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Standard XII

Chemistry

Elevation in Boiling Point

An aqueous so...

Question

An aqueous sol. containing $12.48 g$ of Barium chloride in $1.0 k g$ of water boils at $373.0832 K$ . Calculate the degree of dissociation of barium chloride. [Given $K_{b}$ for $H_{2} O = 0.52 k m^{- 1}$ , molar mass of $B a C l_{2} = 208.34 g m o l^{- 1}$ ]

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Solution

Given that:
$\begin{matrix} M_{2} = 12.4 g M_{^{H_{2} O}} = 1.0 k g T_{b} = 373.0832 K {T_{b}}^{\circ} = 373.15 K K_{b} = 0.52 k m^{- 1} M_{2} = 208.34 α = ? Δ T_{b} = i K_{b} \times m 373.0832 - 373 = i \times \frac{0.52 \times 12.48 \times 1000}{208.34 \times 1} i = 2.67 B a C l_{2} \to B a^{2 +} + 2 C l^{-} n = 3 α = \frac{i - 1}{n - 1} = \frac{2.67 - 1}{3 - 1} = 83.55 % \end{matrix}$

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Similar questions

Q. 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 = 0.52 K K g {m o l}^{- 1}

Molar mass of

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

)

Q. 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)

Q. The valency of barium in barium chloride

({BaCl}_{2})

is _____.

Q. The boiling point of pure water is

100^{\circ}

C. Calculate the boiling point of an aqueous solution containing

0.6 g

of urea (molar mass =

60 {g mol}^{- 1}

) in

100 g

of water.

K_{b}

for water is

0.52 {K kg mol}^{- 1}

10 g of a mixture of barium chloride and anhydrous sodium sulphate is dissolved in water. An excess of barium chloride solution is added and 6.99 g of barium sulphate is precipitated according to the equation given below:

Na₂SO₄ + BaCl₂ → BaSO₄ +2NaCl

[O = 16, Na = 23, S = 32, Ba = 137]

Calculate the percentage of sodium sulphate in the original mixture.

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An aqueous sol. containing 12.48g of Barium chloride in 1.0kg of water boils at 373.0832K. Calculate the degree of dissociation of barium chloride. [Given Kb for H2O=0.52km−1, molar mass of BaCl2=208.34gmol−1]

Given that:M2=12.4gMH2O=1.0kgTb=373.0832KTb∘=373.15KKb=0.52km−1M2=208.34α=?ΔTb=iKb×m373.0832−373=i×0.52×12.48×1000208.34×1i=2.67BaCl2→Ba2++2Cl−n=3α=i−1n−1=2.67−13−1=83.55%

An aqueous sol. containing $12.48 g$ of Barium chloride in $1.0 k g$ of water boils at $373.0832 K$ . Calculate the degree of dissociation of barium chloride. [Given $K_{b}$ for $H_{2} O = 0.52 k m^{- 1}$ , molar mass of $B a C l_{2} = 208.34 g m o l^{- 1}$ ]