Question

A 0.630 m aqueous solution of kbr has a total mass of 86.0 g. what masses of solute and solvent are present?

Answer 1

Solution :

Molality of the solution :  0.630 M

Mass of the KBr (solute) = x grams

Mass of the solvents : 86.0 - x grams

$Molality=\frac{\text{Mass of the solute}}{\text{Molecular mass of the solute}\times \text{Mass of solution in Kg}}$

$0.630=\frac{x}{119 g/mol\times (86.0-x)Kg}$

$x= 6.001grams$

Mass of the solute = 6.001 grams

Mass of the solvent = 86.0 - x grams = 86.0 - 6.001 g = 79.999 grams

Answer 2

The mass of solute in 0.63 m KBr solution is 6.96 grams. The mass of the solvent has been 79.03 grams.

Molality can be defined as the mass of solute present in 1 kg of solvent.

Given, molality = 0.63 m

The total mass of solution = 86 g

Let the mass of solute = x

Mass of solvent = 86 - x

Molality = $\rm \dfrac{mass\;of\;solute}{molecular\;weight\;of\;solute}\;\times\;\dfrac{1000}{Mass\;of\;solvent\;(g)}$

0.63 = $\rm \dfrac{x}{119\;g/mol}\;\times\;\dfrac{1000}{86-x}$

x = 6.96 grams

Thus the mass of solute present in the solution is 6.96 grams.

Mass of solvent = 86 - the mass of solute

Mass of solvent = 86 - 6.96 grams

Mass of solvent = 79.03 grams

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