You visited us 1 times! Enjoying our articles? Unlock Full Access!

Van't Hoff Factor

An aqueous so...

Question

An aqueous solution containing 12.48g of barium chloride ( $B a C l_{2}$ ) in 1000g of water, boils at ${100.0832}^{o}$ C. Calculate the degree of dissociation of barium chloride. ( $K_{b}$ for water = $0.52 K k g m o l^{- 1}$ , at. wt. Ba = 137, Cl = 35.5)

Open in App

Solution

$w_{B a C l_{2}}$ =12.48 g; $w_{H_{2} O}$ = 1000 g; $Δ T_{b}$ = 0.0832 K; $K_{b}$ =0.52 Kkg/mol
$Δ T_{b} = i K_{b} m m = \frac{w_{B a C l_{2}}}{M_{B a C l_{2}}} \times \frac{1000}{w_{H_{2} O}} m = \frac{12.48}{208} \times \frac{1000}{1000} = 0.06 i = \frac{0.0832}{0.52 \times 0.06} = 2.7 A l s o, i = N u m b e r o f m o l e s a t e q u i l l i b r i u m B a {C l}_{2} 1 - x ⇌ {B a}^{2 +} x + 2 {C l}^{-} 2 x i = 1 - x + x + 2 x = 1 + 2 x = 2.7 x = 0.85$
Therefore degree of dissociation is 85 %.

Suggest Corrections

Similar questions

K_{b}

H_{2} O = 0.52 K K g {m o l}^{- 1}

B a C l_{2} = 208.34 g {m o l}^{- 1}

12.48 g

1.0 k g

373.0832 K

K_{b}

H_{2} O = 0.52 k m^{- 1}

B a C l_{2} = 208.34 g m o l^{- 1}

N a_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 N a C l

N a_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 N a C l

0.011 k g

0.1 k g

{100.46}^{\circ} C

K_{b} (w a t e r) = 0.52 K k g m o l^{- 1}

Join BYJU'S Learning Program