Question

The molality of 1 litre solution with% (w/v) of CaCO, is 2. The weight of the solvent present in the solution 450 g then, value of y is : [At. wt : Ca = 40;C = 12 ;O = 16]

Answer 1

Explanation:

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Answer 2

Answer:

The value of y will be 9 where, y is percentage of weight by volume concentration of the CaCO₃ solution.

Explanation:

The percentage w/v of solution of CaCO₃ is y %

Given, The molality of the 1Litre CaCO₃ solution = 2m

The weight of solvent = 450g = 0.45Kg

Molality $=\frac{moles\; of \;CaCO_{3}}{Weight\; of \; solvent (Kg)}$

Moles of CaCO₃ = Molality × weight of solvent

Moles of CaCO₃ = (2)×(0.45) =0.9moles

Molar mass of CaCO₃ = 100gmol⁻¹

Amount of calcium carbonate = (0.9)(100)= 90g

Calcium carbonate present in 1000ml solution = 90g

Amount of CaCO₃ present in 100ml of solution $=\frac{90}{1000}*100=9g$

Therefore, the value of y will be 9.