Question

Addition of $0.643g$ of a compound to $50mL$ of benzene (density : $0.879gm{L}^{-1}$) lower the freezing point from ${5.51}^{\circ }C$ to ${5.03}^{\circ }C$. If $K_f$ for benzene is $5.12Kkgmo{l}^{-1}$, calculate the molar mass of the compound.

Answer 1

Given
$\Delta T=\frac{1000\times k_f^{\prime }\times w}{m\times W}$
$W=wt.ofbenzene=V\times d=50\times 0.879g$
$\therefore 0.48=\frac{1000\times 5.12\times 0.643}{m\times 50\times 0.879}$
$\Delta T=5.51-5.03=0.48$
$\therefore m=156.06$
$w=0.643g,K_f^{\prime }=5.12Kmo{l}^{-1}kg$