Question

An aqueous solution of $BaC{l}_2$ (1.28gm in 100gm of water)boils at ${100.0832}^{\circ }$ at 1 atm. Calculate the degree of dissociation. ( $K_b$ of water is 0.512 k/m)

• A. 0.5
• B. 0.75
• C. 0.85
• D. 0.6

Answer 1

The correct option is C

0.85

No. of moles of $BaC{l}_2=\frac{1.28}{208.34}$
=0.006 moles
$\therefore molality=\frac{0.006\times 1000}{100}=0.06$ moles
$△T_b=i{K}_{b}m\Rightarrow \frac{△T_b}{K_{b}m}=i\Rightarrow \frac{0.083}{0.512\times 0.06}=i\Rightarrow i=2.7$
Now consider the dissociation of $BaC{l}_2$
$BaC{l}_2\to Ba+2ClInitial:C00final:C(1-\alpha )c\alpha 2C\alpha$
$i=\frac{C(1-\alpha )+C\alpha +2C\alpha }{C}=1+2\alpha \Rightarrow i=1+2\alpha \Rightarrow 2.7=1+2\alpha \Rightarrow \alpha =0.85$