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  • The vapour pressure of pure water at 25oC is 23.0 torr. What is the vapour pressure ( in torr) of 100 g  of water to which 100 g of C6H12O<

The vapour pressure of pure water at 25oC is 23.0 torr. What is the vapour pressure ( in torr) of 100 g  of water to which 100 g of C6H12O6 (glucose) has been added?

Option: 1

21.29


Option: 2

27.29


Option: 3

23


Option: 4

18.29


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}}}

-

\text { Raoult's law: }\mathrm P_{\text {solution }}=p_{\mathrm{A}}=p_{\mathrm{A}}^{\circ} \chi_{\mathrm{A}}

\Rightarrow \mathrm P_{\text {solution }}=p_{\mathrm{A}}^{\circ} \frac{n_{\mathrm{A}}}{n_{\mathrm{A}}+n_{\mathrm{B}}}=p_{\mathrm{A}}^{\circ} \frac{\frac{\mathrm W_{\mathrm{A}}}{M \mathrm w_{\mathrm{A}}}}{\frac{W_{\mathrm{A}}}{\mathrm {M w}_{\mathrm{A}}}+\frac{\mathrm W_{\mathrm{B}}}{\mathrm M \mathrm w_{\mathrm{B}}}}

                       =23.00 \times \frac{\frac{100}{18}}{\frac{100}{18}+\frac{100}{180}}


\\ \mathrm{M_W} {\text {Glucose }}=180\: \mathrm{g\: } \mathrm{mol}^{-1} ; p_{\mathrm{A}}^{\circ}=23.0 \text { torr} \\ \\\mathrm P_{\text {solution }}=21.294 \mathrm{\: torr}

 

Posted by

Divya Prakash Singh

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