Solutions and Colligative Properties - Result Question 63
34. Addition of $0.643 $ $g$ of a compound to $50$ $ mL$ of benzene (density : $0.879 $ $g / mL$ ) lowers the freezing point from $5.51^{\circ} C$ to $5.03^{\circ} C$. If $K _f$ for benzene is $5.12$ , calculate the molecular weight of the compound.
(1992,2 M)
Show Answer
Answer:
Correct Answer: 34. $(156$ $ g/mol)$
Solution:
$-\Delta T _f=5.51-5.03=0.48$
$\Rightarrow \quad-\Delta T _f=0.48=K _f \cdot m$
$\Rightarrow \quad 0.48=5.12 \times \dfrac{0.643}{M} \times \dfrac{1000}{50 \times 0.879}$
$\Rightarrow \quad M=156$ $ g / mol$
BETA
