Question

A 4% solution of non volatile solute is isotonic with 0.702% urea solution calculate molar mass of nin volatile solute

Answer 1

Answer:341.88 g/mol

Explanation:Isotonic solution means the solution have same osmotic pressure.

Osmotic pressure, π =nRTV

0.702 % urea solution means 0.702 g urea is dissolved in 100 mL.

mass of urea = 0.702 g

Molar mass of urea = 60 g mol-1

Number of moles = 0.702 / 60

Osmotic pressure of urea solution, π = 0.702×0.082×T60×0.1

For a non-volatile solution

4% solution means 4 g non-volatile solute is dissolved in 100 mL or 0.1 L.

Mass of Non-volatile solute = 4 g

Volume of solution = 0.1 L

Number of moles = 4 / M

where M is molar mass of non-volatile solute.

Osmotic pressure of non-volatile solution,π =4×0.082×TM×0.1

Equating the osmotic pressures

0.702×0.082×T60×0.1=4×0.082×TM×0.1M = 4×600.702M = 341.88 g mol−1

Answer 2

Answer:341.8

Explanation: