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 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-0-76K kg mol-1-

The elevation in boiling point is Δ T_b = K_b. m m = molality =n_2/w_1× 1000 [n_2 = Number of moles of solute, w_1 = Weight of solvent in gram] ⇒ 2=0.76 ×n_2/10 ...

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Byju's Answer

Standard XII

Chemistry

Elevation in Boiling Point

For a dilute ...

Question

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 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 $K_{b}$ =0.76K kg $m o l^{- 1}$ ).

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

The elevation in boiling point is
$Δ T_{b} = K_{b} . m$
$m = m o l a l i t y = \frac{n_{2}}{w_{1}} \times 1000$
[ $n_{2}$ = Number of moles of solute, $w_{1}$ = Weight of solvent in gram]
$\Rightarrow 2 = 0.76 \times \frac{n_{2}}{100} \times 1000$
$\Rightarrow n_{2} = \frac{5}{19}$
Also, from relative lowering of vapour pressure:
$\frac{- Δ p}{p^{0}} = x_{2} = \frac{n_{2}}{n_{1} + n_{2}} \approx \frac{n_{2}}{n_{1}} [∵ n_{1} >> n_{2}]$
$\Rightarrow - Δ P = 760 \times \frac{5}{19} \times \frac{18}{100}$
= 36 mm of Hg
$\Rightarrow$ p = 760 – 36 = 724 mm of Hg

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