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  • Assuming ideal behaviour, calculate the pressure of 1.0 molal solution of a non-volatile molecular solute in water at 50oC. The vapour pressure of the water at 50oC is 0.222 a

Assuming ideal behaviour, calculate the pressure of 1.0 molal solution of a non-volatile molecular solute in water at 50oC. The vapour pressure of the water at 50oC is 0.222 atm.

Option: 1

0.218 atm


Option: 2

0.222 atm


Option: 3

0.982 atm


Option: 4

0.248


Answers (1)

best_answer

As we have learnt,

 

Relation Between Raoult's Law and Dalton's Law -

We have two liquids A and B and their vapour pressures are represented as PA and PB.

According to Raoult's law, we know:

\mathrm{P_{A}=P^{o}_{A} X_{A}\quad\quad\quad\quad\quad............(i)}      

\mathrm{P_{B}=P^{o}_{B} X_{B}\quad\quad\quad\quad\quad............(ii)}

Now, according to Dalton's law of partial pressure, we have:

\mathrm{P_{A}=P_{T} Y_{A}\quad\quad\quad\quad\quad............(iii)}

\mathrm{P_{B}=P_{T} Y_{B}\quad\quad\quad\quad\quad............(iv)} 

Thus, on combining equations (i) with (iii) and (ii) with (iv), we get:

\\\mathrm{P^{0}_{A} X_{A}=P_{T} Y_{A}}\\\\\mathrm{P^{0}_{B} X_{B}=P_{T} Y_{B}}

\\\mathrm{Thus,\: Y_{A}\: =\: \frac{P^{o}_{A}X_{A}}{P_{T}}}\\\\\mathrm{And,\: Y_{B}\: =\: \frac{P^{o}_{B}X_{B}}{P_{T}}}

-

\mathrm P_{\text {solution }}=\mathrm P_{\mathrm{A}}=\mathrm P_{\mathrm{A}}^{\circ} \chi_{\mathrm{A}} \text { (Raoult's law) }
By using the relation between molarity and mole fraction,
we have;
m=\frac{\chi_{B}}{\chi_{A}} \times \frac{1000}{M w_{A}}=\frac{1-\chi_{A}}{\chi_{A}} \times \frac{1000}{M w_{A}}

\Rightarrow \quad \chi_{\mathrm{A}}=\frac{1000}{\mathrm{m} M w_{\mathrm{A}}+1000}

=\frac{1000}{1 \times 18+1000}=\frac{1000}{1018}=0.982

\Rightarrow \mathrm P_{\text {solution }}=P_{A}^{\circ} \chi_{A}=0.222 \times 0.982=0.218 \mathrm{atm}

Therefore, option(1) is correct

Posted by

manish

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