Question

Calculate the mass of KOH in a 35% solution that contains 58.5 g of water.

Answer 1

Answer:

Approximately $\rm 31.5\; g$.

Explanation:

The mass of a solution can be divided into two parts:

• the solute (the material that was dissolved,) and
• the solvent.

In this particular $\rm KOH$ solution in water,

• $\rm KOH$ is the solute, while
• water is the solvent.

The number $35\%$ here likely refers to the concentration of $\rm KOH$ in this solution. That's ratio between the mass of the solute ($\rm KOH$) and the mass of the whole solution (mass of solute plus mass of solvent.) That is:

$\displaystyle \frac{m(\text{KOH})}{m(\text{solution})} = 35\% = 0.35$.

Hence, $m(\mathrm{KOH}) = 0.35\, m(\text{solution})$.

However, since the solution contains only the solute and the solvent, $m(\text{solution}) = m(\text{solute}) + m(\text{solvent})$.

For this solution in particular,

$\begin{aligned}&m(\text{solution})\\&= m(\text{solute}) + m(\text{solvent}) \\ &= m(\text{KOH}) + m(\text{water})\end{aligned}$.

As a result,

$\begin{aligned}&m(\mathrm{KOH})\\ &= 0.35\, m(\text{solution}) \\&= 0.35\, (m(\mathrm{KOH}) + m(\text{water}))\\&= 0.35\, m(\mathrm{KOH}) + 0.35 \, m(\text{water})\end{aligned}$.

Subtract $0.35\, m(\mathrm{KOH})$ from both sides of the equation:

$(1 - 0.35)\, m(\mathrm{KOH}) = 0.35\, m(\text{water})$.

$\begin{aligned} &m(\mathrm{KOH}) \\ &= \left(\frac{0.35}{1 - 0.35}\right)\cdot m(\text{water}) \\ &= \frac{0.35}{0.65} \times 58.5\; \text{g} = 31.5 \; \text{g}\end{aligned}$.

Note, that for this calculation, there's nothing special about this $35\%$ solution of $\mathrm{KOH}$ in water. In general,

$\displaystyle m(\text{solute}) = \left(\frac{\%\text{concentration}}{100\% - \%\text{concentration}}\right)\cdot m(\text{solvent})$.