A solution of $\mathrm{x} \mathrm{moles}$ of sucrose in $100 \mathrm{grams}$ of water freezes at $–0.2º\mathrm{C}$. As ice separates, the freezing point goes down to $– 0.25º\mathrm{C}$. How many grams of ice would have separated?

What volume (in $\mathrm{mL}$) of $90\%$ alcohol by weight $\left(\mathrm{d}=0.8{\mathrm{gmL}}^{-1}\right)$ must be used to prepare $80\mathrm{mL}$  of  $10\%$  alcohol by weight $\left(\mathrm{d}=0.9{\mathrm{gmL}}^{-1}\right)$ ?

The concentrations of solutions can be expressed in number of ways; viz : mass fraction of solute (or mass percent), Molar concentration (Molarity) and Molal concentration (molality). These terms are known as concentration terms and also they are related with each other i.e., knowing one concentration term for the solution, we can find other concentration terms also. The definition of different concentration terms are given below:

Molarity : It is the number of moles of solute present in one litre of the solution.

Molality: It is the number of moles of solute present in one kg of the solvent.

Mole Fraction = $\frac{Moles of a solute}{Moles of solute\left(all\right) + Moles of solvent}$

If molality of the solution is given as a, then mole fraction of the solute can be calculated by

$Mole fraction =\frac{a}{a +{\displaystyle \frac{1000}{M_{solvent}}}}=\frac{a x M_{solvent}}{a x M_{solvent} +1000}$
where a = molality and M_solvent =  Molar mass of solvent.

We can change : Mole fraction $\leftrightarrow$ Molality $\leftrightarrow$ Molarity