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

Options

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

Explanation

To solve the problem of determining 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 can follow these steps:

### Step 1: Understand the relationship between vapor pressure lowering and mole fraction
The lowering of vapor pressure is directly related to the mole fraction of the solute in the solution. For two solutions with the same lowering of vapor pressure, we can set their mole fractions equal to each other.

### Step 2: Define the mole fractions
Let:
- Solution 1 (glucose):
- Moles of glucose = \( n_1 \)
- Moles of water = \( n_{w1} \)

- Solution 2 (urea):
- Moles of urea = \( n_2 \)
- Moles of water = \( n_{w2} \)

The mole fractions can be expressed as:
\[ \frac{n_1}{n_1 + n_{w1}} = \frac{n_2}{n_2 + n_{w2}} \]

### Step 3: Calculate moles of solute and solvent
Using the formula for moles:
\[ \text{Moles} = \frac{\text{mass}}{\text{molar mass}} \]

For glucose (C6H12O6):
- Molar mass = 180 g/mol

For urea (NH2CONH2):
- Molar mass = 60 g/mol

For water (H2O):
- Molar mass = 18 g/mol

### Step 4: Substitute known values into the equation
1. For urea:
- Mass of urea = 1 g
- Mass of water = 50 g
- Moles of urea \( n_2 = \frac{1}{60} \)
- Moles of water \( n_{w2} = \frac{50}{18} \)

2. For glucose:
- Let the mass of glucose be \( W \) grams
- Moles of glucose \( n_1 = \frac{W}{180} \)
- Mass of water = 100 g
- Moles of water \( n_{w1} = \frac{100}{18} \)

### Step 5: Set up the equation
Using the equality of mole fractions:
\[ \frac{\frac{W}{180}}{\frac{W}{180} + \frac{100}{18}} = \frac{\frac{1}{60}}{\frac{1}{60} + \frac{50}{18}} \]

### Step 6: Simplify the equation
Cross-multiplying gives:
\[ \frac{W}{180} \cdot \left(\frac{1}{60} + \frac{50}{18}\right) = \frac{1}{60} \cdot \left(\frac{W}{180} + \frac{100}{18}\right) \]

### Step 7: Calculate the right-hand side
First, calculate \( \frac{1}{60} + \frac{50}{18} \):
- Convert \( \frac{50}{18} \) to a common denominator:
\[ \frac{50}{18} = \frac{50 \times 10}{18 \times 10} = \frac{500}{180} \]
So,
\[ \frac{1}{60} = \frac{3}{180} \]
Thus,
\[ \frac{1}{60} + \frac{50}{18} = \frac{3 + 500}{180} = \frac{503}{180} \]

### Step 8: Substitute back into the equation
Now we can substitute back into the equation:
\[ \frac{W}{180} \cdot \frac{503}{180} = \frac{1}{60} \cdot \left(\frac{W}{180} + \frac{100}{18}\right) \]

### Step 9: Solve for W
After simplification and solving for \( W \):
\[ W = 6 \text{ grams} \]

### Final Answer
The weight of glucose that must be dissolved in 100 grams of water to produce the same lowering of vapor pressure as 1 gram of urea dissolved in 50 grams of water is **6 grams**.

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