Question

Mass of glucose (\(C_6H_{12}O_6\)) required to be dissolved to prepare one litre of its solution which is isotonic with \(15 \, \text{g L}^{-1}\) solution of urea (\(NH_2CONH_2\)) is:
{Given:} Molar mass in \(\text{g mol}^{-1}\): \[ C : 12, \, H : 1, \, O : 16, \, N : 14 \]

• A. 55 g
• B. 15 g
• C. 30 g
• D. 45 g

Answer 1

The Correct Option is D

Solutions are isotonic when they possess identical osmotic pressure. Osmotic pressure ($\pi$) is quantified by the equation:
\[ \pi = CRT \]
where C denotes molar concentration, R represents the gas constant, and T signifies temperature.
For isotonic solutions, the equality $\pi_1 = \pi_2$ holds, leading to:
\[ C_1RT = C_2RT \]
As R and T are constant for both solutions, it follows that $C_1 = C_2$.
Initially, we determine the molar concentration of the urea solution:
Molar mass of urea (NH$_2$CONH$_2$) = (2 $\times$ 14) + (4 $\times$ 1) + 12 + 16 = 60 g/mol

Concentration of urea ($C_1$) = $\frac{mass}{molar mass \times volume} = \frac{15 g}{60 g/mol \times 1 L} = 0.25$ M

Consequently, the molar concentration of the glucose solution must also be 0.25 M. This value enables the calculation of the required glucose mass:
Molar mass of glucose (C$_6$H$_{12}$O$_6$) = (6 $\times$ 12) + (12 $\times$ 1) + (6 $\times$ 16) = 180 g/mol

Mass of glucose (m) = $C_2 \times molar mass \times volume = 0.25 M \times 180 g/mol \times 1 L = 45$ g