How much urea must be dissolved in `10^(-2)m^(3)` water to yield a pressure is `2.03xx10^(5)Nm^(-2)` at 300K?

Options

A. 48.30 g
B. 60.0 g
C. 24.15 g
D. 48.03 g

Explanation

To find out how much urea must be dissolved in \(10^{-2} \, m^3\) of water to yield a pressure of \(2.03 \times 10^5 \, N/m^2\) at 300 K, we will use the formula for osmotic pressure:

\[ \pi = i \cdot C \cdot R \cdot T \]

Where:
- \(\pi\) = osmotic pressure
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