On mixing 3 g of non-volatile solute in 200 mL of water- its boiling point 100- C becomes 100-52- C- If Kb for water is 0-6 K-m then molecular wt- of solute is-

The correct option is D 17.3 g mol^ 1 Δ T = K_b×wm×1000W 0.52 = 0.6×3m×1000200 m = 1.8× 50.52 = 17.3 g mole^ 1 .Hence, the correct option i ...

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Standard XII

Chemistry

Elevation in Boiling Point

On mixing 3...

Question

On mixing $3 g$ of non-volatile solute in $200 m L$ of water, its boiling point $(100^{\circ} C)$ becomes ${100.52}^{\circ} C$ . If $K_{b}$ for water is $0.6 K / m$ then molecular wt. of solute is:

10.5 g m o l^{- 1}

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12.6 g m o l^{- 1}

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15.7 g m o l^{- 1}

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17.3 g m o l^{- 1}

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Solution

The correct option is D $17.3 g m o l^{- 1}$
$Δ T = K_{b} \times \frac{w}{m} \times \frac{1000}{W}$

$0.52 = 0.6 \times \frac{3}{m} \times \frac{1000}{200}$

$m = \frac{1.8 \times 5}{0.52} = 17.3 g {mole}^{- 1}$ .

Hence, the correct option is $D$

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On mixing 3 g of non-volatile solute in 200 mL of water, its boiling point (100∘C) becomes 100.52∘C. If Kb for water is 0.6 K/m then molecular wt. of solute is:

The correct option is D 17.3 g mol−1ΔT=Kb×wm×1000W0.52=0.6×3m×1000200m=1.8×50.52=17.3 g mole−1.Hence, the correct option is D

On mixing $3 g$ of non-volatile solute in $200 m L$ of water, its boiling point $(100^{\circ} C)$ becomes ${100.52}^{\circ} C$ . If $K_{b}$ for water is $0.6 K / m$ then molecular wt. of solute is:

The correct option is D $17.3 g m o l^{- 1}$
$Δ T = K_{b} \times \frac{w}{m} \times \frac{1000}{W}$

$0.52 = 0.6 \times \frac{3}{m} \times \frac{1000}{200}$

$m = \frac{1.8 \times 5}{0.52} = 17.3 g {mole}^{- 1}$ .

Hence, the correct option is $D$