Question

Sea water is found to contains 5.85% $NaCl$ and 9.50% $MgC{l}_2$ by weight of solution. Calculate its normal boiling point assuming 80% ionisation for $NaCl$ and 50% ionisation of $MgC{l}_2\left[K_b\right(H_{2}O)=0.5kgmo{l}^{-1}K]$

• A. $T_b=101.3K$
• B. $T_b=102.{24}^{\circ }C$
• C. $T_b=112K$
• D. $T_b={110}^{\circ }C$

Answer 1

The correct option is B $T_b=102.{24}^{\circ }C$

100 $g$ sea water contains 5.85 $gNaCl\left(0.1mole\right)$ and 9.5 $gMgC{l}_2\left(0.1mole\right)$
$\therefore \text{effective molality}=\frac{0.1\left(1.8\right)+0.12\left(2\right)}{(100-5.85-9.5)}\times 100$
= 4.48
$\therefore \Delta T_b=m_{eff}\times k_b={2.24}^{\circ }C$
$T_b={102.24}^{\circ }C$