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: Molar mass of urea = 60 g mol⁻¹

Molar mass of glucose = 180 g mol⁻¹

Formula: The osmotic pressure formula for both solutions is:

`pi = n/(V)RT`

Both solutions are isotonic, so their osmotic pressures are the same. This means that the equation can be written as:

`n_("urea")/V_("urea") = n_("glucose")/V_("glucose")`

For 1 litre of each solution (where, `V_"urea" = V_"glucose" = 1 L`), this reduces to:

`n_"urea" = n_"glucose"`

The values of R and T don’t need to be there because the volume is the same for both options. Let’s figure out how many moles of urea there are:

`n_"urea" = ("mass of urea")/("molar mass of urea")`

`= 15/60`

= 0.25 mol

There are the same number of moles of glucose as there are moles of urea.

`n_"glucose" = 0.25  "mol"`

Calculating the mass of glucose:

Molar mass of glucose = 180 g mol⁻¹

`"Mass of glucose" = n_"glucose" xx "molar mass of glucose"`

= 0.25 × 180

= 45 g

Answer 2

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