077 \text{ mol}$

$\therefore$ Total no. of moles

$= 1.026 + 0.00077 = 1.02677 \text{ mol}$

$\therefore$ Mole fraction of solute

$= \cfrac{0.00077}{1.02677} = 0.00075$

Vapour pressure of solvent

$= 143 \text{ mm of Hg}$

Noe from equation of relative lowering vapour pressure, we have

$\cfrac{143 - p}{143} = 0.00075$

$\Rightarrow 143 - p = 0.107$

$\Rightarrow p = 142.893 \text{ mm of Hg}$

Hence the vapour pressure of solution is

$142.893 \text{ mm of Hg}$ .

Which of the liquids in each of the following pairs has a higher vapour pressure?

0%
Alcohol, glycerine
0%
Mercury, water
0%
petrol, kerosene
0%
None of these

The vapour pressure of water at

$20$ is 17.5 mm Hg. If 18 g of glucose

${ C }_{ 6 }{ H }_{ 12 }{ O }_{ 6 }$ is added to 178.2 g water

$20$ , the vapour pressure of the resulting solution will be:

0%
17.675 mm Hg
0%
15.750 mm Hg
0%
16.500 mm Hg
0%
17.325 mm Hg

Explanation

No. of moles of glucose=

$\cfrac{18}{180}=0.1$

No. of moles of water=

$\cfrac{178.2}{18}=9.9$

Mole fraction of glucose=

$\cfrac{0.1}{4.9+0.1}=\cfrac{0.1}{10}=0.01$

Using Raoult's law, as solute is non-volatile,

${ x }_{ \beta }=\cfrac { { P }_{ A }^{ o }-{ P }_{ A } }{ { P }_{ A }^{ o } }$

$P_{A}^{o}=$ vapour pressure of pure water

$P_{A}=$ vapour pressure of the solution

$x_{\beta}=$ mole fraction of glucose

$\Longrightarrow 0.01=\cfrac { 17.5-{ P }_{ A } }{ 17.5 } \\ \Longrightarrow { P }_{ A }=17.5-0.175=17.325$

The vapor pressure of the mixture=17.325 mm Hg.

Hence, the correct option is

$D$

${ 20 }^{ 0 }C$ , the vapour pressure of diethyl either is 44mm. When 6.4 g. of a non-volatile solute is dissolved in 50g. of either, the vapour pressure falls to 410mm. The Molecular weight of the solute is:

0%
150
0%
130
0%
160
0%
180

Explanation

$\Delta _{P}=440-410=30mm$

$\therefore \dfrac{\Delta _{P}}{P_{0}}=x_{solute}$

$\Rightarrow \displaystyle \frac{30}{440}=\frac{\frac{6.4}{M}}{\frac{6.4}{M}+\frac{50}{74}}$

$\Rightarrow \displaystyle \frac{19.2}{M}+\frac{150}{74}=\frac{44\times 6.4}{M}$

$\Rightarrow \displaystyle \frac{150}{74}=\frac{6.4(44-3)}{M}$

$\Rightarrow \displaystyle M=\frac{6.4\times 41\times 74}{150}$

$\approx 130$

The vapour pressure of pure

$A$ is

$10$ torr and at the same temperature when

$1\ g$ of

$B$ is dissolved in

$20\ gm$ of

$A$ , its vapour pressure is reduced to

$9.0$ torr. If the molecular mass of

$A$ is

$200$ amu, then the molecular mass of

$B$ is:

0%
$100\ amu$
0%
$90\ amu$
0%
$75\ amu$
0%
$120\ amu$

Explanation

According to Dalton's law of partial pressure,

$p_{total}=\chi_Ap_A+\chi_Bp_B$

Since B is a non-volatile substance,

$p_B=0$

$\Rightarrow p_{total}=p_A\times \dfrac{\dfrac{m_A}{M_A}}{\dfrac{m_A}M_A+\dfrac{m_B}{M_B}}$

$\Rightarrow 9=10\times \dfrac{\dfrac{20}{200}}{\dfrac{20}{200}+\dfrac{1}{M_B}}$

$\Rightarrow M_B=90\ amu$

Hence, option

$B$ is the correct answer.

