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Calculate the boiling point of solution when 4 g of MgSO_4 (M = 120 g "mol"^(-1))was dissolved in 100 g of water, assuming MgSO_4undergoes complete ionisation. |
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Answer» SOLUTION :Apply the relation ` Delta T_b = i xx K_b xx m` For `MgSO_4 , i = 2` Molarity of solution = ` (4//120)/(100) xx 1000 = 4/120 xx 10 = 1/3` Substituting the values in EQUATION (i), we have ` Delta T_b = 2 xx 0.52 xx 1/3 = 0.347` Boiling point of the solution = `100 + 0.347 = 100.347^@C = 373.347 K ` |
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