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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 :(i) Molar mass of the solute.(ii) Vapour pressure of water at 298 K. |
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Answer» Solution :(i) Let, the molar mass of the solute be M g `mol^(-1)` Now, the no. of moles of solvent (water), `n_(1)=(90g)/(18g mol^(-1))=5 mol` And, the no. of solute, `n_(2)=(30g)/(M mol^(-1))=(30)/(M)mol` `p_(1)=2.8` kPa Applying the relation: `(p_(1)^(0)p_(1))/(p_(1)^(0))=(n_(2))/(n_(1)+n_(2))` `therefore (p_(2)^(0)-2.8)/(p_(1)^(0)=((30)/(M))/(5+(30)/(M))` `therefore 1-(2.8)/(p_(1)^(0))=((30)/(M))/((5M+30)/(M))` `therefore 1-(2.8)/(p_(1)^(0))=(30)/(5M+30)` `therefore (2.8)/(p_(1)^(0))=1-(30)/(5M+30)` `therefore (28)/(p_(1)^(0))=(5M+30-30)/(5M+30)` `therefore (2.8)/(p_(1)^(0))=(5M)/(5M+30)` `therefore (p_(1)^(0))/(2.8)=(5M+30)/(5M) ""`....(i) After the addition of 18 g of water : `n_(1)=(90+18)/(18)=6mol` `p_(1)=2.9` kPa Aganin, applying the relation : `(p_(1)^(0)-p_(1))/(p_(1)^(0))=(n_(2))/(n_(1)+n_(2))` `therefore (p_(1)^(0)-2.9)/(p_(1)^(0))=((30)/(M))/(6+(30)/(M))` `therefore 1-(2.9)/(p_(1)^(0))=((30)/(M))/((6M+30)/(M))` `therefore 1-(2.9)/(p_(1)^(0))=(30)/(6M+30)` `therefore (2.9)/(p_(1)^(0))=1-(30)/(6M+30)` `therefore (2.9)/(p_(1)^(0))=(6M+30-30)/(6M+30)` `therefore (2.9)/(p_(1)^(0))=(6M)/(6M+30)` `therefore (p_(1)^(0))/(2.9)=(6M+30)/(6M) ""`....(ii) Dividing equation (i) or (ii), we have : `(2.9)/(2.8)=((5M+30)/(5M))/((6M+30)/(6M))` `(2.9)/(2.8)xx(6M+30)/(6)=(5M+30)/(5)` `2.9xx5xx(6M+30)=2.8xx6xx(5M+30)` `87M+435=84M+504` 3 M = 69 M = 23 Therefore, the molar mass of the solute is 23 g `mol^(-1)`. (ii) Putting the value of M in equation (i), we have : `(p_(1)^(0))/(2.8)=(5xx23+30)/(5xx23)` `therefore (p_(1)^(0))/(2.8)=(145)/(115)` `therefore p_(1)^(0)=3.53` kPa Hence, the VAPOUR pressure of water at 298 K is 3.53 kPa. |
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