The vapour pressure of a solvent decreased by 10 mm of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be 20 mm of mercury?

0%
0.4
0%
0.6
0%
0.8
0%
0.2

Explanation

$\cfrac {\Delta P}{P_o}=\cfrac {n}{n+N}$

$\Rightarrow \cfrac {10}{P_o}=0.2$

$\Rightarrow P_o=50mm$

For other solutions of same solvent,

$\Rightarrow \cfrac {20}{P_o}=\cfrac {n}{n+N}$

$\Rightarrow \cfrac {20}{50}=$ mole fraction of solute=

$0.4$

$\therefore$ Mole fraction of solvent=

$1-0.4=0.6$

Hence, the correct option is

$\text{B}$

Vapours of a liquid can exist only at ?

0%
below boiling point
0%
below critical temperature
0%
below inversion temperature
0%
above critical temperature

When some

$NaCl$ was dissolved in water, the freezing point depression was numerically equal to twice the molal

$K_f$ depression constant. The relative lowering of vapour pressure of the solution is nearly:

0%
$0.036$
0%
$0.018$
0%
$0.0585$
0%
$0.072$

Explanation

$\Delta T_f= K_f \cfrac {W_1}{M_1 \times W_2}\times 1000$ [

$W_2$ is given in grams]

where

$1$ and

$2$ denote solute & solvent respectively.

Given,

$\Delta T_f=2 K_f \Rightarrow \cfrac {W_1 \times 1000}{M_1 \times W_2}= 2$

Relative lowering of vapour pressure=

$\cfrac {W_1 \times M_2}{M_1 \times W_2}$

Here,

$M_2$ = Molecular weight of

$H_2O= 18$

$\therefore$ Relative lowering of vapour pressure=

$\left(\cfrac {W_1}{M_1 \times W_2}\right)\times 18$

$=\cfrac {2}{1000}\times 18$

$= 0.036$

The vapour pressure of pure liquid is

$70$ Torr at

$27^o C$ . The vapour pressure of a solution of this liquid and another liquid (mole fraction 0.2) is

$84$ Torr at

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

$B$ at

$27^o C$ is?

0%
$140$ Torr
0%
$280$ Torr
0%
$160$ Torr
0%
$200$ Torr

Explanation

Vapour Pressure of pure liquid

$A=70$ torr (

$P_{A}^{o}$ )

Mole fraction of liquid

$B=0.2$ (

$x_{B}$ )

Thus mole fraction of liquid

$A=x_{A}=1-x_{B}$

Total vapour pressure

$=84$ torr

$P=x_{A} P_{A}^{o} + x_{B} P_{B}^{o}$

$84=(0.8) \times 70 + 0.2 \times P_{B}^{o}$

$P_{B}^{o}$

$=140$ torr

$=$ vapour pressure of the pure liquid

$B$

Hence, the correct option is

$\text{A}$

${ \Delta }_{ f }{ G }^{ o }$ at

$500\ K$ for substance

$S$ in liquid state and gaseous state are

$+100.7\ kcal\ { mol }^{ -1 }$ and

$+103\ kcal\ { mol }^{ -1 }$ respectively. Vapour pressure of liquid

$S$ at

$500\ K$ is approximately equal to:

$(R= 2\ { cal }\ K^{ -1 }\ { mol }^{ -1 })$

0%
$0.1\ atm$
0%
$1\ atm$
0%
$10\ atm$
0%
$100\ atm$

Explanation

Substance (s)

$\rightarrow$ Substance (g)

$\triangle G^{o}=103-100.7$

$=2.3\ K\ cal\ mol^{-1}$

$K_{p}=$ Vapour pressure of the substance

$K_{p}=P$

$\triangle G=-RT\ In\ K_{p}$

$(\because K_{p}=P)$

$2.3=-2\times 10^{-3}\times 500\ In\ P$

$2.3=-2\times 500\times10^{-3}\times 2.3\ \log P$

$\log P=-1$

$P=0. 1\ atm$

A solution has a

$1:4$ mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at

$20^{\circ}C$ are

$440$ mm of

$Hg$ for pentane and

$120$ mm of

$Hg$ for hexane. The mole fraction of pentane in the vapour phase would be?

