Question

When 10 g of a non-volatile solute is dissolved in 100 g of
benzene, it raises boiling point by 1ºC then molecular mass
of the solute is (Kb for benzene = 2.53 km⁻¹)
(a) 223 g (b) 243 g (c) 233 g (d) 253 g

Answer 1

When 10 g of a non-volatile solute is dissolved in 100 g of benzene, it raises boiling point by 1ºC then the molecular mass of the solute is (Kb for benzene = 2.53 km⁻¹) 253 g.

• Option (d) is correct.
• Given,
• Kb = 2.53 km^-1
• w = 10g
• W = 100g
• Tb = 1 C
• we use the formula,
• $T_b = \dfrac{K_b * w * 1000}{m * W}\\\\m = \dfrac{K_b * w * 1000}{T_b * W}\\\\= \dfrac{2.53 * 10 * 1000}{1 * 100}\\\\=253 g$

Answer 2

The molecular mass of the solute is 253 g.

Option (D) is correct.

Explanation:

We are given that:

• mass of non volatile solute =  10 g
• Mass of benzene = 100 g
• Raise in boiling point = 1ºC
• Kb for benzene = 2.53 km⁻¹

Solution:

ΔTb = Kb × w × 1000 m × W

m = Kb × w × 1000ΔTb × W

m = 2.53 × 10 × 1000 / 1 × 100

m = 253 g

Thus the molecular mass of the solute is 253 g.

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