Question

Molarity of an aqueous solution of urea having mole fraction 0.2 is

Answer 1

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: 14 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n= moles of solute

$V_s$ = volume of solution in ml

Given : Mole fraction of urea is = 0.2, i.e. 0.2 moles of urea are present

Moles of solute (urea) = 0.2

moles of solvent (water) = 0.8

mass of solvent =$moles\times {\text {Molar mass}}=0.8\times 18=14.4g$

Density of solvent (water) = 1g/ml

$Volume=\frac{mass}{density}=\frac{14.4g}{1g/ml}=14.4ml$

$Molarity =\frac{0.2\times 1000}{14.4}=14M$

Thus Molarity of an aqueous solution of urea having mole fraction 0.2 is 14.