For a dilute solution containing 2-5 g of non-volatile non-electrolyte solute in 100 g of water- the elevation in boiling point at 1 atm pressure is 2- C- Assuming concentration of solute is much lower than the concentration of solvent- the vapour pressure mm of Hg of the solution is take Kb - -0-76 Kkgmol -1A- 724B- 736C- 740D- 718

The correct option is A 724Therefore, the freezing point is 2.3 × 10^ 2 ^∘C Δ T_b =2 ^∘C ; W_2 =2.5 g W_1=100 g ; K_b =0.76 K. kg mol^ 1 P_s =? Δ T_b =K_b × m ...

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Standard XII

Chemistry

Colligative Properties

For a dilute ...

Question

For a dilute solution containing 2.5 g of non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is $2^{\circ} C$ . Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take $K b = 0.76 K k g m o l^{- 1})$

724

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740

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736

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718

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Solution

The correct option is A 724
Therefore, the freezing point is $- 2.3 \times 10^{- 2}$ $^{\circ} C$
$Δ T_{b} = 2$ $^{\circ} C$ ; $W_{2} = 2.5 g$
$W_{1} = 100 g$ ; $K_{b} = 0.76 K . k g m o l^{- 1}$
$P_{s}$ =?
$Δ T_{b} = K_{b} \times m$
$2 = 0.76 \times m \Rightarrow m = \frac{2}{0.76}$
We know ;
$m = \frac{n_{1} \times 1000}{n_{2} \times M w_{1}} (M w_{1} (H_{2} O) = 18) \frac{n_{2}}{n_{1}} = x_{2} = \frac{m \times M w_{1}}{1000} = \frac{2 \times 18}{0.76 \times 1000} ∴ \frac{P^{\circ} - P_{s}}{P_{s}} = x_{2} = \frac{2 \times 18}{1.76 \times 1000} = \frac{36}{760} 760 - P_{s} = \frac{36}{760} P_{S} (\frac{36}{760} P_{s} + P_{s}) = 760 P_{S} (\frac{36}{760} + 1) = 760 1.074 P_{s} = 760 P_{s} = \frac{760}{1.047} = 725.6 t o r r \approx 724 t o r r$

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For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in $100 g$ of water, the elevation in boiling point at $1 a t m$ pressure is $2 ℃$ . Assuming that the concentration of solute is much lower than the concentration of solvent, the vapour pressure $(in mm of Hg)$ of the solution is: