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A solution containing 6 g of a solute is dissolved in 25mL of water gave an osmotic pressure of 4.5 atm at 27°C. Calculate the boiling point of the solution. Kb = 0.52Kkg/mol. |
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Answer» Let say molar mass of solute is M g/mol. Then \(\pi = \frac{\frac6M}{25} \times 1000\times R\times T\) \(\pi = 4.5 = \frac6{M\times25} \times 1000\times0.082 \times300\) \(M = \frac6{4.5} \times 40 \times 0.082 \times 300\) \(m = 1312 g/ mol\) ∴ Molality of solution \(= \frac{\frac6{1312}}{25} \times 1000 = 0.183 m\) ∴ \(\triangle T_b = K_bm\) \(T - 100 = 0.52 \times 0.183\) \(T = 100.095° C\) Hence, boiling point of solution becomes 100.095°C. |
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