A solution containing 2.5 × 10-3 kg of a solute dissolved in 75 × 10-3 kg of water boils at 373.535 K. The molar mass of the solute is g mol -1. [nearest integer] (Given: K b ( H 2 O )=0.52 K Kg mol -1, boiling point of water =373.15 K )

Question

A solution containing 2.5 × 10-3 kg of a solute dissolved in 75 × 10-3 kg of water boils at 373.535 K. The molar mass of the solute is g mol -1. [nearest integer] (Given: K b ( H 2 O )=0.52 K Kg mol -1, boiling point of water =373.15 K ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

W =2.5 g K b =0.52 W text solvent =75 g M = Mol. Wt. of solute T B prime=373.535 K T B o =373.15 K Δ T B =0.385= K b molality 0.385=0.52 ×((2.5/ M ) × (1000/75)) M =45 g mol -1

Q. A solution containing $2.5 \times 1 0^{- 3} k g$ of a solute dissolved in $75 \times 1 0^{- 3} k g$ of water boils at $373.535 K$ . The molar mass of the solute is ___ $g m o l^{- 1}$ . [nearest integer] (Given: $K_{b} (H_{2} O) = 0.52 K K g m o l^{- 1}$ , boiling point of water $= 373.15 K$ )

$W = 2.5 g$
$K_{b} = 0.52$
$W_{solvent} = 75 g$
$M =$ Mol. Wt. of solute
$T_{B}^{'} = 373.535 K$
$T_{B}^{o} = 373.15 K$
$Δ T_{B} = 0.385 = K_{b}$ molality
$0.385 = 0.52 \times (\frac{2.5}{M} \times \frac{1000}{75})$
$M = 45 g m o l^{- 1}$

Q. A solution containing 2.5×10−3kg of a solute dissolved in 75×10−3kg of water boils at 373.535K. The molar mass of the solute is ___gmol−1. [nearest integer] (Given: Kb​(H2​O)=0.52KKgmol−1, boiling point of water =373.15K )

W=2.5g Kb​=0.52 Wsolvent ​=75g M= Mol. Wt. of solute TB′​=373.535K TBo​=373.15K ΔTB​=0.385=Kb​ molality 0.385=0.52×(M2.5​×751000​) M=45gmol−1

Q. A solution containing $2.5 \times 1 0^{- 3} k g$ of a solute dissolved in $75 \times 1 0^{- 3} k g$ of water boils at $373.535 K$ . The molar mass of the solute is ___ $g m o l^{- 1}$ . [nearest integer] (Given: $K_{b} (H_{2} O) = 0.52 K K g m o l^{- 1}$ , boiling point of water $= 373.15 K$ )

$W = 2.5 g$
$K_{b} = 0.52$
$W_{solvent} = 75 g$
$M =$ Mol. Wt. of solute
$T_{B}^{'} = 373.535 K$
$T_{B}^{o} = 373.15 K$
$Δ T_{B} = 0.385 = K_{b}$ molality
$0.385 = 0.52 \times (\frac{2.5}{M} \times \frac{1000}{75})$
$M = 45 g m o l^{- 1}$