NEET Previous Year Questions Solved

Q 1 :
Mass of glucose $(C_{6} H_{12} O_{6})$ required to be dissolved to prepare one litre of its solution which is isotonic with 15 g $L^{- 1}$ solution of urea $({NH}_{2} {CONH}_{2})$ is

(Given: Molar mass in g ${mol}^{- 1}$ C : 12, H : 1, O : 16, N : 14) [2024]

55 g
15 g
30 g
45 g

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(4)

As the two solutions of glucose and urea are isotonic, it means they have the same osmotic pressure at a given temperature.

$π_{1} (Glucose) = π_{2} (Urea)$

$C_{1} R T = C_{2} R T (∵ π = C R T)$

$\frac{n_{1}}{V_{1}} R T = \frac{n_{2}}{V_{2}} R T$ $\Rightarrow \frac{n_{1}}{V_{1}} = \frac{n_{2}}{V_{2}}$ $\Rightarrow \frac{w_{1}}{M_{1} V_{1}} = \frac{w_{2}}{M_{2} V_{2}}$

$w_{1} = \frac{15 \times 180}{60} = 45$ $g$

(Molar mass; glucose = 180 g ${mol}^{- 1}$ , urea = 60 g ${mol}^{- 1}$ )

Q 2 :
The plot of osmotic pressure ( $π$ ) vs concentration (mol $L^{- 1}$ ) for a solution gives a straight line with slope 25.73 L bar ${mol}^{- 1}$ . The temperature at which the osmotic pressure measurement is done is (Use R = 0.083 L bar ${mol}^{- 1}$ $K^{- 1}$ ) [2024]

37 °C
310 °C
25.73 °C
12.05 °C

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(1)

As we know, $π = C R T$

The plot of $π$ (osmotic pressure) vs concentration will give

(On comparing with straight line equation $y = m x + C$ )

Slope = $(m) = R T$

$25.73 = R T$

$T = \frac{25.73}{R} = \frac{25.73}{0.083} = 310$ $K$

$T = 310$ $K$ $or$ $37 ° C$

Q 3 :
The following solutions were prepared by dissolving 10 g of glucose $(C_{6} H_{12} O_{6})$ in 250 mL of water ( $P_{1}$ ), 10 g of urea $(C H_{4} N_{2} O)$ in 250 mL of water ( $P_{2}$ ) and 10 g of sucrose $(C_{12} H_{22} O_{11})$ in 250 mL of water ( $P_{3}$ ). The right option for the decreasing order of osmotic pressure of these solutions is [2021]

$P_{3} > P_{1} > P_{2}$
$P_{2} > P_{1} > P_{3}$
$P_{1} > P_{2} > P_{3}$
$P_{2} > P_{3} > P_{1}$

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(2)

Osmotic pressure $(π) = C R T$

$∴$ $π \propto C$

For glucose solution, $C_{1} = \frac{10}{180} \times \frac{1000}{250} = 0.22$ $M$

For urea solution, $C_{2} = \frac{10}{60} \times \frac{1000}{250} = 0.66$ $M$

For sucrose solution, $C_{3} = \frac{10}{342} \times \frac{1000}{250} = 0.117$ $M$

Hence, order of osmotic pressure is $P_{2} > P_{1} > P_{3}$

Q 4 :
The freezing point depression constant ( $K_{f}$ ) of benzene is 5.12 K kg ${mol}^{- 1}$ . The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off up to two decimal places) [2020]

0.20 K
0.80 K
0.40 K
0.60 K

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(3)

Given: $K_{f} = 5.12$ $K$ $kg$ ${mol}^{- 1}, m = 0.078$ $m$

$Δ T_{f} = K_{f} \times m = 5.12 \times 0.078 = 0.39936 \approx 0.40$ $K$

Q 5 :
If molality of the dilute solution is doubled, the value of molal depression constant ( $K_{f}$ ) will be [2017]

halved
tripled
unchanged
doubled

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(3)

The value of molal depression constant, $K_{f}$ is constant for a particular solvent, thus, it will be unchanged when molality of the dilute solution is doubled.

