Question

214.2 g of sugar syrup contains 34.2 g of sugar.

Calculate
(i) molarity of the solution
(ii) mole fraction of sugar in the syrup.

Answer 1

Analyzing the given data:

Given Mass of sugar $=34.2\mathrm{g}$

Molar Mass of sugar $=342\mathrm{g}{\mathrm{mol}}^{-1}$

No. of moles of sugar = $\frac{\mathrm{Given}\mathrm{mass}}{\mathrm{Molecular}\mathrm{mass}}=\frac{34.2}{342}=0.1\mathrm{mole}$

Sugar syrup contains sugar in water.

So, $\mathrm{mass}\mathrm{of}\mathrm{water}=(214.2-34.2)\mathrm{g}=180\mathrm{g}$

Molar Mass of Water $=18\mathrm{g}{\mathrm{mol}}^{-1}$
No. of moles of water = $\frac{\mathrm{Given}\mathrm{mass}}{\mathrm{Molecular}\mathrm{mass}}=\frac{180}{18}=10\mathrm{mole}$

Part (i): Molarity of solution

Molarity(M): The number of moles of a solute dissolved in 1 L (1000 ml) of the solution is known as the molarity of the solution.

$\mathrm{Molarity}\left(\mathrm{M}\right)=\frac{\mathrm{Mole}\mathrm{of}\mathrm{solute}}{\mathrm{Volume}\mathrm{of}\mathrm{solution}(\mathrm{in}\mathrm{litre})}$

$\mathrm{M}=\frac{0.1}{180}\times 1000=0.555\mathrm{M}$

Thus, the molarity of sugar is 0.555 M

Part (ii): Mole fraction of sugar

Total number of moles of solution = mole of solute + mole of solvent $=0.1+10=10.1$

$\mathrm{Mole}\mathrm{fraction}\mathrm{of}\mathrm{sugar}=\frac{\mathrm{Mole}\mathrm{of}\mathrm{sugar}}{\mathrm{Total}\mathrm{mole}\mathrm{of}\mathrm{solution}}=\frac{0.1}{10.1}=0.0099$

Thus, mole fraction of sugar in the syrup is 0.0099.