MCQ Questions for Class 11 Engineering Chemistry States Of Matter Quiz 15

Two liquids A and B form an ideal solution. the solution has a vapor pressure of 700 torrs at

$80^0 C$ . it is distilled till 2/3rd of the solution is collected as condensate. the composition of the condensate

$x'_A = 0.75$ . and that of residue is

$x''_A = 0.30$ if the vapor pressure of the residue at

$80^0 C$ is 600 torr. which of the following is/are true?

0%
The composition of the original liquid was $x_A = 0.6$
0%
The composition of the original liquid was $x_A = 0.4$
0%
$P^0_A = \frac {2500}{3} Torr$
0%
$P^0_B = 500 Torr$

Explanation

$x_AP^0_A + x_B P^0_B = 700 ......(i)$

$X_A "P^0_A + X^"_B P^0_B = 0.30 P^0_A + 0.70 P^0_B = 600 .....(ii)$
If moles of A ^b initially are C & Y then

$X = 0.75 \times \frac {2}{3} (x +y) + 0.30 \times \frac {1}{3}( x +y)......(iii)$

$x_A = \frac {X}{x +y} \quad or \quad x_B = \frac {y}{x +y} ......(iv)$
solving gives

$X_A = 0.6 \quad x_B = 0.4 , P^0_A = \frac { 2500}{3} torr$ &

$P^0_B = 500 torr$

The reading of pressure gauge at which only liquid phase exists

0%
499
0%
399
0%
299
0%
none

Explanation

$X_A = \dfrac34=0.75; \quad X_B = 0.25$

$P_{\text{bubble point}} = X_A P^0_A + X_B P^0_B$

$0.75 +400 + 0.25 \times 800 = 500 mm$

Above 500 mm Hg only liquid phase exists.

$y_A = 0.75; \quad \quad y_B = 0.5$

At dew point:

$\dfrac {1}{P_T } = \dfrac {Y_A }{P^0A} + \dfrac {Y_B}{P^0_B } \Rightarrow \dfrac {1}{ P_T} = \dfrac { 0.75}{400} + \dfrac { 0.25}{800} = \dfrac { 1.5 + 0.25}{800}$

$\Rightarrow P_T = \dfrac {800}{1.75} = 457.14\ mm Hg$

Below the dew point, only the vapour phase exists.

Vapour pressure of a solution is

0%
Directly proportional to the mole fraction of the solvent
0%
Inversely proportional to the mole fraction of the solute
0%
Inversely proportional to the mole fraction of the solvent
0%
Directly proportional to the mole fraction of the solute

Which of the following solution in water possesses the lowest vapour pressure?

0%
$0.1 (M) NaCl$
0%
$0.1 (M) BaCl_2$
0%
$0.1 (M) KCl$
0%
None of these

To which of the following gaseous mixtures is Dalton's law not applicable?

0%
$Ne + He + SO_{2}$
0%
$NH_{3} + HCl + HBr$
0%
$O_{2} + N_{2} + CO_{2}$
0%
$N_{2} + H_{2} + O_{2}$

Explanation

$NH_{3}$ and

$HCl$ &

$HBr$ is a reacting gas mixture to produce

$NH_{4}Cl$ &

$NH_{4}Br$ .

Therefore, Dalton's law is not applicable.

$60 \,gm$ of Urea (Mol. wt 60) was dissolved in

$9.9$ moles of water. If the vapour pressure of pure water is

$P_0$ , the vapour pressure of solution is

0%
$0.10 \,P_0$
0%
$1.10 \,P_0$
0%
$0.90 \,P_0$
0%
$0.99 \,P_0$

Explanation

Given that,

Moles of water.

$N=9.9$ moles

Moles of urea,

$n=\dfrac{w}{M}=\dfrac{60}{60}=1$

Now,

$\dfrac {P^{0} - P_{S}}{P^{0}} = \dfrac {n}{N} \Rightarrow \dfrac {P^{0} - P_{S}}{P^{0}} = \dfrac {1}{9.9} \Rightarrow 9.9 \,P^{0} - 9.9 \,P_s = P^{0}$

$\Rightarrow 8.9 \,P^{0} = 9.9 \,P_s \Rightarrow P_s = \dfrac {8.9}{9.9} P^{0} \approx 0.90 \,P^{0}$

The vapour pressure lowering caused by the addition of

$100 \,g$ of sucrose (molecular mass = 342) to

$1000 \,g$ of water if the vapour pressure of pure water at

$25 ^\circ C$ is

$23.8 \,mm \,Hg$

0%
$1.25 \,mm \,Hg$
0%
$0.125 \,mm \,Hg$
0%
$1.15 \,mm \,Hg$
0%
$00.12 \,mm \,Hg$

