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 ~kg ~mol^ -1 . The molecular mass of

Correct answer: A. 58 ~g ~mol^ -1 T_b^ =353.23 ~K, W_B=1.8 ~g , W_A=90 ~g_ T_b=354.11 ~K , K_b=2.53 ~kg mol ^ - 1 Δ T_b=T_b-T_b ^ =354.11-353.23=0.88 ~K M_B=(W_B × K_b × 1000)/(Δ T_b × W_A)=(1.8 × 2.53 × 1000)/(0.88 × 90) =57.5 58 ~g ~mol^ -1

AIIMS 2013

Exam: AIIMS
Year: 2013
Subject: Chemistry
Chapter: Solutions

AIIMS 2013 · Chemistry · Solutions

Boiling point of benzene is 353.23 ~K . When 1.8 ~g of non-volatile...

Boiling point of benzene is

353.23 \mathrm{~K}

. When

1.8 \mathrm{~g}

of non-volatile solute is dissolved in

90 \mathrm{~g}

of benzene. Then boiling point is raised to

354.11 \mathrm{~K}

. Given

K_b

(benzene)

=2.53 \mathrm{~kg} \mathrm{~mol}^{-1}

. The molecular mass of non-volatile substance is

Options

A $58 \mathrm{~g} \mathrm{~mol}^{-1}$
B $120 \mathrm{~g} \mathrm{~mol}^{-1}$
C $116 \mathrm{~g} \mathrm{~mol}^{-1}$
D $60 \mathrm{~g} \mathrm{~mol}^{-1}$

Correct answer

A. $58 \mathrm{~g} \mathrm{~mol}^{-1}$

Step-by-step solution

T_b^{\circ}=353.23 \mathrm{~K}, W_B=1.8 \mathrm{~g}

W_A=90 \mathrm{~g}_{\prime} T_b=354.11 \mathrm{~K}

K_b=2.53 \mathrm{~kg} \mathrm{\textrm {mol } ^ { - 1 }}

\Delta T_b=T_b-T_b{ }^{\circ}=354.11-353.23=0.88 \mathrm{~K}

M_B=\frac{W_B \times K_b \times 1000}{\Delta T_b \times W_A}=\frac{1.8 \times 2.53 \times 1000}{0.88 \times 90}

=57.5 \approx 58 \mathrm{~g} \mathrm{~mol}^{-1}

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