Question

The teacher instructed three students A, B, and C to prepare a 50% (mass by volume) solution of sodium hydroxide $\left(\mathrm{NaOH}\right)$. Student A dissolved 50 g of $\mathrm{NaOH}$ in 100 mL of water, Student B dissolved 50 g of $\mathrm{NaOH}$ in 100 g of water and Student C dissolved 50 g of $\mathrm{NaOH}$ in water to make 100 mL solution. Which one of them has made the solution of desired concentration and why?

Answer 1

There are different ways present to represent the concentration of solution and one of them is mass by volume percentage

Mass by volume percentage

It is defined as the amount of solute in parts by mass present in 100 parts by volume of the solution

$\mathrm{Mass}\mathrm{by}\mathrm{volume}\%=(\mathrm{Mass}\mathrm{of}\mathrm{solute}(\mathrm{in}\mathrm{g})/\mathrm{Mass}\mathrm{of}\mathrm{solution}(\mathrm{in}\mathrm{L}\left)\right)\mathrm{x}100$

Here it is given that students want to prepare 50% mass by volume of sodium hydroxide solution

That means 50 g of $\mathrm{NaOH}$ dissolved in 100 mL of the solution

• Student A has taken 100 mL of water, that will not possible because water act as the solvent here and to prepare 50% sodium hydroxide, 100 mL of solution is only required. Here the solvent itself constitutes the volume of 100 mL.
• Student B has taken 100 g of water, i.e the mass not volume (mL) and that will not constitute mass by volume %.
• Student C has taken 50 g of $\mathrm{NaOH}$ in 100 mL solution which is the correct way to prepare 50% (mass by volume) of $\mathrm{NaOH}$ solution.

Therefore, Student C made the desired concentration of solution.