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A 4% Solution(W/W) of Sucrose (M = 342 G Mol−1) in Water Has a Freezing Point of 271.15 K. - Chemistry

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प्रश्न

A 4% solution(w/w) of sucrose (M = 342 g mol⁻¹) in water has a freezing point of 271.15 K. Calculate the freezing point of 5% glucose (M = 180 g mol⁻¹)in water.
(Given: Freezing point of pure water = 273.15 K)

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उत्तर

4% solution(w/w) of sucrose ⇒ 4 g sucrose in 96 g water

w₂ (solute) = 4 g
w₁ (solvent) = 96 g
M₂(solute) = 342 g mol⁻¹

∆T_f = k_f m
T_f = 271.15
$T_{f}^{\circ} = 273 \cdot 15 ∆ T_{f} = T_{f}^{\circ} - T_{f} = (273 \cdot 15 - 271 \cdot 15) K = 2 \cdot 0 K k_{f} = \frac{∆ T_{f}}{m} m = \frac{w_{2}}{M_{2} \times w_{1}} \times 1000 = \frac{4 \times 1000}{96 \times 342} = 0 \cdot 122 m k_{f} = \frac{2}{0 \cdot 122} = 16 \cdot 39 {Km}^{- 1}$

$T_{f}^{\circ} = 273 \cdot 15 ∆ T_{f} = T_{f}^{\circ} - T_{f} = (273 \cdot 15 - 271 \cdot 15) K = 2 \cdot 0 K k_{f} = \frac{∆ T_{f}}{m} m = \frac{w_{2}}{M_{2} \times w_{1}} \times 1000 = \frac{4 \times 1000}{96 \times 342} = 0 \cdot 122 m k_{f} = \frac{2}{0 \cdot 122} = 16 \cdot 39 {Km}^{- 1}$

$T_{f}^{\circ} = 273 \cdot 15 ∆ T_{f} = T_{f}^{\circ} - T_{f} = (273 \cdot 15 - 271 \cdot 15) K = 2 \cdot 0 K k_{f} = \frac{∆ T_{f}}{m} m = \frac{w_{2}}{M_{2} \times w_{1}} \times 1000 = \frac{4 \times 1000}{96 \times 342} = 0 \cdot 122 m k_{f} = \frac{2}{0 \cdot 122} = 16 \cdot 39 {Km}^{- 1}$

$T_{f}^{\circ} = 273 \cdot 15 ∆ T_{f} = T_{f}^{\circ} - T_{f} = (273 \cdot 15 - 271 \cdot 15) K = 2 \cdot 0 K k_{f} = \frac{∆ T_{f}}{m} m = \frac{w_{2}}{M_{2} \times w_{1}} \times 1000 = \frac{4 \times 1000}{96 \times 342} = 0 \cdot 122 m k_{f} = \frac{2}{0 \cdot 122} = 16 \cdot 39 {Km}^{- 1}$

$T_{f}^{\circ} = 273 \cdot 15 ∆ T_{f} = T_{f}^{\circ} - T_{f} = (273 \cdot 15 - 271 \cdot 15) K = 2 \cdot 0 K k_{f} = \frac{∆ T_{f}}{m} m = \frac{w_{2}}{M_{2} \times w_{1}} \times 1000 = \frac{4 \times 1000}{96 \times 342} = 0 \cdot 122 m k_{f} = \frac{2}{0 \cdot 122} = 16 \cdot 39 {Km}^{- 1}$

$T_{f}^{\circ} = 273 \cdot 15 ∆ T_{f} = T_{f}^{\circ} - T_{f} = (273 \cdot 15 - 271 \cdot 15) K = 2 \cdot 0 K k_{f} = \frac{∆ T_{f}}{m} m = \frac{w_{2}}{M_{2} \times w_{1}} \times 1000 = \frac{4 \times 1000}{96 \times 342} = 0 \cdot 122 m k_{f} = \frac{2}{0 \cdot 122} = 16 \cdot 39 {Km}^{- 1}$

$T_{f}^{\circ} = 273 \cdot 15 ∆ T_{f} = T_{f}^{\circ} - T_{f} = (273 \cdot 15 - 271 \cdot 15) K = 2 \cdot 0 K k_{f} = \frac{∆ T_{f}}{m} m = \frac{w_{2}}{M_{2} \times w_{1}} \times 1000 = \frac{4 \times 1000}{96 \times 342} = 0 \cdot 122 m k_{f} = \frac{2}{0 \cdot 122} = 16 \cdot 39 {Km}^{- 1}$

