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A 100 cm3 solution of sodium carbonate is prepared by dissolving 8.653 g of the salt in water. The density of solution is 1.0816 g per mililitre. What are the molality and molarity of the solution? |
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Answer» We have given, density of solution = 1.0816 g/ml \(\therefore\) Numbers of moles of Na2CO3 = \(\frac{8.653}{100}\) = 0.08 mol \(\because\) 1cm3 = 1ml \(\therefore\) 100 cm3 = 100 ml \(\therefore\) molarity of Na2CO3 solution = \(\frac{0.08}{100}\times1000\) = 0.8 m weight of solution = 100 ml x 1.816 g/ml = 108.16 g weight of solvent = weight of solution - weight of solute = 108.16 g - 8.653 = 99.507 g \(\therefore\) Molality of Na2CO3 solution = \(\frac{0.08}{99.507}\times1000\) = 0.8039 m Hence, molality and molality of Na2CO3 solution will be 0.8039 m and 0.8 m respectively. |
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