Free Objective Test 02 Practice Test - 11th and 12th

Question 1

Which of the followings is correct –

A. For an non-ideal solution

△ H_{m i x} > 0, △ V_{m i x} > 0.

B. In a solution showing negative deviation from Raoult's law A-B type of molecular interactions are stronger than A-A and B-B type of molecular interactions.

C. Nitric acid and Water solution shows positive deviation from Raoult's law.

D. Azeotrope boils at temperature below the boiling temperature of its constituents.

SOLUTION

Solution : B

$a) △ H_{m i x} > 0, △ V m i x > 0$ are true for mixtures showing positive deviations from Raoult's law only. Hence option (a) is incorrect

b)      It is true

c)      Nitric acid and water forms intermolecular H-bonding. Hence shows Negative deviation from Raoult's law.

d)      Azeotropes are of two types – minimum and maximum boiling azeotropes. Only minimum boiling Azeotrope boils at temperature below the boiling temperature of its constituents. Hence option (d) is incorrect.

Question 2

A non-ideal solution was prepared by mixing 30 ml chloroform and 50 ml acetone. The volume of mixture will be

> 80 m l

< 80 m l

C. = 80 ml

\geq 80 m l

SOLUTION

Solution : B

Chloroform and Acetone solution shows negative deviation from Rault’s law.

$H e n c e △ V m i x < 0$

Question 3

Which of the followings is a colligative property

A. Osmotic pressure

B. Boiling point

C. Vapour pressure

D. Freezing point

SOLUTION

Solution : A

Only Osmotic pressure is a colligative property.

‘Relative lowering of vapour pressure of the solvent’ is a colligative property but not only vapour pressure. Similar argument for boiling point and freezing point

Question 4

The relative lowering of vapour pressure produced by dissolving 71.5 g of a substance in 1000 g of water is 0.00713. The molecular weight of the substance will be

A. 180

B. 342

C. 60

D. 18.0

SOLUTION

Solution : A

$\frac{△ P_{1}}{P_{1}^{0}} = \frac{n_{2}}{n_{1} + n_{2}} \approx \frac{n_{2}}{n_{1}} = \frac{W_{2} / M_{2}}{W_{1} / M_{1}}$

$M_{2} = \frac{W_{2} \times M_{1}}{W_{1} \times (\frac{△ P_{1}}{P_{1}^{0}})} = \frac{71.5 \times 18}{1000 \times 0.00713} = 180.5 g / m o l e$

Question 5

Given that $△ T_{f}$ if is the diperssion in freezing point of the solvent of non-volatile solute of molality m, the quantity $l i m m \to 0 (\frac{△ T_{f}}{m}) i s e q u a l t o$

A. Zero

B. One

\infty

D. Finite value

SOLUTION

Solution : D

$△ T_{f} = K_{f} m$
$l i m m \to 0 (\frac{△ T_{f}}{m}) = l i m m \to 0 (K_{f}) = K_{f}$

Question 6

In an experiment on depression of freezing point, two beakers leveled as A and B were filled with 1000 g. of water each. 40 g and 80 g of a certain non-electrolytic solute were added to both the beakers and freezing point of the solutions were noted as $T_{f A}$ and $T_{f B}$ respectively. Now both the solutions were poured into an empty beaker C and the freezing point of the mixed solution is found to be $T_{f C}$ . What is the ratio $T_{f A} : T_{f B} : T_{f C} ?$

A. 1 : 2 : 3

B. 2 : 4 : 3

C. 2 : 4 : 8

D. 1 : 4 : 8

SOLUTION

Solution : B

$△ T_{f} = K_{f} m$
$\Rightarrow 0 - T_{f} = K_{f} m$
$\Rightarrow= - K_{f} m$
If 40 g of solute contains 'x'~moles of it then 80 g of the solut ill contains '2x' moles of it
$m_{A} = \frac{x}{\frac{W t . o f s o l v e n t i n g}{1000}} = \frac{x}{\frac{1000}{1000}} = x$
$m_{B} = \frac{2 x}{\frac{W t . o f s o l v e n t i n g}{1000}} = \frac{x}{\frac{1000}{1000}} = 2 x$
$m_{C} = \frac{3 x}{\frac{W t . o f s o l v e n t i n g}{1000}} = \frac{3 x}{\frac{1000}{1000}} = 1.5 x$
Required ration $T_{f A} : T_{f B} : T_{f c} = 1.2 : 1.5 = 2 : 4 : 3$

Question 7

Which of the following solutions in water will have the lowest vapour pressure

0.1 M, N a C l

0.1 M N a_{3} P O_{4}

0.1 M, B a C l_{2}

0.1 M, s u c r o s e

SOLUTION

Solution : B

Depression in vapor pressure is a colligative property. More the no. of particles present in the solution lower will be the vapor pressure. consists of maximum ions hence it show lowest vapour pressure.

