Question

an aqueous solution containing 1.248 g of BaCl2(molar mass =208.34). in 100 g of water boils at 100.0832°c. calculate degree of dissociation of BaCl2

Answer 1

Is the answer correct?

Answer 2

Answer:The degree of dissociation is 88.5 %

Explanation:-

Weight of solvent(water) = 100 g = 0.1 kg (1 kg=1000 g)

Molar mass of $BaCl_2$ = 208.34 g/mol

Mass of  $BaCl_2$  added = 1.248 g

$\Delta T_b=i\times k_f\times \frac{\text{mass}}{\text {molar mass}\times {\text {weight of solvent in kg}}}$

$\Delta T_b$ = elevation in boiling point

$k_f$ = boiling point elevation constant

$\Delta T_b=T_b-T^{o}_b=(100.0832-100)^0C=0.0832^0C$

$0.0832=i\times 0.512^0Ckg/mol\frac{1.248g}{208.34g/mol\times 0.1kg}$

$i=2.77$

$BaCl_2\rightarrow Ba^{2+}+2Cl^-$

$\alpha=\frac{i-1}{n-1}$

$\alpha$ =degree of dissociation

i= vant hoff factor

n= number of ions produced if complete dissociation is there = 3

$\alpha=\frac{2.77-1}{3-1}=0.885=88.5\%$