A mixture of carbon monoxide and carbon dioxide is found to have a density of 1.7g/lit at S.T.P. The mole fraction of carbon monoxide is

Options

A. 0.37
B. `0.40`
C. `0.30`
D. `0.50`

Explanation

To solve the problem of finding the mole fraction of carbon monoxide (CO) in a mixture of carbon monoxide and carbon dioxide (CO2) with a given density of 1.7 g/L at standard temperature and pressure (STP), we can follow these steps:

### Step 1: Calculate the Average Molar Mass
Using the formula for average molar mass at STP:
\[ \text{Average Molar Mass} = \frac{\rho \cdot RT}{P} \]
Where:
- \(\rho\) = density of the gas mixture = 1.7 g/L
- \(R\) = ideal gas constant = 0.0821 L·atm/(K·mol)
- \(T\) = temperature at STP = 273 K
- \(P\) = pressure at STP = 1 atm

Substituting the values:
\[ \text{Average Molar Mass} = \frac{1.7 \cdot 0.0821 \cdot 273}{1} \]
Calculating this gives:
\[ \text{Average Molar Mass} \approx 38.1 \text{ g/mol} \]

### Step 2: Set Up the Mole Fraction Equation
Let \(X\) be the mole fraction of CO. Then, the mole fraction of CO2 will be \(1 - X\).

The molar masses are:
- Molar mass of CO = 28 g/mol
- Molar mass of CO2 = 44 g/mol

The formula for average molar mass in terms of mole fractions is:
\[ \text{Average Molar Mass} = (X \cdot 28) + ((1 - X) \cdot 44) \]

### Step 3: Substitute and Solve for \(X\)
Setting the average molar mass equal to 38.1 g/mol:
\[ 38.1 = 28X + 44(1 - X) \]

Expanding and simplifying:
\[ 38.1 = 28X + 44 - 44X \]
\[ 38.1 = 44 - 16X \]
\[ 16X = 44 - 38.1 \]
\[ 16X = 5.9 \]
\[ X = \frac{5.9}{16} \approx 0.36875 \]

### Step 4: Final Calculation of Mole Fraction
Rounding the value:
\[ X \approx 0.37 \]

Thus, the mole fraction of carbon monoxide (CO) in the mixture is approximately **0.37**.

### Conclusion
The mole fraction of carbon monoxide (CO) in the mixture is **0.37**.

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