1.00 ~g of a non-electrolyte solute (molar mass .250 ~g ~mol^ -1 ) was dissolved in 51.2 ~g of benzene. If the freezing point depression constant, K_ f of benzene is 5.12 ~K ~kg ~mol^ -1 , the freezing point of benzene

Correct answer: C. 0.4 K Δ T_ f =K_ f m=5.12 × (1)/(250) × (1000)/(51.2)=0.4 K

BITSAT 2011

BITSAT 2011 · Chemistry · Solutions

1.00 ~g of a non-electrolyte solute (molar mass .250 ~g ~mol^ -1 )...

1.00 \mathrm{~g}

of a non-electrolyte solute (molar mass

\left.250 \mathrm{~g} \mathrm{~mol}^{-1}\right)

was dissolved in

51.2 \mathrm{~g}

of benzene. If the freezing point depression constant,

\mathrm{K}_{\mathrm{f}}

of benzene is

5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}

, the freezing point of benzene will be lowered by

C. $0.4 \mathrm{ K}$

\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \mathrm{m}=5.12 \times \frac{1}{250} \times \frac{1000}{51.2}=0.4 \mathrm{ K}