Question

Boiling point of water at $750mmHg\text{is}{99.63}^{0}C$. How much sucrose is to be added to $500g$ of water such that it boils at ${100}^{0}C$.

Answer 1

Given:
Mass of water $\left(W_A\right)=500g$
Boiling point of pure water at $750mmHg\left(T_b^0\right)={99.63}^{0}C$
Boiling point of mixture $\left(T_b\right)={100}^{0}C$
Molar mass $\left(M_B\right)$ of sucrose $\left(C_{12}{H}_{22}{O}_{11}\right)$
$=12\times 12+1\times 22+16\times 11$
$=342gmo{l}^{-1}$


Elevation in boiling point $\left(\Delta T_b\right)=T_b-T_b^0$
$\Delta T_b=100-99.63={0.37}^{0}CorK$

Amount of sucrose.

We know, molal elevation constant for water $\left(K_b\right)=0.52Kkgmo{l}^{-1}$
We know, $\Delta T_b=K_b\times molality\left(m\right)$
We also know, molality(m)$=\frac{\text{mass of solute(w\_B)}}{\text{molar mass of solute(M\_g)}\times \text{mass of solvent(w\_A)}\left(inkg\right)}$

So, $\Delta T_b=\frac{K_b\times w_B}{M_B\times w_A}$
$0.37K=\frac{0.52Kkgmo{l}^{-1}\times w_B}{342gmo{l}^{-1}\times 500\times {10}^{-3}kg}$

$\Rightarrow w_B=\frac{0.37\times 500\times {10}^{-3}\times 342}{0.52}$
$=121.67g$

Thus, $121.67g$ of sucrose needs to be
added.