Question

The freezing point of an aqueous solution of a non-electrolyte having an osmotic pressure $2.0$ atm at $300$K ($K_f=1.86Kmo{l}^{-1}kg$ and $R=0.08251Latm{K}^{-1}mo{l}^{-1}$), assuming molarity and molality are same is:

• A. $-{0.151}^{o}C$
• B. $-{0.453}^{o}C$
• C. $-{0.261}^{o}C$
• D. none of these

Answer 1

The correct option is A $-{0.151}^{o}C$

$\because \pi =2atm,S=0.0821Latm{K}^{-1}mo{l}^{-1},T=300K$

Also, $\pi V=nST$

$\therefore$ Molarity, $n/V=\pi /ST=2/(0.0821\times 300)=0.0812mol{L}^{-1}$

For dilute solutions, Molality $=$ Molarity $=0.0812m$

$\because \Delta T=K_f^{\prime }\times molality=1.86\times 0.0812=0.151$

$\therefore$ Freezing pt. $=T_0-\Delta T=0-0.151=-{0.151}^{o}C$

Option A is correct.