Question

An aqueous solution containing 12.48g of barium chloride in 1kg of water boils at 373.0832K . Calculate the degree of dissociation of barium chloride.

($K_b$ for $H_{2}O=0.52KKg{mol}^{-1}$ Molar mass of $BaC{l}_2=208.34g{mol}^{-1}$)

• A. 0.165
• B. 0.835
• C. 0.785
• D. None of these

Answer 1

The correct option is B 0.835

Current boiling pt $=373.0832k$

Actual boiling pt $=378k$

$\Delta T=373.0832-373$

$=0.0832$

Now,

n of $Bac{l}_2=\frac{mass}{molarmass}$

$=\frac{12.48}{208.34}$

$\approx 0.0599,i.e0.06$

$\therefore molality=\frac{nofmoles}{manofsales\left(inkg\right)}$

$=0.06$

Now,

$\Delta T_b=k_b\times m\times i$ (i vant- half factor)

$\Rightarrow 0.0832=0.52\times 0.06\times i$

$\Rightarrow i\approx \frac{0.0832}{0.52\times 0.06}$

$\approx 2.67$

$\Rightarrow 1+(n-1)\times \alpha =2.67$ ($\alpha$ degree of diss)

$\Rightarrow 1+2\alpha =2.67$

$2\alpha =1.67$

$\alpha \approx 0.835$