Question

How much water should be added to $10\mathrm{g}$ of common salt to obtain $10\%$ salt solution?

Answer 1

Mass percentage of solute: It is defined as the mass of solute in grams present in $100\mathrm{g}$ of solution.

Step 1: Given:

Mass of solute$=10\mathrm{g}$

Step 2: Formula:

$Mass\%ofsolute=\frac{Massofsolute}{Massofsolution}\times 100$ , where, $(\mathrm{mass}\mathrm{of}\mathrm{solution}=\mathrm{mass}\mathrm{of}\mathrm{solute}+\mathrm{mass}\mathrm{of}\mathrm{solvent})$.

Step 3: Finding the amount of water to be added:

Let the mass of solvent be $\mathrm{x}\mathrm{g}$.

$\mathrm{Mass}\mathrm{of}\mathrm{solution}=\mathrm{Mass}\mathrm{of}\mathrm{solute}+\mathrm{Mass}\mathrm{of}\mathrm{solvent}$

Then, the mass of solution$=(10+\mathrm{x})\mathrm{g}$.

We have to find the mass of water so that the concentration of salt in the solution is $10\%$.

$Mass\%ofsolute=\frac{Massofsolute}{Massofsolution}\times 100$

$10=\frac{10\mathrm{g}}{(10+\mathrm{x})\mathrm{g}}\times 100$

$10+\mathrm{x}=100$

Adding $-10$ to both sides of the above equation.

$$\begin{gathered} \mathrm{x}=100-10 \\ \mathrm{x}=90\mathrm{g} \end{gathered}$$

Hence, the amount of water that should be added is $90\mathrm{g}$.