Water — Question 6

Given,
Mass of solvent = 50 g

Solubility of NaNO₃ at 50°C = 114 g

Solubility of NaNO₃ at 30°C = 86 g

Solubility = $\dfrac{\text{Mass of solute}}{\text{Mass of solvent}}$ x 100

Rearranging the above equation

Mass of solute = $\dfrac{\text{Solubility x Mass of solvent}}{100}$

∴ Mass of solute at 50°C = $\dfrac{114}{100}$ x 50

= 57 g

∴ Mass of solute at 30°C = $\dfrac{86}{100}$ x 50

= 43 g

∴ Amount of NaNO₃ that separate = 57 g - 43 g = 14 g