Question

A sugar of weight 214.2 gram contain 34.2 kg of sugar in it find molality and mole fraction of sugar in the syrup. Find answers

Answer 1

(i) Molal concentration = Molality = Moles of solute/Mass of solvent in Kg

Weight of sugar syrup = 214.2 g

Weight of sugar in the syrup = 34.2 g

∴ Weight of water in the syrup = 214.2 – 34.2 = 180.0 g
Mol. wt. of sugar, C12H22O11 = 342
∴Molal concentration = 34.2 *1000/342 *180 = 0.56



(ii) Mole fraction of sugar = Moles of sugar/Total moles in solution

Mol. wt. water, H2O = 18

∴ Mole fraction of sugar = 34.2/342 /180/18+34.2/342
= 0.1/10+0.1 = 0.1/10.1
= 0.0099
Hope it helps you...
Please do comment if there are any mistakes or problems in the steps..

Answer 2

Correct Question :

$\implies$ A Sugar syrup of weight 214.2 gram contain 34.2 gram of sugar in it find (i) Molality.

(¡¡) Mole fraction

${\bold{\underline{\underline{Solution:}}}}$

$\implies$ Weight of sugar syrup = 214.2 g

$\implies$ weight of water in syrup = 214.2 - 34.2 = 180.0 g

${\bold{\underline{ we\: know\:that :}}}}$

$\implies$ Molar mass = 342

$\implies$ Now, Moles of sugar = $\sf\cancel\dfrac{34.2}{342}=0.1$

$\implies$ Molality = $\frac{0.1}{180} \times 1000 = <strong>0</strong><strong>.</strong><strong>5</strong><strong>6</strong> g<strong>.</strong><strong> </strong>$

$\implies$ (ii) Molal Concentration

$\implies$ Moles of sugar = $\sf\cancel\dfrac{34.2}{342}=0.1$

Moles of water = $\sf\cancel\dfrac{180}{18}= 10$

$\implies$ Now, mole fraction of sugar

= $\frac{0.1}{10 + 0.1} = <strong>0</strong><strong>.</strong><strong>0</strong><strong>0</strong><strong>9</strong><strong>9</strong><strong>.</strong>$