0%
$0.549$
0%
$0.200$
0%
$0.786$
0%
$0.478$

Explanation

Mole fraction of pentane solution=

$\cfrac {1}{1+4}=\cfrac {1}{5}$

Mole fraction of hexane solution=

$\cfrac {4}{1+4}=\cfrac {4}{5}$

Total pressure of solution,

$P_S=X_{P}P_P^O+X_HP_H^O$

$=\cfrac {1}{5}\times 440\ mm\ Hg+\cfrac {4}{5}\times 120\ mm\ Hg$

$=184\ mm\ Hg$

$\therefore$ The fraction of pentane in the vapour phase

$=\cfrac {X_PP_P^O}{P_S}=\cfrac {88}{184}=0.478$ .

Hence, the correct option is

$D$

Vapour pressure of pure benzene is 119 torr and of toluene is 37.0 torr at the same temperature mole fraction of toluene in vapour phase which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene 0.50, will be:

0%
$0.137$
0%
$0.205$
0%
$0.237$
0%
$0.435$

The amount of solute (mol mass=

$60$ ) that must be added to

$180g$ of water so that the vapour pressure of water is lowered by

$10 \%$ is?

0%
$30$ g
0%
$60$ g
0%
$120$ g
0%
$12$ g

$20$ ml of

$1M\ BaCl_2$ solution and

$25$ ml of

$0.1$ M

$Na_2SO_4$ are mixed to form

$BaSO_4$ . The amount of

$BaSO_4$ precipitated are:

0%
$2.5$ g
0%
$2.5$ moles
0%
$2.5$ millimoles
0%
$0.05$ millimoles

Explanation

$BaCl_2+Na_2SO_4\longrightarrow BaSO_4\downarrow +2NaCl$

$1$ mole of

$BaCl_2$ and

$1$ mole of

$Na_2SO_4$ mix to form

$1$ mole of

$BaSO_4$

Moles present of

$BaCl_2=20\times 1\times 10^{-3}=20\ mmoles$ .

Moles of

$Na_2SO_4=25\times 0.1\times 10^{-3}=2.5\ mmoles.$

$\therefore$ Moles of

$BaSO_4$ ppt

$=2.5\ mmoles.$

A mixture of ethanol and propanol has vapor pressure 279 mm of Hg at 300 K. The mol fraction of ethanol in the solution is 0.6 and vapour pressure of pure propanol is 210 mm of Hg. Vapour pressure of pure ethanol will be:

0%
$336.6$ mm
0%
$325$ mm
0%
$375$ mm
0%
$415$ mm

Explanation

$p=p_1+p_2$

$\Rightarrow p={p_1}^0x_1+{p_2}^0x_2$ ( Raoult's law )

$p={p_1}^0\left( 1-x_2\right) +{p_2}^0x_2$

$p={p_1}^0+\left( {p_2}^0-{p_1}^0\right) x_2$

$279mm={p_1}^0+\left( {p_2}^0-{p_1}^0\right) x_2$

Let

$2,$ ethanol

$1,$ propanol

$\therefore 279mm=210mm+\left( {p_2}^0-210\right) 0.6$

$69=\left( {p_2}^0-210\right)0.6$

${p_2}^0-210=115$

$\Rightarrow {p_2}^0=325mm$

Hence, the answer is

$325mm.$

The vapour pressure of a pure liquid A is 70 torr at 300 K. It forms an ideal solution with another liquid B. The mole fraction of B in the solution is 0.2 and total pressure of solution is 84 torr at 300 K. The vapour pressure of pure liquid B at 27C is ?

