An isotonic solution of glucose with urea contains $18 g$ in $1 dm^{3}$ . Calculate the mass of urea in $500 mL$ solution.
[Molar mass of urea $= 60 g mol^{- 1}$ . Molar mass of glucose $= 180 g mol^{- 1}$ ]

Question

An isotonic solution of glucose with urea contains 18 g in 1 dm 3 . Calculate the mass of urea in 500 mL solution. [Molar mass of urea = 60 g mol − 1 . Molar mass of glucose = 180 g mol − 1 ]

TestHub · Accepted Answer

For isotonic solutions, the molar concentrations of the solutes are equal.

$C_{glucose} = C_{urea}$

The concentration of the glucose solution is given by:

$C_{glucose} = \frac{18}{180 \times 1} = 0.1 mol dm^{- 3}$

Let $W$ be the mass of urea in $500 mL$ ( $0.5 dm^{3}$ ) of the solution. The concentration of the urea solution is:

$C_{urea} = \frac{W}{60 \times 0.5}$

Equating the concentrations:

$\frac{W}{30} = 0.1$

$W = 3 g$

Answer: $3 g$ .

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