Question

A solution containing 15 g urea (molar mass = 60 g mol^–1) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol^–1) in water. Calculate the mass of glucose present in one litre of its solution.

Answer 1

Given: W_A (of urea) = 15 g

Molar mass of urea = 60 g mol⁻¹

Molar mass of glucose = 180 g mol⁻¹

To find: W_B (of glucose) = ?

Formula: `"Moles of urea" = "wt. of urea"/"molar mass"`

= `15/60`

= 0.25

For an isotonic solution,

π = CRT

(where, C is concentration, R is a constant, and T is temperature)

`π_"urea" = π_"glucose"`

`C_"urea" RT = C_"glucose" RT`   ...(RT is common)

0.25 = n/1L   ...(Taking C = n/V)

n glucose = 0.25

`"Moles of glucose" = "wt. of glucose"/"molar mass"`

0.25 = `W_B/180`

W_B = 0.25 × 180

= 45 g