Explanation

Given molecular mass of sucrose

$= 342$

Moles of sucrose

$= \dfrac {100}{342} = 0.292 \,mole$

Moles of water

$N = \dfrac {1000}{18} = 55.5 \,moles$ and

Vapour pressure of pure water

$P^{0} = 23.8 \,mm \,Hg$

According to Raoult's law

$\dfrac {\Delta P}{P^{0}} = \dfrac {n}{n + N} \Rightarrow \dfrac {\Delta P}{23.8} = \dfrac {0.292}{0.292 + 55.5}$

$\Delta P = \dfrac {23.8 \times 0.292}{55.792} = 0.125 \,mm \,Hg$

The pressure P of a gas is plotted against its absolute temperature

$T$ for two different constant volumes,

$V_{1}$ and

$V_{2}$ . when

$V_{1}$ >

$V_{2}$ , the:

0%
Curves have the same slope and do not intersect
0%
Curves must intersect at some point other than $T = 0$
0%
Curve for $V_{2}$ has a greater slope than that for $V_{1}$
0%
Curve for $V_{1}$ has a greater slope than that for $V_{2}$

Explanation

At constant volumes

$P \alpha T$

$P=$ constant

$T; PV = nRT \because P= \dfrac {nR} {V} T$

slope

$= m = \dfrac{nR}{V} \because V_{2} < V_{1}$

$\dfrac{m_{1}} {m_{2}} = \dfrac{V_{2}} {V_{1}} \because m_{1} < m_{2}$ is curve for

$V.$ has a greater slope than for

$V.$

Which has maximum vapour pressure?

0%
$HI$
0%
$HBr$
0%
$HCl$
0%
$HF$

Explanation

The lower is boiling point more is the vapour pressure. Boiling point order is

$HCl < HBr < HI < HF$

Correct gas equation is:

0%
$\dfrac {V_{1} T_{2}}{P_{1} }= \dfrac {V_{2} T_{1}}{P_{2}}$
0%
$\dfrac {P_{1} V_{1}}{P_{2}V_{2} }= \dfrac {T_{1}} {T_{2}}$
0%
$\dfrac {P_{1} T_{2}}{V_{1} }= \dfrac {P_{2} V_{1}}{T_{2}}$
0%
$\dfrac {V_{1} V_{2}}{T_{1}T_{2} }= P_{1}P_{2}$

Explanation

Ideal gas law equation:

$PV = nRT$

$\dfrac {P_{1} V_{1}} {T_{1} }=\dfrac {P_{2}V_{2}}{T}$

$\because$

$\dfrac{P_{1}V_{1}}{P_{2}V_{2}}=$

$\dfrac{T_{1}}{T_{2}}$

Hence, option B is correct.

Two which of the following the Dalton's law of particle pressure is not applicable?

0%
$H_2$ and $He$
0%
$NH_3$ and $HCl$
0%
$N_2$ and $H_2$
0%
$N_2$ and $O_2$

Explanation

Dalton's law of partial pressure is applicable to non-reaction gases

$NH_3$ and

$HCl$ are reacting gases, so Dalton's law will not be applicable for them.

Choose the correct answer

The graph of

$PV$ vs

$P$ for a gas is:

0%
parabolic
0%
hyperbolic
0%
a straight line parallel to X-axis
0%
a straight line passing through origin

Directions: In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice:

Assertion: The vapour pressure of an aqueous solution is less than 1.013 bar at 373.15 K.
Reason: Vapour pressure of water is 1.013 bar at 373.15K.

0%
If both assertion and reason are true and reason is the correct explanation of assertion.
0%
If both assertion and reason are true but reason is not the correct explanation of assertion
0%
If assertion is true but reason is false.
0%
If both assertion and reason are false.

Which of the following hydrogen halide is most volatile ?

0%
$HF$
0%
$HCl$
0%
$HBr$
0%
$HI$

Explanation

Most volatile hydrogen halide is the one which has least boling point viz

$HCl.$

Which of the following is not a colligative property?

0%
Vapour pressure
0%
Depression in f. pt.
0%
Elevation in b. pt.
0%
Osmotic pressure

Explanation

$\text { Auswer:- (a) vapour pressure. }$

Colligative properties are those properties of solutions that depend on the

ratio of number of solute particles to number of solvent particles in a solution.

There are 4 colligative properties:-

-vapour pressure Lowering

-osmotic pressure

-freezing point depression

-boiling point elevation

CBSE Questions for Class 11 Engineering Chemistry States Of Matter Quiz 15 - MCQExams.com

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Practice Class 11 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Chemistry States Of Matter Quiz 15 - MCQExams.com

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Practice Class 11 Engineering Chemistry Quiz Questions and Answers

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