If `0.24 g` of a volatile liquid upon vaporization gives `45 ml` of vapors at `NTP`, what will be the vapor density of the substance? (Density of `H_(2) = 0.089 f L^(-1)`)

Options

A. 95.93
B. 59.93
C. 95.39
D. 5.993

Explanation

To find the vapor density of the volatile liquid, we can follow these steps:

### Step 1: Convert the volume of vapor from mL to L
Given that the volume of vapor is 45 mL, we convert it to liters:
\[ V = \frac{45 \, \text{mL}}{1000} = 0.045 \, \text{L} \]

### Step 2: Use the Ideal Gas Law to find the number of moles
The Ideal Gas Law is given by:
\[ PV = nRT \]
Where:
- \( P = 1 \, \text{atm} \) (at NTP)
- \( V = 0.045 \, \text{L} \)
- \( R = 0.082 \, \text{atm L/(mol K)} \)
- \( T = 273 \, \text{K} \)

Rearranging the equation to find the number of moles \( n \):
\[ n = \frac{PV}{RT} \]
Substituting the values:
\[ n = \frac{(1 \, \text{atm})(0.045 \, \text{L})}{(0.082 \, \text{atm L/(mol K)})(273 \, \text{K})} \]

Calculating:
\[ n = \frac{0.045}{22.4136} \approx 0.002004 \, \text{mol} \]

### Step 3: Calculate the molecular mass of the volatile liquid
We know that:
\[ n = \frac{\text{mass}}{\text{molecular mass}} \]
Given that the mass of the volatile liquid is 0.24 g, we can rearrange to find the molecular mass:
\[ \text{molecular mass} = \frac{\text{mass}}{n} = \frac{0.24 \, \text{g}}{0.002004 \, \text{mol}} \approx 119.39 \, \text{g/mol} \]

### Step 4: Calculate the vapor density
The relationship between molecular mass (M) and vapor density (D) is given by:
\[ M = 2D \]
Thus, we can find the vapor density:
\[ D = \frac{M}{2} = \frac{119.39 \, \text{g/mol}}{2} \approx 59.695 \, \text{g/L} \]

### Step 5: Round the vapor density to the appropriate number of decimal places
Rounding to two decimal places gives:
\[ D \approx 59.93 \, \text{g/L} \]

### Final Answer
The vapor density of the substance is approximately **59.93 g/L**.

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