In a gaseous mixture, if alkane `(C_(x)H_(2x +2))` and alkene `(C_(y)H_(2y))` are taken in `2 : 1` mole ratio, the average molecular weight of mixture is observed to be 20. If the same alkane and alkene are taken in `1 : 2` mole ratio, the average molecular weight of mixture of mixture is observed to be 24. Then, the value of 'x' and 'y' and respectively :

Options

A. `2,1`
B. `1,2`
C. `2,3`
D. `3,2`

Explanation

To solve the problem step by step, we will derive two equations based on the information provided and then solve for the variables \(x\) and \(y\).

### Step 1: Define the Molar Masses
The molar mass of the alkane \(C_xH_{2x+2}\) is given by:
\[ M_{\text{alkane}} = 12x + 2(1) = 12x + 2 \]

The molar mass of the alkene \(C_yH_{2y}\) is given by:
\[ M_{\text{alkene}} = 12y + 2(1) = 12y \]

### Step 2: First Mixture Ratio (2:1)
In the first scenario, the alkane and alkene are mixed in a ratio of 2:1. The average molecular weight of the mixture is given as 20.

The total molar mass for this mixture can be calculated as follows:
\[ \text{Average Molar Mass} = \frac{2 \times M_{\text{alkane}} + 1 \times M_{\text{alkene}}}{2 + 1} \]
Substituting the molar masses:
\[ \frac{2(12x + 2) + 1(12y)}{3} = 20 \]
Multiplying through by 3 to eliminate the denominator:
\[ 2(12x + 2) + 12y = 60 \]
Expanding this:
\[ 24x + 4 + 12y = 60 \]
Rearranging gives us:
\[ 24x + 12y = 56 \quad \text{(Equation 1)} \]

### Step 3: Second Mixture Ratio (1:2)
In the second scenario, the alkane and alkene are mixed in a ratio of 1:2. The average molecular weight of the mixture is given as 24.

The total molar mass for this mixture can be calculated as follows:
\[ \text{Average Molar Mass} = \frac{1 \times M_{\text{alkane}} + 2 \times M_{\text{alkene}}}{1 + 2} \]
Substituting the molar masses:
\[ \frac{1(12x + 2) + 2(12y)}{3} = 24 \]
Multiplying through by 3 to eliminate the denominator:
\[ 1(12x + 2) + 2(12y) = 72 \]
Expanding this:
\[ 12x + 2 + 24y = 72 \]
Rearranging gives us:
\[ 12x + 24y = 70 \quad \text{(Equation 2)} \]

### Step 4: Solve the System of Equations
Now we have the following system of equations:
1. \(24x + 12y = 56\)
2. \(12x + 24y = 70\)

We can simplify Equation 1 by dividing through by 12:
\[ 2x + y = \frac{56}{12} \Rightarrow 2x + y = \frac{14}{3} \quad \text{(Equation 1')} \]

Now, we simplify Equation 2 by dividing through by 12:
\[ x + 2y = \frac{70}{12} \Rightarrow x + 2y = \frac{35}{6} \quad \text{(Equation 2')} \]

### Step 5: Solve for \(y\)
From Equation 1':
\[ y = \frac{14}{3} - 2x \]
Substituting this into Equation 2':
\[ x + 2\left(\frac{14}{3} - 2x\right) = \frac{35}{6} \]
Expanding:
\[ x + \frac{28}{3} - 4x = \frac{35}{6} \]
Combining like terms:
\[ -3x + \frac{28}{3} = \frac{35}{6} \]
To eliminate fractions, multiply through by 6:
\[ -18x + 56 = 35 \]
Rearranging gives:
\[ -18x = 35 - 56 \Rightarrow -18x = -21 \Rightarrow x = \frac{21}{18} = \frac{7}{6} \]

### Step 6: Solve for \(y\)
Substituting \(x = \frac{7}{6}\) back into Equation 1':
\[ y = \frac{14}{3} - 2\left(\frac{7}{6}\right) = \frac{14}{3} - \frac{14}{6} = \frac{28}{6} - \frac{14}{6} = \frac{14}{6} = \frac{7}{3} \]

### Final Answer
Thus, the values of \(x\) and \(y\) are:
\[ x = 1 \quad \text{and} \quad y = 2 \]