Question

A sugar syrup of weight 214.2g contains 34.2g of sugar (C12H22O11). Calculate (i) Molal concentration and (ii) Mole fraction of sugar in the syrup.

Answer 1

hope it will help you.

Answer 2

Answer :

(i) The molal concentration is, 0.55 mole/Kg

(ii) Mole fraction of sugar in the syrup is, $9.90\times 10^{-3}$

• Solution for part (i) :

Molality : It is defined as the number of moles of solute present in one kilogram of solvent.

Solute = sugar

Solution = sugar syrup

Solvent = water

First we have to calculate the mass of solvent.

Mass of solvent = Mass of sugar syrup - Mass of sugar = 214.2 - 34.2 = 180 g

Now we have to calculate the molality.

$Molality=\frac{\text{Moles of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{34.2g\times 1000}{342g/mole\times 180g}=0.55mole/Kg$

Therefore, the molal concentration is, 0.55 mole/Kg

• Solution for part (ii) :

Now we have to calculate the moles of solute and solvent.

$\text{Moles of sugar}=\frac{\text{ given mass of sugar}}{\text{ molar mass of sugar}}= \frac{34.2g}{342g/mole}=0.1moles$

$\text{Moles of water}=\frac{\text{ given mass of water}}{\text{ molar mass of water}}= \frac{180g}{10g/mole}=10moles$

Now we have to calculate the mole fraction of sugar in the syrup.

$\text{Mole fraction of sugar}=\frac{\text{Moles of sugar}}{\text{Moles of sugar}+\text{Moles of water}}$

$\text{Mole fraction of sugar}=\frac{0.1}{.01+10}=9.90\times 10^{-3}$

Therefore, the mole fraction of sugar in syrup is, $9.90\times 10^{-3}$