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Current Question (ID: 8142)
Question:
$\text{A mixture of CO and CO}_2 \text{ is found to have a density } 1.5\text{ g/litre at } 30^{\circ}\text{C and } 730\text{ mm. The composition of the mixture is-}$
Options:
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1. $\text{CO: } 44\%; \text{ CO}_2 = 56\%$
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2. $\text{CO: } 32\%; \text{ CO}_2 = 68\%$
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3. $\text{CO: } 70\%; \text{ CO}_2 = 30\%$
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4. $\text{CO: } 9\%; \text{ CO}_2 = 91\%$
Solution:
$\text{Hint: } \text{P} = \frac{\text{w}}{\text{Vm}} \text{RT} \text{ (This is a simplified version of Ideal Gas Law, it should be } \text{P} = \frac{\text{m}}{\text{M V}} \text{RT or } \text{P} = \text{d} \frac{\text{RT}}{\text{M}} \text{ where M is molar mass).}
\text{Step 1:}
\text{The given values are as follows:}
\text{For a mixture of CO and CO}_2\text{, d} = 1.50\text{ g/litre}
\text{P} = \frac{730}{760}\text{ atm}
\text{T} = 30^{\circ}\text{C} + 273 = 303\text{ K}
\text{Calculate the molecular weight of gaseous mixture is as follows:}
\text{Using the Ideal Gas Law: } \text{PV} = \text{nRT}
\text{Since } \text{n} = \frac{\text{mass (w)}}{\text{molar mass (M)}}
\text{PV} = \frac{\text{w}}{\text{M}}\text{RT}
\text{P} = \frac{\text{w}}{\text{V}}\frac{\text{RT}}{\text{M}}
\text{Since density } \text{d} = \frac{\text{w}}{\text{V}}
\text{P} = \text{d} \frac{\text{RT}}{\text{M}}
\text{Rearranging for M: } \text{M} = \frac{\text{dRT}}{\text{P}}
\text{M} = \frac{1.5 \times 0.0821 \times 303}{730/760}
\text{M} = \frac{1.5 \times 0.0821 \times 303}{0.9605}
\text{M} = 38.85
\text{i.e. molecular weight of mixture of CO and CO}_2 = 38.85
\text{Step 2:}
\text{Let } \text{a}% \text{ of mole of CO be in mixture then}
\text{Average molecular weight} = \frac{\text{a} \times \text{M}_{\text{CO}} + (100 - \text{a}) \times \text{M}_{\text{CO}_2}}{100}
\text{Molecular weight of CO } (\text{M}_{\text{CO}}) = 12 + 16 = 28
\text{Molecular weight of CO}_2 \text{ (M}_{\text{CO}_2}) = 12 + 2 \times 16 = 44
38.85 = \frac{\text{a} \times 28 + (100 - \text{a}) \times 44}{100}
38.85 \times 100 = 28\text{a} + 4400 - 44\text{a}
3885 = 4400 - 16\text{a}
16\text{a} = 4400 - 3885
16\text{a} = 515
\text{a} = \frac{515}{16}
\text{a} = 32.19
\text{Mole % of CO} = 32.19\%
\text{Mole % of CO}_2 = 100 - 32.19 = 67.81\%
\text{Therefore, the composition is approximately CO: } 32\%; \text{ CO}_2 = 68\%$
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