0%
0.14 torr
0%
560 torr
0%
140 torr
0%
70 torr

Explanation

Solution:- (C)

$140 \text{ torr.}$

Mole fraction of

$B = 0.2$

As we know that,

${x}_{A} + {x}_{B} = 1$

$\Rightarrow {x}_{A} = 1 - 0.2 = 0.8$

${{p}^{°}}_{A} = 70 \text{ torr.}$

${{p}^{°}}_{B} = ?$

${p}_{total} = 84 \text{ torr.}$

From Dalton's law of partial pressure,

${p}_{total} = {p}_{A} + {p}_{B} ..... \left( 1 \right)$

From Raoult's law of vapour pressure,

${p}_{A} = {{p}^{°}}_{A} \cdot {x}_{A}$

$\Rightarrow {p}_{A} = 70 \times 0.8 = 56$

Again, from Raoult's law of vapour pressure,

${p}_{B} = {{p}^{°}}_{B} \cdot {x}_{B}$

${p}_{B} = {{p}^{°}}_{B} \times 0.2$

Substituting these values in equation

$\left( 1 \right)$ , we have

$84 = 56 + 0.2 \times {{p}^{°}}_{B}$

$\Rightarrow {{p}^{°}}_{B} = \cfrac{84 - 56}{0.2} = 140 \text{ torr.}$

Hence the vapour pressure of pure liquid

$B$ is

$140 \text{ torr.}$ .

The boiling point of

$C_{6}H_{6},CH_{3}OH, C_{6}H_{5}NH_{2}$ and

$C_{6}H_{5}NO_{2}$ are

$80^{\circ},\ 65 ^{\circ},\ 184^{\circ}$ and

$212^{\circ}$ respectively. Which will show highest vapourr pressure at room temperature?

0%
$C_{6}H_{6}$
0%
$CH_{3}OH$
0%
$C_{6}H_{5}NH_{2}$
0%
$C_{6}H_{5}NO_{2}$

Explanation

Boiling point is the temperature at which the vapor pressure of a liquid become equal to atmospheric pressure.

$\rightarrow$ So, the compounds which has low boiling point will have high vapor pressure.

$\rightarrow$ Given

${ CH }_{ 3 }OH$ has boiling point of

${ 65 }^{ 0 }$ .

$\rightarrow$ It has highest vapor among given with

$0.166$ atm.

Ans :- Option B.

For a non-electrolytic solution, van't Hoff factor is equal to:

0%
$0$
0%
$1$
0%
$2$
0%
$0\quad <\quad i\quad \le \quad 1$

One mole of non volatile solute is dissolved in two moles of water. The vapour pressure of the solution relative to that of water is:

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

Explanation

Relative lowering in vapor pressure

$=\cfrac{P^{\circ}-P_s}{P^{\circ}}=x_A$

Where

$x_A=$ mole fraction of non-volatile solute.

$\therefore \cfrac{P^{\circ}-P_s}{P^{\circ}}=\cfrac{1}{3}\\ \therefore \cfrac{P_s}{P^{\circ}}=1-\cfrac{1}{3}=\cfrac{2}{3}$

Hence, the correct option is

$A$

$P^o$ and

$P^s$ are the vapour pressure of the solvent and solution respectively.

$n_ 1$ and

$n_2$ are the mole fraction of the solvent and solute respectively, then?

0%
$P^S=P^o n_1$
0%
$P^S=P^o n_2$
0%
$P^o=P^S n_2$
0%
$P^S=P^o \dfrac { { n }_{ 1 } }{ { n }_{ 2 } }$

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

$290mm$ at

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

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

$0.6$ its vapour presssure at same temperature will be:-

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

Explanation

$P_{sol} = X_{sol} . P^o_{sol}$

$X \rightarrow$ mole fraction

$P^o \rightarrow$ pure vapour pressure of solution

$290 mm =X_{ethanal} . P^o_{ethanal} + X_{propanal} . P^o_{propanal}$

$290 mm =0.6 \times P^o_{ethanal} + (1-0.6) \times 200 mm$
on solving

$P^o_{ethanal} =350$

$nm$

Vapour pressure diagram of some liquids plotted against temperature are shown below most volatile liquid is

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

Two liquids, A and B, form an ideal solution. At the specified temperature, the vapour pressure of pure A is 200 mm Hg while that of pure B is 75 mm Hg. If the vapour over the mixture consists of 50 mol percent A, what is the mole percent A in the liquid?

