What weight of glucose dissolved in `100g` of water will produce the same lowering of vapour pressure as one gram of urea dissolved in `50g` of water at the same temperature

Options

A. `3g`
B. `5g`
C. `6g`
D. `4g`

Explanation

To solve the problem, we need to find the weight of glucose that will produce the same lowering of vapor pressure as 1 gram of urea dissolved in 50 grams of water. We will use the concept of colligative properties, specifically the lowering of vapor pressure, which depends on the mole fraction of the solute.

### Step-by-step Solution:

1. **Calculate the moles of urea:**
- The molar mass of urea (NH₂CONH₂) is approximately 60 g/mol.
- Moles of urea = mass of urea / molar mass of urea
\[ \text{Moles of urea} = \frac{1 \text{ g}}{60 \text{ g/mol}} = \frac{1}{60} \text{ mol} \]

2. **Calculate the moles of water in the solution:**
- The molar mass of water (H₂O) is approximately 18 g/mol.
- Moles of water = mass of water / molar mass of water
\[ \text{Moles of water} = \frac{50 \text{ g}}{18 \text{ g/mol}} \approx 2.78 \text{ mol} \]

3. **Calculate the mole fraction of urea:**
- Mole fraction of urea (X_urea) is given by:
\[ X_{\text{urea}} = \frac{\text{moles of urea}}{\text{moles of urea} + \text{moles of water}} = \frac{\frac{1}{60}}{\frac{1}{60} + 2.78} \]
- Calculate the denominator:
\[ \frac{1}{60} + 2.78 \approx 2.78 + 0.01667 \approx 2.79667 \]
- Now calculate X_urea:
\[ X_{\text{urea}} \approx \frac{0.01667}{2.79667} \approx 0.00596 \]

4. **Set up the equation for glucose:**
- Let the weight of glucose be \( x \) grams. The molar mass of glucose (C₆H₁₂O₆) is approximately 180 g/mol.
- Moles of glucose = \( \frac{x}{180} \)
- The mass of water is 100 g, so the moles of water in this case are:
\[ \text{Moles of water} = \frac{100 \text{ g}}{18 \text{ g/mol}} \approx 5.56 \text{ mol} \]

5. **Calculate the mole fraction of glucose:**
- Mole fraction of glucose (X_glucose) is given by:
\[ X_{\text{glucose}} = \frac{\text{moles of glucose}}{\text{moles of glucose} + \text{moles of water}} = \frac{\frac{x}{180}}{\frac{x}{180} + 5.56} \]

6. **Equate the mole fractions:**
- Since the lowering of vapor pressure is the same for both solutions:
\[ X_{\text{urea}} = X_{\text{glucose}} \]
- Thus,
\[ \frac{0.01667}{2.79667} = \frac{\frac{x}{180}}{\frac{x}{180} + 5.56} \]

7. **Cross-multiply and solve for \( x \):**
\[ 0.01667 \left( \frac{x}{180} + 5.56 \right) = \frac{x}{180} \cdot 2.79667 \]
- Simplifying this equation leads to:
\[ 0.01667 \cdot 5.56 + 0.01667 \cdot \frac{x}{180} = \frac{2.79667x}{180} \]
- Rearranging gives:
\[ 0.0927 + 0.0000927x = 0.0155x \]
- Solving for \( x \) gives:
\[ 0.0927 = 0.0155x - 0.0000927x \]
- Thus,
\[ x \approx \frac{0.0927}{0.0155 - 0.0000927} \approx 5.97 \text{ g} \]

### Final Answer:
The weight of glucose that must be dissolved in 100 g of water to produce the same lowering of vapor pressure as 1 g of urea in 50 g of water is approximately **5.97 grams**.