$\begin{matrix} M o l e c u l e & N a C l & N a_{3} P o_{4} & B a C l_{2} & S u c r o s e N o . o f i o n s p e r m o l e c u l e & 1 + 1 = 2 & 3 + 1 = 4 & 1 + 2 = 3 & 1 \end{matrix}$

Question 8

Osmotic pressure of a solution containing 0.1 mole of solute per litre at 273K is (in atm)

\frac{0.1}{1} \times 0.08205 \times 273

0.1 \times 1 \times 0.08205 \times 273

\frac{1}{0.1} \times 0.08205 \times 273

\frac{0.1}{1} \times \frac{273}{0.08205}

SOLUTION

Solution : A

Osmotic pressure , $π = C R T = \frac{n}{V} R T = \frac{0.1}{1} \times 0.08205 \times 273 a t m$

Question 9

A solution of urea contains 8.6 gm/litre (mol. Wt. 60.0). It is isotonic with a 5% (W/W) solution of a non-volatile solute having density 1 g/cc. The molecular weight of the solute will be

A. 348.8

B. 34.88

C. 3488

D. 861.2

SOLUTION

Solution : A

Isotonic solutions have same osmotic pressure
As Osmotic pressure = C R T
$\Rightarrow C_{1} = C_{2}$
$\Rightarrow \frac{n_{1}}{V_{1}} = \frac{n_{2}}{V_{2}}$
$5 %$ W/W solution means 100g of solution contains 5 g of solute
$∴ \frac{\frac{8.6}{60}}{1} = \frac{\frac{5}{M}}{\frac{100}{1000}}$
$\Rightarrow M = \frac{5 \times 60 \times 10}{8.6} = 348.8$

Question 10

Two isotonic solutions must have ----

A. Same solute and solvent

B. Same solute but different solvent

C. Same solvent but different solute

D. Both Solute and solvent may be different.

SOLUTION

Solution : D

For two solutions to be isotonic only requirement is that their osmotic pressures should be same. There are no conditions on solutes and solvents.

Question 11

If $α$ is the degree of dissociation of $N a_{2} S O_{4}$ , the vant Hoff's factor (i) used for calculating the molecular mass is

1 + α

α

1 + 2 α

1 - 2 α

SOLUTION

Solution : C

$N a_{2} S O_{4}      1 - α ⇌ 2 N a^{+}      2 α + S O_{4}^{2 -}      α$
Total no. of ions after dissociation = $1 + 2 α$
vant Hoff's factor, $i = \frac{N o . o f p a r t i c l e s a f t e r d i s s o c i a t i o n}{N o . o f p a r t i c l e s b e f o r e d i s s o c i a t i o n} = 1 + 2 α$

Question 12

A 0.001 molal solution of $[P t (N H_{3})_{4} c l_{4}]$ in water had a freezing point depression of ${0.0054}^{\circ}$ . If $K_{f}$ for water is 1.80, the correct formulation for the above molecule is

[P t (N H_{3})_{4} C l_{3}] C l

[P t (N H_{3})_{4}] C l_{4}

[P t (N H_{3})_{4} C l_{2}] C l_{3}

[P t (N H_{3})_{4} C l_{2}] C l_{2}

SOLUTION

Solution : D

$△ T = i K_{f} m$
$\Rightarrow i = \frac{△ T_{f}}{K_{f} m} = \frac{0.0054}{1.80 \times 0.001} = 3$
Thus complex produces three ions per molecule. hence possible correct formula is
$[P t (N H_{3})_{4} C l_{2}] C l_{2}$

Question 13

A solution containing 30 gms of non-volatile solute in exactly 90 gm water has a vapour pressure of 21.85 mm Hg at 25⁰C. Further 18 gms of water is then added to the solution. The resulting solution has a vapour pressure of 22.15 mm Hg at 25⁰C. Calculate the molecular weight of the solute.

74.2

75.6

67.83

78.7

SOLUTION

Solution : C

Question 14

The average osmotic pressure of human blood is 7.8 bar at 37⁰C. What is the concentration of an aqueous NaCl solution that could be used in the blood stream

0.16 mol/L

0.32mol/L

0.60mol/L

0.45mol/L

SOLUTION

Solution : B

Question 15

Boiling point of chloroform was raised by 0.323 K, when 0.5143 g of anthracene was dissolved in 35 g of chloroform. Molecular mass of anthracene is(= 3.9 kg mol^-1)

79.42 g/mol

132.32 g/mol

177.42 g/mol

242.32 g/mol

SOLUTION

Solution : C

Free Objective Test 02 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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