0%
20
0%
76
0%
27
0%
65

The relative decrease in vapour pressure of an aqua solution containing 2 moles of

$4NaCl$ in 3 moles of

${H}2_{O}$ is 0.On reaction with

$AgNO_3$ this solution will form:

0%
1 mol of $AgCl$
0%
0.25 mol of $AgCl$
0%
2 mol of $AgCl$
0%
0.4 mol of $AgCl$

An ideal solution contains two volatile liquids

$A\left( p=100\ torr \right)$ and

$B\left( p=200\ torr \right)$ . If mixture contain 1 mole of

$A$ and 4 mole of

$B$ then total vapour pressure of the distillate is:

0%
$150$
0%
$180$
0%
$188.88$
0%
$198.88$

Explanation

Solution:- (B)

$180$

Given:-

No. of moles of A

$\left( {n}_{A} \right) = 1 \text{ mole}$

No. of moles of B

$\left( {n}_{B} \right) = 4 \text{ moles}$

As we know that,

$\text{mole fraction} = \cfrac{\text{no. of moles}}{\text{total moles}}$

Therefore,

${x}_{A} = \cfrac{1}{1 + 4} = 0.2$

${x}_{B} = \cfrac{4}{1+4} = 0.8$

Again, as we know that,

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

$\Rightarrow {P}_{T} = \left( {x}_{A} \times {{p}^{°}}_{A} \right) + \left( {x}_{B} \times {{p}^{°}}_{B} \right)$

Given:-

${{p}^{°}}_{A} = 100 \text{ torr.}$

${{p}^{°}}_{B} = 200 \text{ torr.}$

$\therefore {P}_{T} = \left( 0.2 \times 100 \right) + \left( 0.8 \times 200 \right)$

$\Rightarrow {P}_{T} = 20 + 160 = 180 \text{ torr.}$

Hence the total vapour pressure of the distillate is

$180 \text{ torr.}$ .

The vapour pressure of water is

$12.3k \, Pa$ at

$300K$ . Calculate the vapour pressure

$1$ molal solution of a non-volatile solute in it.

0%
$12.08 \ kPa$
0%
$1.208 \ Ppa$
0%
$2.4 \ kPa$
0%
$0.4 kPa$

Explanation

$1$ molal solution means 1 mol of the solute is present in

$1000 g$ of the solvent (water).

The molar mass of water

$= 18 g$

$mol^{- 1}$

Number of moles present in

$1000$ g of water

$= 1000/18$

$= 55.56 \ mol$

Therefore, the mole fraction of the solute in the solution is

$x_2 = 1 / (1+55.56)$

$= 0.0177$

It is given that,

Vapour pressure of water,

$P^0_1$

$= 12.3 kPa$

Applying the relation,

$(P^0_1 - P_1) / P^0_1 = X_2$

$= (12.3 - P_1) / 12.3$

$=0.0177$

$= 12.3 - P_1$

$= 0.2177$

$= P_1$

$= 12.0823$

$= 12.08 kPa$

An ideal solution if obtained by mixing

$CH_3 CHBr_2$ and

$CH_2Br - CH_2Br$ in the ratio

$2:1$

$P^o_{CH_3CHBr_2} = 173, P^o_{CH_2BrCH_2Br} = 127$ . Thus total vapour pressure is:

0%
$158$
0%
$257$
0%
$137$
0%
$197$

Explanation

$P = Po_1x_1 + Po_2x_2$

$x_1 = \dfrac{2}{2 +1} = \dfrac{2}{3}$

$x_2 = \dfrac{1}{3}$

Total pressure

$= P$
Partial pressure of

$CH_3CHBr_2 = Po_1$
Partial pressure of

$CH_3BrCH_2Br = Po_2$

Mole fraction

$CH_3CHBr_2 = x_1$
Mole fraction

$CH_3CHBrCH_2Br = x_2$

$P = 173 \times \dfrac{2}{3} + 127 \times \dfrac{1}{3}$

$P = 158$ .

