Question

On mixing 3 g of non - volatile solute in 200 mL of water, its
boiling point (100°) becomes 100.52°C. If Kb for water is 0.6
K/m then molecular wt. of solute is :
(a) 10.5 gmol⁻¹ (b) 12.6 gmol⁻¹
(c) 15.7 gmol⁻¹ (d) 17.3 gmol⁻¹

Answer 1

answer : option (d) 17.3 g/mol

given, mass of non volatile solute = 3g

let molecular wt of solute is M.

so number of mole of solute = 3/M

volume of solvent (water) = 200 mL

we know, density of water = 1g/mL

so, mass of water = 200ml × 1g/ml = 200g

now molality, m = no of mole of solute/mass of solvent in Kg

= (3/M)/(200/1000)

= 15/M

using formula, ∆T_b = k_b × m

⇒(100.52°C - 100°C) = 0.6 × 15/M

⇒0.52 = 9/M

⇒M = 9/0.52 = 17.3 g/mol

hence molecular wt of solute is 17.3 g/mol

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