Boiling point of benzene is 353.23 K. When 1.8 g of non-volatile solute is dissolved in 90 g of benzene. Then boiling point is raised to 354.11 K. Given Kb (benzene) = 2.53 kg mol-1. The molecular mass of non-volatile substance is

Question

Boiling point of benzene is 353.23 K. When 1.8 g of non-volatile solute is dissolved in 90 g of benzene. Then boiling point is raised to 354.11 K. Given Kb (benzene) = 2.53 kg mol-1. The molecular mass of non-volatile substance is (A) 58 g mol-1 (B) 120 g mol-1 (C) 116 g mol-1 (D) 60 g mol-1

Tardigrade Admin · Accepted Answer

Tb°=353.23 K, WB=1.8 g, WA=90 g, Tb=354.11 K, Kb=2.53 kg mol -1 Δ Tb=Tb-Tb°=354.11-353.23=0.88 K MB=(WB× Kb×1000/Δ Tb× WA)=(1.8×2.53×1000/0.88×90) =57.5≈ 58 g mol -1

Q. Boiling point of benzene is $353.23 K$ . When $1.8 g$ of non-volatile solute is dissolved in $90 g$ of benzene. Then boiling point is raised to $354.11 K$ . Given $K_{b}$ (benzene) = $2.53 k g m o l^{- 1}$ . The molecular mass of non-volatile substance is

$58 g m o l^{- 1}$

$120 g m o l^{- 1}$

$116 g m o l^{- 1}$

$60 g m o l^{- 1}$

Q. Boiling point of benzene is 353.23K. When 1.8g of non-volatile solute is dissolved in 90g of benzene. Then boiling point is raised to 354.11K. Given Kb​ (benzene) = 2.53kgmol−1. The molecular mass of non-volatile substance is

58gmol−1

120gmol−1

116gmol−1

60gmol−1

Tb∘​=353.23K,WB​=1.8g, WA​=90g,Tb​=354.11K, Kb​=2.53kgmol−1 ΔTb​=Tb​−Tb∘​=354.11−353.23=0.88K MB​=ΔTb​×WA​WB​×Kb​×1000​=0.88×901.8×2.53×1000​ =57.5≈58gmol−1

Q. Boiling point of benzene is $353.23 K$ . When $1.8 g$ of non-volatile solute is dissolved in $90 g$ of benzene. Then boiling point is raised to $354.11 K$ . Given $K_{b}$ (benzene) = $2.53 k g m o l^{- 1}$ . The molecular mass of non-volatile substance is

$58 g m o l^{- 1}$

$120 g m o l^{- 1}$

$116 g m o l^{- 1}$

$60 g m o l^{- 1}$