Question

Calculate the normal boiling point of a sample of sea water found to contain $3.5\%$ of $NaCl$ and $0.13\%$ of $MgC{l}_2$ by mass. The normal boiling point of water is $100^{\circ }C$ and $K_b$ (water) $=0.51Kkgmo{l}^{-1}$ Assume that both the salts are completely ionised

• A. $100.655^{\circ }C$
• B. $99.655^{\circ }C$
• C. $101.655^{\circ }C$
• D. $102.655^{\circ }C$

Answer 1

Answer: (A)

Mass of NaCl = 3.5 g
No. of moles of $NaCl=\frac{3.5}{58.5}$
Number of ions furnished by one molecule of NaCl is 2
So, actual number of moles of particles furnished by sodium chloride $=2\times \frac{3.5}{58.5}$
Similarly, actual number of moles of particles furnished by magnesium chloride $=3\times \frac{0.13}{95}$
Total number of moles of particles $=\left(2\times \frac{3.5}{58.5}+3\times \frac{0.13}{95}\right)$
$=0.1238$
Mass of water $=\left(100-3.5-0.13\right)$
$=96.37g=\frac{96.37}{1000}kg$
Molality $=\frac{0.1238}{96.37}\times 1000=1.2846$
$\Delta T_b$ =Molality $\times K_b=1.2846\times 0.51=0.655K$
Hence, boiling point of sea water $=373.655K$ or $100.655^{\circ }C$