Two liquids A and B have

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

${P}_{B}^{o}$ in the ratio

$1:3$ and the ratio of number of moles of

$A$ and

$B$ in liquid phase are

$1:3$ . Then mole fraction of A in vapour phase in equilibrium with the solution is equal to:

0%
$0.1$
0%
$0.2$
0%
$0.5$
0%
$1.0$

Explanation

Partial pressure of

$A=K\times\cfrac{1}{4},$ where

$K$ is ratio constant.)

Partial pressure of

$B=3K\times \cfrac{3}{4}$

Total pressure is

$\cfrac{10K}{4}$ .

Vapor phase mole fraction of

$A$ is,

$\cfrac{\text{Partial pressure of }A}{\text{Total pressure}}=\cfrac{K\times \cfrac{1}{4}}{\cfrac{10K}{4}}=\cfrac{1}{10}=0.1$

In an ideal solution of non-volatile solute B in solvent A in 2: 5 molar ratio has vapour pressure 250 mm. If another solution in ratio 3: 4 prepared then vapour pressure above this solution is:

0%
200 mm
0%
250 mm
0%
350 mm
0%
400 mm

Explanation

Given molar ratio

$=2:5$ v.p

$=250\ mm\left(\text{one solute}\right)$

Moral ratio

$=3:4\quad v.p=? \quad\left(\text{another solute}\right)$

Mole fraction of solven

$=\dfrac{5}{2+5}=\dfrac{5}{4}=0.71\quad \left({1}^{st}{\text{solute}}\right)$

According to Raoult's law;

$p$ (solution)

$\times$ mole fraction of solven

$\Rightarrow250=p\left(\text{pure solvent}\right)0.71$

$p\left(\text{pure solvent}\right)= \dfrac{250}{0.71}=352.11\ mm$

mole fraction of solvent

$=\dfrac{4}{3+4}=\dfrac{4}{7}=0.57$

$p\left(\text{solution}\right)=352.11\times 0.57=200.7\approx 200\ mm$

Hence, the correct option is

$A$

Calculate the vapour pressure of a solution at

$100^oC$ containing 3 g of cane sugar in 33 g of water. (Atomic weight of

$C =12\ u,H=1\ u,O=16\ u$ )

0%
760 mm Hg
0%
756.9 mm Hg
0%
758.3 mm Hg
0%
None of these

A sample of drinking water was found to be severely contaminated with chloroform,

${CHCl}_{3}$ , supposed to be a carcinogen. The level of contamination was

$15$ ppm (by mass). Express this in percent by mass.

0%
$1.5\times {10}^{-3}$
0%
$4.0\times {10}^{-3}$
0%
$6\times {10}^{-3}$
0%
$0.75\times {10}^{-3}$

Explanation

15 ppm means

$15$ parts in million

$(10^6)$ parts

$\%\,by\,mass=\dfrac{15\times 100}{10^{6}}=15\times 10^{-4}$

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

Hence, the correct option is

$\text{A}$

In steam distillation, the vapour pressure of the volatile organic compound is:

0%
equal to atmospheric pressure
0%
less than atmospheric pressure
0%
more than atmosphetic pressure
0%
none of the above

A sample of 16g charcoal was brought into contact with

$C{ H}_{ 4}$ gas contained in a vessel of 1 litre at

$2{ 7 }^{0 }C$ . The pressure of gas was found to fall from 760 to 608 torr. The density of charcoal sample is 1.6g/

$c{ m}^{ 3 }$ . What is the volume of the

$C{ H}_{ 4}$ gas adsorbed per gram of the adsorbent at 608 torr and

$2{ 7 }^{0 }C$ ?

0%
125 mL/g
0%
16.25 mL/g
0%
26 mL/g
0%
None of these

Explanation

Final volume of gas, at

$608$ torr pressure

$V_{2} = \dfrac{P_1V_1}{P_2}$

$= \dfrac{760 \times 1}{608}$

$\Rightarrow$

$1.25$ or

$V_2 = 1250\,mL$

Volume occupied by gas = Volume of vessel - Volume occupied by charcoal

$= 1000 - \dfrac{16}{1.6} = 990 \,mL$

Difference of volume is due to adsorption of gas by charcoal.

