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  • At 100°C the v.p of a solution of 6.5g of a solute in 100g water is 732mm.If kb=0.52k the boiling point of this solution will be?

At 100°C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If Kb = 0.52, the boiling point of this solution will be:

Option 1)  101°C

Option 2)  100°C

Option 3)  102°C

Option 4)  103°C    

Answers (1)

Mathematical Expression -

\Delta T_{b}= K_{b}\: m

Using of K_{b}=\frac{K-K_{g}}{mole}

\Delta T_{b}= Elevation\: in \: boiling\: point
 

- wherein

K_{b}= Boiling \: point \: elevation \: constant

m= molality

Let's find the molar mass of the solute first By Roalt's law of partial pressure

\Rightarrow \frac{760-732}{732}=\frac{6.5 \times 18}{M \times 100}

\Delta M = 30.6

Then we know that \Delta T_{b}= 0.52 \times \frac{6.5}{30.6} \times \frac{1000}{100}=1.1

\therefore boiling \ point = 100 +1.1 \simeq 101.1^{\circ}C

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lovekush

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