Question

If $18\mathrm{g}$ glucose is present in $1000\mathrm{g}$ of solvent, the solution is said to be?

Answer 1

Step 1: Given data

Mass of glucose$=18\mathrm{g}$

Mass of solvent$=1000\mathrm{g}$

Step 2: Calculating molality

Molality$=\frac{\mathrm{moles}\mathrm{of}\mathrm{solute}}{\mathrm{Mass}\mathrm{of}\mathrm{solvent}\mathrm{in}\mathrm{Kg}}$

Moles of glucose$=\frac{\mathrm{given}\mathrm{mass}}{\mathrm{Molecular}\mathrm{mass}}\Rightarrow \frac{18}{180}\mathrm{mol}$

Molality$=\frac{18}{180\times 1}\Rightarrow 0.1$molal.

Also, the concentration of solvent is more than solute therefore, the given solution is hypotonic.

Therefore. the given solution is hypotonic with molality$=0.1$m