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

Isotonic solutions have the same osmotic pressure. The osmotic pressure ($\pi$) is given by:
\[ \pi = CRT \]
where C is the molar concentration, R is the gas constant, and T is the temperature.
For isotonic solutions, $\pi_1 = \pi_2$, so:
\[ C_1RT = C_2RT \]
Since R and T are the same for both solutions, we have $C_1 = C_2$.
First, let's calculate 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

Now, we know the molar concentration of the glucose solution must also be 0.25 M. We can use this to calculate the mass of glucose needed: 

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