Question

A 6.90 M solution of KOH in water solution contains 30% by mass of KOH. Calculate the density of the solution.
(A) $1.288g\text{ m}{{\text{L}}^{-1}}$
(B) $1.578g\text{ m}{{\text{L}}^{-1}}$
(C) $1.008g\text{ m}{{\text{L}}^{-1}}$
(D) $1.608g\text{ m}{{\text{L}}^{-1}}$

Answer 1

Hint: density can be calculated by taking the ratio of Mass and density. When one mole of solute is dissolved in 1 litre of solution, 1 Molar solution is obtained. 30% by mass of KOH indicates 30g of KOH is dissolved in 100g of solution.

Complete step by step solution:
-When one mole of solute is dissolved in 1 Liter of the solution, one Molar of the solution is obtained.
So one Molar solution of KOH is obtained, one mole of KOH is dissolved in 1 litre of solution, so 6.9M solution will have 6.9 moles of KOH present in 1 litre of solution.
-One mole of KOH is equal to the Molar mass of KOH. The molar mass of KOH is equal to 56g
So, As one mole of KOH is equal to 56g of KOH, so 6.9 moles of KOH will be 386.4g.
30% solution indicates 30g of KOH is present in 100g of solution.
-AS 30g of KOH is present in $100g$ of the solution, so $386.4g$ of $KOH$ will be present in 1288g of solution.
The density of the KOH solution can be calculated by dividing mass to volume.
So,
Density of KOH=\[\dfrac{Mass}{Volume}=\dfrac{1288}{1000}=1.288g/c{{m}^{3}}\]

The density of the solution (A) $1.288g\text{ m}{{\text{L}}^{-1}}$.

Note: 30% by mass of KOH indicates 30g of KOH is dissolved in 100g of solution. Density can be also calculated using this method. In 100g of the solution, 30g of KOH is present. So $\dfrac{100}{d}mL$ solution will contain $\dfrac{30}{56}$ mol KOH. So, density can be calculated as
Molarity=6.90 = $\dfrac{30}{56}\times \dfrac{d}{100}\times 1000$