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A solution containing 30 g of non-volati...

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 this solution, the new vapour pressure becomes 2.9 kPa at 298 K. Calculate (i) the molar mass of solute, (ii) vapour pressure of water at 298 K.

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Case I `(p_A^0 -p_A)/(p_A^0) = (w_B //M_B)/(w_A // M_A)`
Substituting the values, we have
`(p_A^0 - 2.8)/(p_A^0) = (30 // M_B)/(90 //18) = (6)/(M_B) " or " (p_A^0 - 2.8)/(p_A^0) = (6)/(M_B)` ....(i)
Case II `(p_A^0 -2.9)/(p_A^0) = (30//M_B)/(108//18) = (5)/(M_B) " or " (p_A^0 - 2.9)/(p_A^0) = (5)/(M_B)` ...(ii)
Dividing equation (i) by (ii), we get
`(p_A^0 - 2.8)/(p_A^0 - 2.9) = 6/5`
On solving we get `p_A^0 = 3.4 kPa`
ubstituting the value of `p_A^0` in equation (i), we get
`(3.4-2.8)/(3.4) = ( 6)/(M_B) " or " M_B = 34` g/mol
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