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2.19    A solution containing 30\; g of non-volatile solute exactly in 90\; g of water has a vapour pressure of  2.8\; kPa at 298 \; K Further, 18 \; g of water is then added to the solution and the new vapour pressure becomes 2.9 \; kPa at 298 \; K. Calculate

               (ii) vapour pressure of water at 298 K.

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In the previous part we have calculated the value of molar mass the Raoul's law equation.

We had :-

                             \frac{p_w^{\circ} }{2.8} = \frac{5M+30}{5M}

Putting M = 23 u in the above equation we get, 

                            \frac{p_w^{\circ} }{2.8} = \frac{5(23)+30}{5(23) } =\frac{145}{115}

or                         p_w^{\circ} =3.53

Thus vapour pressure of water = 3.53 kPa.

Posted by

Devendra Khairwa

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