The 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. If the freezing point constant, Kf for benzene is 5.12 K kg mol -1, the molar mass of the compound is approximately

Question

The 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. If the freezing point constant, Kf for benzene is 5.12 K kg mol -1, the molar mass of the compound is approximately (A) 156 g mol-1 (B) 88 g mol-1 (C) 60 g mol-1 (D) 312 g mol-1

Tardigrade Admin · Accepted Answer

Given, weight of compound, w2=0.643 g kf=5.12 K kg mol-1 Δ Tf=T°f-Tf 278.51-278.03=0.48 K Volume of benzene = 50 mL Density of benzene = 0.879 g mL-1 ∴ Weight of benzene, w1 = 50×0.879 = 43.95 g Molar mass of a compound can be calculated by using the following formula M2=(kf× w2×1000/Δ Tf× w1) =(5.12×0.643×1000/0.48×43.95) =156.05 g mol-1

Q. The 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 $5.0 3^{\circ} C$ . If the freezing point constant, $K_{f}$ for benzene is $5.12 K k g m o l^{- 1}$ , the molar mass of the compound is approximately

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

$88 g m o l^{- 1}$

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

$312 g m o l^{- 1}$

Q. The 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. If the freezing point constant, Kf​ for benzene is 5.12Kkgmol−1, the molar mass of the compound is approximately

156gmol−1

88gmol−1

60gmol−1

312gmol−1

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

$88 g m o l^{- 1}$

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

$312 g m o l^{- 1}$