510 milli g of a liquid on vapourisation in Victor Mayer 's appartus displaces 510 c c. of dry air at NTP. The molecular weight of liquid is -

Options

A. 0.224
B. 2.24
C. 224
D. 22.4

Answer

To find the molecular weight of the liquid based on the given data, we can follow these steps:

### Step 1: Understand the given data
We have:
- Mass of the liquid = 510 mg
- Volume of dry air displaced = 510 cc (cubic centimeters)
- Conditions are at NTP (Normal Temperature and Pressure)

### Step 2: Convert mass from milligrams to grams
Since 1 mg = 10^-3 g, we convert the mass:
\[ \text{Mass of the liquid} = 510 \, \text{mg} = 510 \times 10^{-3} \, \text{g} = 0.510 \, \text{g} \]

### Step 3: Convert the volume from cc to liters
Since 1 cc = 1 mL and 1 L = 1000 mL, we convert the volume:
\[ \text{Volume of air displaced} = 510 \, \text{cc} = 510 \, \text{mL} = 510 \times 10^{-3} \, \text{L} = 0.510 \, \text{L} \]

### Step 4: Use the ideal gas law to find the number of moles of air displaced
At NTP, 1 mole of gas occupies 22.4 L. Therefore, we can calculate the number of moles (n) of air displaced:
\[ n = \frac{\text{Volume}}{\text{Molar Volume at NTP}} = \frac{0.510 \, \text{L}}{22.4 \, \text{L/mol}} \]
Calculating this gives:
\[ n = \frac{0.510}{22.4} \approx 0.0227 \, \text{mol} \]

### Step 5: Relate the number of moles to the mass and molecular weight
The number of moles (n) is also given by the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} \implies \text{molar mass} = \frac{\text{mass}}{n} \]
Substituting the values we have:
\[ \text{molar mass} = \frac{0.510 \, \text{g}}{0.0227 \, \text{mol}} \approx 22.4 \, \text{g/mol} \]

### Conclusion
The molecular weight of the liquid is approximately 22.4 g/mol.