An aqueous solution of BaCl 2 1-28gm in 100 gm of waterboils at 100-0832- at 1 atm- Calculate the degree of dissociation- K b of water is 0-512 k - m A- 0-5B- 0-75C- 0-85D- 0-6

The correct option is C 0.85 No. of moles of BaCl_2 =1.28/208.34 =0.006 moles ∴ molality=0.006 × 1000/100=0.06 moles △ ...

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Standard XII

Chemistry

Degree of Dissociation

An aqueous so...

Question

An aqueous solution of $B a C l_{2}$ (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)

0.5

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0.75

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0.85

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0.6

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Solution

The correct option is C
0.85

No. of moles of $B a C l_{2} = \frac{1.28}{208.34}$
=0.006 moles
$∴ m o l a l i t y = \frac{0.006 \times 1000}{100} = 0.06$ 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 $B a C l_{2}$
$B a C l_{2} \to B a + 2 C l I n i t i a l : C 00 f i n a l : C (1 - α) c α 2 C α$
$i = \frac{C (1 - α) + C α + 2 C α}{C} = 1 + 2 α \Rightarrow i = 1 + 2 α \Rightarrow 2.7 = 1 + 2 α \Rightarrow α = 0.85$

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An aqueous solution of $B a C l_{2}$ (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)

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An aqueous solution of BaCl2 (1.28gm in 100gm of water)boils at 100.0832∘ at 1 atm. Calculate the degree of dissociation. ( Kb of water is 0.512 k/m)

An aqueous solution of $B a C l_{2}$ (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)