$\therefore$ Volume of gas adsorbed by charcoal

$= 1250 - 990$

$\Rightarrow 260\,mL$
Volume of the gas adsorbed per gram of charcoal

$= \dfrac{260}{16} = 16.25\,mL / g$ at

$608$ torr and

$27^{\circ}C$ .

The gas Z in the above fraction is ?

0%
Nitrous oxide
0%
Nitric Oxide
0%
Dinitrogen tetroxide
0%
Nitrogen dioxide

$100\ ml$ of an aqueous solution contains

$6.0\times {10}^{21}$ solute molecules. The solution is diluted to

$1$ lit. The number of solute molecules present in

$10\ ml$ of the dilute solution is:

0%
$6.0\times {10}^{20}$
0%
$6.0\times {10}^{19}$
0%
$6.0\times {10}^{18}$
0%
$6.0\times {10}^{17}$

Explanation

100 ml solution diluted to 1 liters (1000) ml contains

$6.0 \times 10^{21}$ solute molecular

No of molecules present in 10 ml

$= \dfrac{10 \times 6.0 \times 10 ^{21}}{1000} = 6 \times 10 ^{19}$

Lowering in vapour pressure is highest for:

0%
$0.2m$ urea
0%
$0.1m$ glucose
0%
$0.1m \, MgSO_4$
0%
$0.1m \, BaCl_2$

Explanation

Lowering in vapour pressure is directly proportional

to molality which in turn is directly proportional

to n factor (no. of ions it gives)

The concentration of urea, i.e,

$0.2 \ M$ is much

more than glucose,

$MgSO_{4}$ &

$BaCl_{2}$

Therefore urea will have greater molality &

therefore it will have highest vapour pressure.

Ans is A)

$0.2 M$ urea

Normal boiling point of a liquid is that temperature at which vapour of the liquid pressure of the liquid is equal to:

0%
zero
0%
380mm of Hg
0%
760mm of Hg
0%
100mm of Hg

A solution X contains 30 mole percent of A and is in equilibrium with its vapour that contains 40 mole percent of B. The ratio of vapour pressure of pure A and B will be

0%
2 : 7
0%
7 : 2
0%
3 : 4
0%
4 : 3

Which of the following is an ideal solution

0%
Water + ethanol
0%
Chloroform + carbon tetrachloride
0%
Benzene + toluene
0%
Water + hydrochloric acid

The concentration of a salt solution in terms of mass by mass percentage is

$20\%$ and the mass of the solution is

$550 g.$ What is the mass of solute present in the solution?

0%
$110g$
0%
$105g$
0%
$205g$
0%
$210g$

Explanation

Given that:

The mass percentage of the solution is 20%.

The mass of the solution is 550g.

The mass of the solute can be calculated by using the formula given below.

${\rm{Mass}}\;{\rm{percentage}} = \dfrac{{{\rm{Mass}}\;{\rm{of}}\;{\rm{solute}}}}{{{\rm{Mass}}\;{\rm{of}}\;{\rm{solution}}}} \times 100$

The mass of solute can be calculated as,

$\begin{align*}{\rm{Mass}}\;{\rm{of}}\;{\rm{solute}} &= \dfrac{{{\rm{Mass}}\;{\rm{percentage}}}}{{{\rm{Mass}}\;{\rm{of}}\;{\rm{solution}}}} \times 100\\ &= \dfrac{{20}}{{100}} \times 550\;{\rm{g}}\\ &= 110\;{\rm{g}}\end{align*}$

Therefore, the mass of solute is 110 g.

