1.

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?

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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