A gaseous mixture contains three gases: Carbon monoxide, Carbon dioxide, and another unknown gas 'X'. If the percent by mass of CO2 is 44% in the mixture, and that of CO is 28% in... A gaseous mixture contains three gases: Carbon monoxide, Carbon dioxide, and another unknown gas 'X'. If the percent by mass of CO2 is 44% in the mixture, and that of CO is 28% in the mixture, find out the vapor density of the mixture. [Molar mass of X = 14 g mole^-1] Read more

Steps to Solve

1. Calculate the mass percentage of gas X

   Given:

   • Mass percentage of CO2 = 44%
   • Mass percentage of CO = 28%
   • Therefore, mass percentage of gas X = 100% - (44% + 28%) = 28%

2. Determine molar masses of the gases

   Molar masses:

   • CO2 = 44 g/mol
   • CO = 28 g/mol
   • Gas X = 14 g/mol

3. Calculate the average molar mass of the mixture

   Using the formula for the average molar mass of the mixture:

   $$ M_{mix} = \frac{(mass_{CO2} \cdot M_{CO2}) + (mass_{CO} \cdot M_{CO}) + (mass_{X} \cdot M_{X})}{100} $$

   Substitute the values:

   $$ M_{mix} = \frac{(44 \cdot 44) + (28 \cdot 28) + (28 \cdot 14)}{100} $$

4. Calculate each component in the equation

   Now calculate:

   • $44 \cdot 44 = 1936$
   • $28 \cdot 28 = 784$
   • $28 \cdot 14 = 392$

   Add them together: $$ 1936 + 784 + 392 = 3112 $$

5. Find the average molar mass

   Divide by 100: $$ M_{mix} = \frac{3112}{100} = 31.12, g/mol $$

6. Calculate the vapor density of the mixture

   The vapor density (VD) is given by the formula:

   $$ VD = \frac{M_{mix}}{R} $$

   Using the ideal gas constant $R = 0.0821, L \cdot atm/(K \cdot mol)$, we find:

   Assuming standard conditions (R = 1 g/L), we approximate: $$ VD \approx 31.12 \text{ g/L (often simplified in exams)} $$

7. Final adjustments based on the choices

   Convert to the closest value from the options given.