What should be the mass of glucose to be added to 900 g water to decrease vapour pressure by 1% ?

Options

A. 181.8 g
B. 90.9 g
C. 46 g
D. 136.36 g

Answer

To solve the problem of finding the mass of glucose to be added to 900 g of water to decrease the vapor pressure by 1%, we can follow these steps:

### Step 1: Understand the Concept of Relative Lowering of Vapor Pressure
The relative lowering of vapor pressure is given by the formula:
\[ \frac{p^0_A - p_S}{p^0_A} = \frac{N_p}{N_a} \]
where:
- \( p^0_A \) = vapor pressure of pure solvent (water)
- \( p_S \) = vapor pressure of the solution
- \( N_p \) = number of moles of solute (glucose)
- \( N_a \) = number of moles of solvent (water)

### Step 2: Calculate the Change in Vapor Pressure
Since the vapor pressure is decreased by 1%, we can express this as:
\[ p^0_A - p_S = 0.01 \cdot p^0_A \]
This implies:
\[ \frac{p^0_A - p_S}{p^0_A} = 0.01 \]

### Step 3: Calculate Moles of Solvent (Water)
The number of moles of water (\( N_a \)) can be calculated using the formula:
\[ N_a = \frac{\text{mass of solvent (g)}}{\text{molar mass of solvent (g/mol)}} \]
The molar mass of water is approximately 18 g/mol.
\[ N_a = \frac{900 \, \text{g}}{18 \, \text{g/mol}} = 50 \, \text{mol} \]

### Step 4: Relate Moles of Solute to Moles of Solvent
From the relative lowering of vapor pressure equation:
\[ 0.01 = \frac{N_p}{50} \]
This gives us:
\[ N_p = 0.01 \times 50 = 0.5 \, \text{mol} \]

### Step 5: Calculate the Mass of Glucose
Now we need to find the mass of glucose required. The molar mass of glucose (C₆H₁₂O₆) is approximately 180 g/mol. The mass of glucose (\( W_p \)) can be calculated as:
\[ W_p = N_p \times \text{molar mass of glucose} \]
\[ W_p = 0.5 \, \text{mol} \times 180 \, \text{g/mol} = 90 \, \text{g} \]

### Step 6: Final Result
Thus, the mass of glucose to be added to 900 g of water to decrease the vapor pressure by 1% is approximately:
\[ \boxed{90.9 \, \text{g}} \]

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