Addition of 0.643 g of a compound to 50 mL of benzene (density =0.879 g mL -1 ) lowers the freezing point from 5.51° C to 5.03° C (Kf. of benzene .=5.12 K mol -1 kg ) . Molar mass of the compound is

Question

Addition of 0.643 g of a compound to 50 mL of benzene (density =0.879 g mL -1 ) lowers the freezing point from 5.51° C to 5.03° C (Kf. of benzene .=5.12 K mol -1 kg ) . Molar mass of the compound is (A) 162 g mol -1 (B) -156 g mol -1 (C) 145 g mol -1 (D) 156 g mol -1

Tardigrade Admin · Accepted Answer

On adding ((w1/m1)) mole of solute to w2 g of solvent, depression in freezing point is Δ Tf=(1000 Kf w1/m1 w2) (15.51-5.03)=(1000 × 5.12 × 0.643/m1 ×(50 × 0.879)) m1=156 g mol -1

Q. Addition of $0.643 g$ of a compound to $50 m L$ of benzene (density $= 0.879 g m L^{- 1}$ ) lowers the freezing point from $5.5 1^{\circ} C$ to $5.0 3^{\circ} C$ $(K_{f}$ of benzene $= 5.12 K m o l^{- 1} k g) .$ Molar mass of the compound is

$162 g m o l^{- 1}$

$- 156 g m o l^{- 1}$

$145 g m o l^{- 1}$

$156 g m o l^{- 1}$

On adding $(\frac{w _{1}}{m _{1}})$ mole of solute to $w_{2} g$ of solvent, depression in freezing point is

$Δ T_{f} = \frac{1000 K _{f} w _{1}}{m _{1} w _{2}}$

$(15.51 - 5.03) = \frac{1000 \times 5.12 \times 0.643}{m _{1} \times ( 50 \times 0.879 )}$

$m_{1} = 156 g m o l^{- 1}$

Q. Addition of 0.643g of a compound to 50mL of benzene (density =0.879gmL−1 ) lowers the freezing point from 5.51∘C to 5.03∘C (Kf​ of benzene =5.12Kmol−1kg). Molar mass of the compound is

162gmol−1

−156gmol−1

145gmol−1

156gmol−1