A solution is prepared by dissolving 0.6 g of urea (molar mass = 60 g mol$$^{-1}$$) and 1.8 g of glucose (molar mass = 180 g mol$$^{-1}$$) in 100 mL of water at 27°C. The osmotic pressure of the solution is:
(R = 0.08206 L atm K$$^{-1}$$ mol$$^{-1}$$)

We first recall the van ’t Hoff relation for osmotic pressure. For a dilute nonelectrolyte solution the relation is stated as

$$\Pi \;=\; C \, R \, T,$$

where $$\Pi$$ is the osmotic pressure, $$C$$ is the molar concentration of the total solute (in mol L$$^{-1}$$), $$R$$ is the gas constant and $$T$$ is the absolute temperature.

We have two nonelectrolyte solutes, urea and glucose. The number of moles of each solute is obtained from

$$n = \frac{m\;(\text{in g})}{M\;(\text{in g mol}^{-1})}.$$

For urea:

$$n_{\text{urea}} = \frac{0.6}{60} = 0.010\;\text{mol}.$$

For glucose:

$$n_{\text{glucose}} = \frac{1.8}{180} = 0.010\;\text{mol}.$$

The total number of moles of solute present is therefore

$$n_{\text{total}} = n_{\text{urea}} + n_{\text{glucose}} = 0.010 + 0.010 = 0.020\;\text{mol}.$$

The solution is prepared in 100 mL of water. Converting this volume to litres gives

$$V = 100\;\text{mL} = 0.100\;\text{L}.$$

Now the molar concentration of the solute is

$$C = \frac{n_{\text{total}}}{V} = \frac{0.020\;\text{mol}}{0.100\;\text{L}} = 0.20\;\text{mol L}^{-1}.$$

The absolute temperature corresponding to 27 °C is

$$T = 27 + 273 = 300\;\text{K}.$$

Substituting the values of $$C$$, $$R$$, and $$T$$ in the van ’t Hoff equation, we get

$$\Pi = (0.20\;\text{mol L}^{-1})(0.08206\;\text{L atm K}^{-1}\text{ mol}^{-1})(300\;\text{K}).$$

Carrying out the multiplication step by step,

$$0.08206 \times 300 = 24.618,$$

and then

$$\Pi = 0.20 \times 24.618 = 4.9236\;\text{atm}.$$

Rounding to three significant figures,

$$\Pi \approx 4.92\;\text{atm}.$$

Hence, the correct answer is Option C.