Question

He mole percent of urea in 0.5 m aqueous solution of urea is

Answer 1

Answer:

Explanation:

Mole % in Urea = 0.5m ( Given)

Molarity is defined as the number of  moles of a solute dissolved in 1 kg of the solvent.  

Molarity (m) = Number of moles of solute / Mass of solvent in kg

The mole fraction of a component in a solution  is  defines as the total  ratio of number of moles of the component to the total number of moles of all the components present in the solution.

Thus, for a binary solution of A and B,

Mole fraction of A (XA) = (nA/nA+nB)

Mole fraction of B (XB) = (nB/nA+nB)

The total mole fraction of all the components of a solution = 1, thus  XA + XB = 1

XA/XB = nA/nB = (because moles of solute/moles of solvent) = (wA mB)/(wBmA)

1000XA/(XBmB) = (1000wA)/(wBmA) = m

= m = 1000XA/ mB(1 - XA)

For urea, mB = 60 g/mol and  XA = 0.5

Therefore, molarity (m) = 100/6 mol/kg.

Mole percent of urea -

= 0.5/56 × 100

= 0.89 %

Answer 2

molality of aqueous solution of urea is 0.5m

it means, 0.5 mole of urea present in 1 kg of water (solvent).

mass of water = 1kg or 1000g

molar mass of water = 18g/mol

so, number of mole of water = 1000/18 = 55.5 mol

so, no of mole of solution = no of mole of urea + no of mole of water

= 0.5 mol + 55.5 mol

= 56 mol

mole- percent of urea = no of mole of urea in solution/(no of mole of solution ) × 100

= 0.5/56 × 100

= 50/56

≈ 0.89 %

hence, 0.89 % of urea present in 0.5m aqueous solution of urea.