A solvent X has normal boiling point 350 K, molar mass 100 g mol^-1 and Delta H_vap = 30.0 kJ mol^-1. Use K_b = (M R T_b^2)/(1000 Delta H_vap) (M in g mol^-1, Delta H_vap in J mol^-1, R = 8.314 J mol^-1 K^-1). Find the elevation of boiling point when 4.0 g of a non-electrolyte (molar mass 80 g mol^-1) is dissolved in 250 g of X.

JEE Advanced levelSingle correct optionDifficulty 3 of 3so072

A solvent X has normal boiling point $350 K$ , molar mass $100 g mo l^{- 1}$ and $Δ H_{v a p} = 30.0 kJ mo l^{- 1}$ . Use $K_{b} = \frac{MR T _{b}^{2}}{1000 Δ H _{v a p}}$ ( $M$ in $g mo l^{- 1}$ , $Δ H_{v a p}$ in $J mo l^{- 1}$ , $R = 8.314 J mo l^{- 1} K^{- 1}$ ). Find the elevation of boiling point when $4.0 g$ of a non-electrolyte (molar mass $80 g mo l^{- 1}$ ) is dissolved in $250 g$ of X.

K_{b} = \frac{100 \times 8.314 \times 35 0 ^{2}}{1000 \times 30000} = 3.395 K kg mo l^{- 1}

. Molality

= (4.0/80) /0.250 = 0.200 mol k g^{- 1}

Δ T_{b} = 3.395 \times 0.200 = 0.679 K