$T_{f}^{\circ} = 273 \cdot 15 ∆ T_{f} = T_{f}^{\circ} - T_{f} = (273 \cdot 15 - 271 \cdot 15) K = 2 \cdot 0 K k_{f} = \frac{∆ T_{f}}{m} m = \frac{w_{2}}{M_{2} \times w_{1}} \times 1000 = \frac{4 \times 1000}{96 \times 342} = 0 \cdot 122 m k_{f} = \frac{2}{0 \cdot 122} = 16 \cdot 39 {Km}^{- 1}$

$T_{f}^{\circ} = 273 \cdot 15 ∆ T_{f} = T_{f}^{\circ} - T_{f} = (273 \cdot 15 - 271 \cdot 15) K = 2 \cdot 0 K k_{f} = \frac{∆ T_{f}}{m} m = \frac{w_{2}}{M_{2} \times w_{1}} \times 1000 = \frac{4 \times 1000}{96 \times 342} = 0 \cdot 122 m k_{f} = \frac{2}{0 \cdot 122} = 16 \cdot 39 {Km}^{- 1}$

$T_{f}^{\circ} = 273 \cdot 15 ∆ T_{f} = T_{f}^{\circ} - T_{f} = (273 \cdot 15 - 271 \cdot 15) K = 2 \cdot 0 K k_{f} = \frac{∆ T_{f}}{m} m = \frac{w_{2}}{M_{2} \times w_{1}} \times 1000 = \frac{4 \times 1000}{96 \times 342} = 0 \cdot 122 m k_{f} = \frac{2}{0 \cdot 122} = 16 \cdot 39 {Km}^{- 1}$

$T_{f}^{\circ} = 273 \cdot 15 ∆ T_{f} = T_{f}^{\circ} - T_{f} = (273 \cdot 15 - 271 \cdot 15) K = 2 \cdot 0 K k_{f} = \frac{∆ T_{f}}{m} m = \frac{w_{2}}{M_{2} \times w_{1}} \times 1000 = \frac{4 \times 1000}{96 \times 342} = 0 \cdot 122 m k_{f} = \frac{2}{0 \cdot 122} = 16 \cdot 39 {Km}^{- 1}$

$T_{f}^{\circ} = 273 \cdot 15 ∆ T_{f} = T_{f}^{\circ} - T_{f} = (273 \cdot 15 - 271 \cdot 15) K = 2 \cdot 0 K k_{f} = \frac{∆ T_{f}}{m} m = \frac{w_{2}}{M_{2} \times w_{1}} \times 1000 = \frac{4 \times 1000}{96 \times 342} = 0 \cdot 122 m k_{f} = \frac{2}{0 \cdot 122} = 16 \cdot 39 {Km}^{- 1}$

$T_{f}^{\circ} = 273 \cdot 15 ∆ T_{f} = T_{f}^{\circ} - T_{f} = (273 \cdot 15 - 271 \cdot 15) K = 2 \cdot 0 K k_{f} = \frac{∆ T_{f}}{m} m = \frac{w_{2}}{M_{2} \times w_{1}} \times 1000 = \frac{4 \times 1000}{96 \times 342} = 0 \cdot 122 m k_{f} = \frac{2}{0 \cdot 122} = 16 \cdot 39 {Km}^{- 1}$

$T_{f}^{\circ} = 273 \cdot 15 ∆ T_{f} = T_{f}^{\circ} - T_{f} = (273 \cdot 15 - 271 \cdot 15) K = 2 \cdot 0 K k_{f} = \frac{∆ T_{f}}{m} m = \frac{w_{2}}{M_{2} \times w_{1}} \times 1000 = \frac{4 \times 1000}{96 \times 342} = 0 \cdot 122 m k_{f} = \frac{2}{0 \cdot 122} = 16 \cdot 39 {Km}^{- 1}$

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Colligative Properties and Determination of Molar Mass - Depression of Freezing Point

क्या इस प्रश्न या उत्तर में कोई त्रुटि है?

2018-2019 (March) 56/1/3

APPEARS IN

2018-2019 (March) 56/1/3 (with solutions)

Q 15 | 3 marks

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Colligative Properties and Determination of Molar Mass - Depression of Freezing Point

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