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The vapour pressure of water is 12.3 kPa at 300 K. Calculate the vapour pressure of 1 molalsolution of a non-volatile solute in it |
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Answer» Solution :1 MOLAL solution CONTAINS 1 mol of the SOLUTE in 1 kg of water ` therefore ` Mole fraction of solute =`(1)/(1 + (1000//18)) = (1)/(1 + 55.5)= 0.0177` Applying Raoult.s LAW, we have `(p^0 -p_x)/(p^0)- x_2 " or " (12.3 -p_s)/(12.3) = 0.0177` `p_s = 12.08 kPa` |
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