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.
Answers
Answer:
23 g/mol and 3.53 kPa.
Explanation:
The number of moles of solute (n1) will be 30/M if we take the molar mass of the solute to be M g/mol. The moles of solvent (n2) will be 90/18 = 5 moles.
So, the partial pressure will be Po - Pa/Po=(30/M)/(5+30/M) which on solving we will get that 2.8/Po=1- (30/M)/(5+30/M) or Po/2.8= (5+30M)/5 or Po/2.8= 1+6/M.
Now, when 18 gram of water is added then Po-2.9/Po = (30/M)/(6+30/M) which on solving we 2.9/Po = 1- (30/M)/(6+30/M) or Po/2.9= 1+5/M.
Now, if we divide both of the equations then we will get 2.8/2.8=(1+6/M)/(1+5/M) which on solving we will get that 2.3/M=0.1 or M will be 23 g/mol.
Now, if we put the M=23 in any of the equation Po/2.8=1+6/23 which on solving we will get the value of Po = (29/23)*2.8 = 3.53 kPa.
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