Question

Sea water is found to contain 5.85% NaCl and 9.50% $MgC{l}_2$ by weight of solution. Calculate it's boiling point assuming 80% ionisation for NaCl and 50% ionisation of $MgC{l}_2$$\left(K_b\right(H_{2}O)$ = 0.51 kgm :

• A. $T_b={101.9}^{\circ }C$
• B. $T_b={102.3}^{\circ }C$
• C. $T_b={108.5}^{\circ }C$
• D. $T_b={110.3}^{\circ }C$

Answer 1

The correct option is B $T_b={102.3}^{\circ }C$

Suppose $100gm$ of sea water

mass of $NaCl=5.85g$

No. of mole $=\frac{5.85}{58.5}=0.1mole$

Mass of $MgC{l}_2=9.5g$

No.of mole $=\frac{9.5}{95}=0.1mole$

$NaCl\to N{a}^{+}+C{l}^{-}$

$0.1$ $0$ $0$

$0.1-\frac{0.1\times 80}{100}$ $\frac{0.1\times 80}{100}$ $\frac{0.1\times 80}{100}$

$0.02$ $0.08$ $0.08$

$=0.18$

$MgC{l}_2\to M{g}^{2+}+2C{l}^{-}$

$0.1$ $0$ $0$

$0.1-\frac{0.1\times 50}{100}$ $\frac{0.1\times 50}{100}$ $\frac{0.1\times 50}{100}\times 2$

$0.05$ $0.05$ $0.10$

No of mole $=0.05+0.05+0.10=0.20$

total number of mole after dissociation

in sea water $=0.81+0.20=0.38m\%$

molarity $=\frac{0.38}{100-5.85-9.5}=\frac{0.38}{89.65\times {10}^{-3}}$

molality $=4.4mol/kg$

$\Delta T_6=k_{6}m=0.51\times 4.4=2.289\simeq 2.3$

$T_6=T_6+O{T}_6=100+2.3={102.3}^{o}C$