To compare the properties of the given solutions, calculate their molarity and use this information to determine the osmotic pressure, which is given by the formula:

$$ \Pi = iCRT $$

where $ \Pi $ is the osmotic pressure, $ i $ is the van't Hoff factor (1 for non-electrolytes like urea and sucrose), $ C $ is the molarity, $ R $ is the ideal gas constant, and $ T $ is the temperature in Kelvin.

Calculations

Molarity of Urea Solution:

Given mass of urea $ = 6 \, \text{g} \text{ L}^{-1} $.

Molar mass of urea $ = 60 \, \text{g} \text{ mol}^{-1} $.

Molarity ($ C_{\text{urea}} $):

$$ C_{\text{urea}} = \frac{6 \, \text{g L}^{-1}}{60 \, \text{g mol}^{-1}} = 0.1 \, \text{mol L}^{-1} $$

Molarity of Sucrose Solution:

Given mass of sucrose $ = 17.12 \, \text{g} \text{ L}^{-1} $.

Molar mass of sucrose $ = 342 \, \text{g} \text{ mol}^{-1} $.

Molarity ($ C_{\text{sucrose}} $):

$$ C_{\text{sucrose}} = \frac{17.12 \, \text{g L}^{-1}}{342 \, \text{g mol}^{-1}} \approx 0.05 \, \text{mol L}^{-1} $$

Comparison of Osmotic Pressures

Since both solutions are at the same temperature and are non-electrolytes, the osmotic pressure directly depends on molarity.

Urea solution has a molarity of $ 0.1 \, \text{mol L}^{-1} $.

Sucrose solution has a molarity of $ 0.05 \, \text{mol L}^{-1} $.

Therefore, the osmotic pressure of the urea solution is higher than that of the sucrose solution.

Conclusion

Option A is incorrect because the urea solution has a higher osmotic pressure than the sucrose solution.

Option B is correct because urea solution, being more concentrated, is hypertonic to the sucrose solution.

Option C is incorrect because the solutions have different osmotic pressures.

Option D is incorrect because doubling the concentration of the sucrose solution to $ 0.1 \, \text{mol L}^{-1} $ would make it isotonic to the original urea solution, not hypertonic.

Therefore, the correct answer is Option B: Urea solution is hypertonic to sucrose solution.

Liquid Solution

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