Question

The mole fraction of glucose in aqueous solution is 0.2, then molality of solution will be

Answer 1

Explanation:

Answer:

Given:

Mole fraction of glucose= 0.2

Molefraction of solution = 1

So, mole fraction of water= 1-0.2

= 0.8

Molality = moles of solute/ kg solvent

Molar mass H2O = 18 g/kg

Mass of H2O = (0.8×18)/1000

=0.014 kg

Molality = 0.2 / 0.014

= 14 M

Answer 2

Answer:

The correct answer is 14 M. The molality of the given solution is 14 M.

Explanation:

The given solution has a mole fraction of 0.2, to calculate the molality of the glucose in an aqueous solution, following methods are to be followed.

$Mole fraction of glucose= 0.2$

$Mole fraction of solution = 1$

To calculate the mole fraction of water, we need to subtract the mole fraction of solution from the given mole fraction of the glucose.

$So, mole fraction of water= 1-0.2$

$Mole fraction of water = 0.8$

$Molality = \frac{moles of solute}{kg solvent}$

$Molar mass H_{2}O = 18 g/kg$

$Mass H_{2}O = (0.8\times18)/1000 = 0.014 kg$

$Molality = 0.2 \div 0.014= 14 M$

Therefore, the molality of the solution is 14 M.