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Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C. - CBSE (Science) Class 12 - Chemistry

ConceptColligative Properties and Determination of Molar Mass Elevation of Boiling Point

Question

Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C. Molal elevation constant for water is 0.52 K kg mol−1.

Solution 1

Here, elevation of boiling point ΔTb = (100 + 273) − (99.63 + 273)

= 0.37 K

Mass of water, wl = 500 g

Molar mass of sucrose (C12H22O11), M2 = 11 × 12 + 22 × 1 + 11 × 16

= 342 g mol−1

Molal elevation constant, Kb = 0.52 K kg mol−1

We know that:

triangleT_b = (K_bxx1000xxw_2)/(M_2xxw_1)

=>w_2 = (triangleT_bxxM_2xxw_1)/(K_bxx1000)

= (0.37xx342xx500)/(0.52xx1000)

= 121.67 g (approximately)

Hence, 121.67 g of sucrose is to be added.

Solution 2

Given ΔTb = 100 - 96.63 = 3.37º

Mass of water, w1 = 500g

Molar mass of water, M1 = 18g mol-1

Molar mass of sucrose, M2 = 342g mol-1

To find :Mass of sucrose, w2 = ?

Solution: we know, ΔT= Kb x m

= K_bxxw_2/M_2xx1000/w_1

=>w_2 = (M_2xxw_1xxtriangleT_B)/(1000xxK_b) = (342xx500xx3.37)/(1000xx0.52)

w2 = 1108.2g

Mass of solute w2 = 1.11 kg

Solution 3

Here, elevation of boiling point ΔTb = (100 + 273) − (99.63 + 273)

= 0.37 K

Mass of water, wl = 500 g

Molar mass of sucrose (C12H22O11), M2 = 11 × 12 + 22 × 1 + 11 × 16

= 342 g mol−1

Molal elevation constant, Kb = 0.52 K kg mol−1

We know that:

triangleT_b = (K_bxx1000xxw_2)/(M_2xxw_1)

=>w_2 = (triangleT_bxxM_2xxw_1)/(K_bxx1000)

= (0.37xx342xx500)/(0.52xx1000)

= 121.67 g (approximately)

Hence, 121.67 g of sucrose is to be added.

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Solution Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C. Concept: Colligative Properties and Determination of Molar Mass - Elevation of Boiling Point.
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