10 g of a non-volatile solute when dissolved in 100 g of benzene raises its boiling point by 1° . What is the molecular mass of the solute? ( Kb for benzene =2.53 km-1 )

Question

10 g of a non-volatile solute when dissolved in 100 g of benzene raises its boiling point by 1° . What is the molecular mass of the solute? ( Kb for benzene =2.53 km-1 ) (A)  352  (B)  253  (C)  275  (D)  310

Tardigrade Admin · Accepted Answer

w1=100 g, w2=10 g, Δ Tb=1°, Kb=2.53 Km -1 M2=(1000 Kb × w2/w1 × Δ Tb) =(1000 × 2.53 × 10/100 × 1) = 253 g

Q. $10 g$ of a non-volatile solute when dissolved in $100 g$ of benzene raises its boiling point by $1^{\circ}$ . What is the molecular mass of the solute? ( $K_{b}$ for benzene $= 2.53 k m^{- 1}$ )

$352$

$253$

$275$

$310$

$w_{1} = 100 g, w_{2} = 10 g$ ,
$Δ T_{b} = 1^{\circ}, K_{b} = 2.53 K m^{- 1}$
$M_{2} = \frac{1000 K _{b} \times w _{2}}{w _{1} \times Δ T _{b}}$
$= \frac{1000 \times 2.53 \times 10}{100 \times 1}$
$= 253 g$

Q. 10g of a non-volatile solute when dissolved in 100g of benzene raises its boiling point by 1∘ . What is the molecular mass of the solute? ( Kb​ for benzene =2.53km−1 )

352

253

275

310