Question

The boiling point of a solution containing $68.4$g of sucrose (molar mass$=342$g $mo{l}^{-1}$) in $100$ g of water is:

[$K_b$ for water $=0.512$K kg $mo{l}^{-1}$]

• A. ${101.02}^o$C
• B. ${100.512}^o$C
• C. ${100.02}^o$C
• D. ${98.98}^o$C

Answer 1

Answer: (A)

${101.02}^o$C

Number of moles of sucrose is the ratio of mass to molar mass.
$n=\frac{68.4}{342}=0.2$ moles

Molality of solution is the ratio of the number of moles of sucrose to the volume of water (in kg) is
$m=\frac{0.2}{0.1}=2$

The elevation in the boiling point of the solution is
$\Delta T_b=K_{b}m=0.512\times 2={1.024}^{o}C$

The boiling point of the solution is $100+1.024={101.02}^{o}C$