Q 6 :
At 100°C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If $K_{b}$ = 0.52, the boiling point of this solution will be [2016]

102 °C
103 °C
101 °C
100 °C

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(3)

Given: $W_{B} = 6.5$ $g, W_{A} = 100$ $g$

$p_{s} = 732$ $mm, K_{b} = 0.52, T_{b}^{0} = 100 ° C, p^{0} = 760$ $mm$

$\frac{p^{o} - p_{s}}{p^{o}} = \frac{n_{2}}{n_{1}} \Rightarrow \frac{760 - 732}{760} = \frac{n_{2}}{100 / 18}$

$\Rightarrow n_{2} = \frac{28 \times 100}{760 \times 18} = 0.2046$ $mol$

$Δ T_{b} = K_{b} \times m$

$T_{b} - T_{b}^{o} = K_{b} \times \frac{n_{2} \times 1000}{W_{A} (g)}$

$T_{b} - 100 ° C = \frac{0.52 \times 0.2046 \times 1000}{100} = 1.06$

$T_{b} = 100 + 1.06 = 101.06 ° C$

Topic Question Set

Q 1 :
Mass of glucose $(C_{6} H_{12} O_{6})$ required to be dissolved to prepare one litre of its solution which is isotonic with 15 g $L^{- 1}$ solution of urea $({NH}_{2} {CONH}_{2})$ is

(Given: Molar mass in g ${mol}^{- 1}$ C : 12, H : 1, O : 16, N : 14) [2024]

Q 2 :
The plot of osmotic pressure ( $π$ ) vs concentration (mol $L^{- 1}$ ) for a solution gives a straight line with slope 25.73 L bar ${mol}^{- 1}$ . The temperature at which the osmotic pressure measurement is done is (Use R = 0.083 L bar ${mol}^{- 1}$ $K^{- 1}$ ) [2024]

Q 4 :
The freezing point depression constant ( $K_{f}$ ) of benzene is 5.12 K kg ${mol}^{- 1}$ . The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off up to two decimal places) [2020]

Q 5 :
If molality of the dilute solution is doubled, the value of molal depression constant ( $K_{f}$ ) will be [2017]

Q 6 :
At 100°C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If $K_{b}$ = 0.52, the boiling point of this solution will be [2016]

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Topic Question Set

Q 1 : Mass of glucose (C6H12O6) required to be dissolved to prepare one litre of its solution which is isotonic with 15 g L-1 solution of urea (NH2CONH2) is (Given: Molar mass in g mol-1 C : 12, H : 1, O : 16, N : 14) [2024]

Q 2 : The plot of osmotic pressure (π) vs concentration (mol L-1) for a solution gives a straight line with slope 25.73 L bar mol-1. The temperature at which the osmotic pressure measurement is done is (Use R = 0.083 L bar mol-1 K-1) [2024]

Q 4 : The freezing point depression constant (Kf) of benzene is 5.12 K kg mol-1. The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off up to two decimal places) [2020]

Q 5 : If molality of the dilute solution is doubled, the value of molal depression constant (Kf) will be [2017]

Q 6 : At 100°C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If Kb = 0.52, the boiling point of this solution will be [2016]

Want Us to Email you About Special Offers & Updates?

Q 1 :
Mass of glucose $(C_{6} H_{12} O_{6})$ required to be dissolved to prepare one litre of its solution which is isotonic with 15 g $L^{- 1}$ solution of urea $({NH}_{2} {CONH}_{2})$ is

(Given: Molar mass in g ${mol}^{- 1}$ C : 12, H : 1, O : 16, N : 14) [2024]

Q 2 :
The plot of osmotic pressure ( $π$ ) vs concentration (mol $L^{- 1}$ ) for a solution gives a straight line with slope 25.73 L bar ${mol}^{- 1}$ . The temperature at which the osmotic pressure measurement is done is (Use R = 0.083 L bar ${mol}^{- 1}$ $K^{- 1}$ ) [2024]

Q 4 :
The freezing point depression constant ( $K_{f}$ ) of benzene is 5.12 K kg ${mol}^{- 1}$ . The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off up to two decimal places) [2020]

Q 5 :
If molality of the dilute solution is doubled, the value of molal depression constant ( $K_{f}$ ) will be [2017]

Q 6 :
At 100°C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If $K_{b}$ = 0.52, the boiling point of this solution will be [2016]