The vapour pressure of a solution containing 5 g of a non electrolyte in 100 g of water at a particular temperature is

$2985 N{m}^{-2}$ . If the vapour pressure of pure water is

$3000 N{m}^{-2}$ , the molecular weight of solute is:

0%
$60$
0%
$120$
0%
$180$
0%
$380$

Explanation

we have formula of

$\text{Relative lowering of vapour pressure}$ which is

$\dfrac{P_o-P}{P_o}=X_{solute}=\dfrac{n_{solute}}{n_{solute}+n_{solvent}}\approx\dfrac{n_{solute}}{n_{solvent}}$

Now, pute the values

$\dfrac{3000-2985}{3000}=\dfrac{\dfrac{5}{M}}{\dfrac{100}{18}}$

$\dfrac{15}{3000}=\dfrac{5\times 18}{100\times M}$

$M=180$

correct answer is C.

The temperature and pressure at which ice, liquid. water and water vapour can exist together are ?

0%
${ 0 }^{ o }C,\quad 1 atm$
0%
${ 2 }^{ o }C,\quad 4.7atm$
0%
${ 0 }^{ o }C,\quad 4.7mm$
0%
${ -2 }^{ o }C,\quad 4.7mm$

Explanation

The temperature and pressure at which ice, liquid, water and water vapour exist together at triple point of water.

Triple point of water is

$0^\circ C \ and \ 4.7mm$

Hence, Option "C" is the correct answer.

All form ideal solution except

0%
$C_2H_5Br$ and $C_2H_5I$
0%
$C_6H_5Cl$ and $C_6H_5Br$
0%
$C_6H_6$ and $C_6H_5CH_3$
0%
$C_2H_5I$ and $C_2H_5OH$

Explanation

The following forms an ideal solution

$C_2H_5Br + C_2H_5I$
$C_6H_5Cl + C_6H_5Br$
$C_6H_6 + C_6H_5CH_3$

$\text{As we know,} C_2H_5OH \text{ shows H-bonding as well as it is polar so it will not make an ideal solution. }$

thus, the correct answer is D.

Vapour pressure is _________ dependent property.

0%
Temperature
0%
Pressure
0%
Volume
0%
Density

The solubility of a specific non-volatile salt is

$4\mathrm { g }$ in

$100\mathrm { g }$ of water at

$25 ^ { \circ } \mathrm { C } .$ If

$2.0 g,$

$4.0\mathrm g$ and

$6.0 g$ of the salt added of

$100 g$ of water at

$25 ^ { \circ } \mathrm { C } ,$ in system

$X , Y$ and

$Z .$ The vapour pressure would be in the order:

0%
$X < Y < Z$
0%
$X > Y > Z$
0%
$Z > X = Y$
0%
$x > Y = z$

$250^{o}C$ and

$1$ atmosphere pressure, the vapour density of

$PCI_{5}$ is

$57.9$ . What will be the dissociation of

$PCI_{5}-$

0%
$1.00$
0%
$0.90$
0%
$0.80$
0%
$0.65$

A solution

$X$ of

$A$ and

$B$ contains

$30$ mole

$\%$ of

$A$ & is in equilibrium with its vapour that contains

$40$ mole

$\%$ of

$B$ . The ratio of

$V.P$ of pure,

$A$ and

$B$ will be?

0%
$2:7$
0%
$7:2$
0%
$3:4$
0%
$4:3$

The vapour pressure of water at

$20^0C$ is 17.54 mm. When 20 g of non-ionic, substance is dissolved in 100 g of water, the vapour pressure is lowered by 0.30 mm.What is the molecular weight of the substances.

0%
210.2
0%
208.16
0%
215.2
0%
200.8

The boiling points of

$C_6H_6, CH_3OH, C_6H_5 NH_2$ and

$C_6H_5NO_2$ are

$80^0C, \ 65^0C$ and

$212^0C$ respectively. Which of the following will have highest vapour pressure at room temperature?

0%
$C_6H_6$
0%
$CH_3OH$
0%
$C_6H_5NH_2$
0%
$C_6H_5NO_2$

Explanation

(B)

$CH_{3}OH$

Because boiling point is the temp.

at which the vapour pressure of

liquid becomes equal to atm. pressure

methanol have lowest boiling pt.

$\therefore$ will show highest vapour pressure.

1148776_1249578_ans_5e85a67554eb43c388760d12458fc066.jpg

CBSE Questions for Class 12 Engineering Chemistry Solutions Quiz 7 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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