Question

The molal elevation constant of water $= 0.52^{o}C$. The boiling point of 1.0 molal aqueous $KCl$ solution (assuming complete dissociation of $KCl$), therefore, should be.

• A. $100.52^{o}C$
• B. $101.04^{o}C$
• C. $99.48^{o}C$
• D. $98.96^{o}C$

Answer 1

Answer: (B)

$101.04^{o}C$

Answer 2

$\Delta T_{b} = imk_{b} = 0.52 \times 1 \times 2 = 1.04$.

∴ $T_{b} = 100 + 1.04 = 101.04^{o}C$.