Question

The freezing point of a solution containing 2.40 g of a compound in 60.0 g of benzene is ${0.10}^{\circ }C$ lower than that of pure benzene. What is the molecular weight of the compound? $(K_{f}is{5.12}^{\circ }C/m$ for benzene)

• A. $2050$ g/mol
• B. $2500$ g/mol
• C. $2005$ g/mol
• D. $1500$ g/mol

Answer 1

The correct option is A $2050$ g/mol

Let $M$ be the molecular weight of the compound.
Number of moles is the ratio of mass to molecular weight.
Number of moles of solute $\frac{2.40}{M}$
Molality of solution is the number of moles of solute in $1000$ g of water.
Molality $=\frac{2.40\times 1000}{M\times 60.0}=\frac{40}{M}$
The depression in the freezing point $=\Delta T_f=K_{f}m⟹0.10=5.12\times \frac{40}{M}⟹M=2050g/mol$