Question

Addition of 0.643 g of a compound to 50 mL of benzene (density 0.879 g\mL) lowers the freezing point from ${5.51}^{o}C$ to ${5.03}^{o}C$. If $K_f$ for benzene is 5.12, calculate the molecular mass of the compound.

• A. 126
• B. 166
• C. 156
• D. 146

Answer 1

The correct option is C 156

we have,

$\Delta T_f=K_f\ast m$

density of benzene = 0.879 g/ml

volume of benzene = 50 ml

so mass of benzene = volume * density = 0.879 * 50 = 43.95 g

so, molality of solute = $\frac{massofsolute}{molarmassofsolute\ast massofsolvent}\ast 1000$

molality of solute = $\frac{0.643}{molarmassofsolute\ast 43.95}\ast 1000$

so,

$\Delta T_f=K_f\ast$$\frac{0.643}{molarmassofsolute\ast 43.95}\ast 1000$

5.51-5.03 = 5.12 * $\frac{0.643}{molarmassofsolute\ast 43.95}\ast 1000$

hence, molecular mass of solute = 156