x gram of urea and y gram of glucose produce a freezing point depression of `0.186^(@)C` when both are dissolved in 100 ml water. If ration of `x:y=1:2` then which is correct ? [ Molal freezing point depression constant for water `=1.86^(@)C//m &` density of water `=1` gram`//ml]`

Options

A. `x=0.36` gram, `y=0.72` gram
B. `x=0.36` gram,`y=1.04` gram
C. `x=0.52` gram, `y=0.72`gram
D. `x=0.52` gram`,y=1.04`gram

Explanation

To solve the problem step by step, we will use the formula for freezing point depression and the given information.

### Step 1: Understand the Formula
The freezing point depression (\( \Delta T_f \)) is given by the formula:
\[ \Delta T_f = K_f \cdot m \]
where:
- \( K_f \) is the molal freezing point depression constant,
- \( m \) is the molality of the solution.

### Step 2: Identify Given Values
From the question, we have:
- \( \Delta T_f = 0.186 \, ^\circ C \)
- \( K_f = 1.86 \, ^\circ C/m \)
- The density of water is \( 1 \, \text{g/ml} \), so 100 ml of water has a mass of 100 g.

### Step 3: Calculate Molality
Molality (\( m \)) is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
In this case, the mass of the solvent (water) is 100 g or 0.1 kg.

### Step 4: Express Moles of Solute
Let \( x \) be the mass of urea and \( y \) be the mass of glucose. The molar masses are:
- Molar mass of urea (NH₂CONH₂) = 60 g/mol
- Molar mass of glucose (C₆H₁₂O₆) = 180 g/mol

The moles of urea and glucose can be expressed as:
\[ \text{moles of urea} = \frac{x}{60} \]
\[ \text{moles of glucose} = \frac{y}{180} \]

### Step 5: Total Moles of Solute
The total moles of solute is:
\[ \text{Total moles} = \frac{x}{60} + \frac{y}{180} \]

### Step 6: Substitute into Molality Formula
Now, substituting into the molality formula:
\[ m = \frac{\frac{x}{60} + \frac{y}{180}}{0.1} \]

### Step 7: Substitute into Freezing Point Depression Formula
Now, substituting \( m \) into the freezing point depression formula:
\[ 0.186 = 1.86 \cdot \left(\frac{\frac{x}{60} + \frac{y}{180}}{0.1}\right) \]

### Step 8: Simplify the Equation
Rearranging gives:
\[ 0.186 = 18.6 \cdot \left(\frac{x}{60} + \frac{y}{180}\right) \]
\[ \frac{x}{60} + \frac{y}{180} = \frac{0.186}{18.6} \]
\[ \frac{x}{60} + \frac{y}{180} = 0.01 \]

### Step 9: Use the Ratio \( x:y = 1:2 \)
Given that \( x:y = 1:2 \), we can express \( y \) in terms of \( x \):
\[ y = 2x \]

### Step 10: Substitute \( y \) into the Equation
Substituting \( y \) into the equation:
\[ \frac{x}{60} + \frac{2x}{180} = 0.01 \]
Finding a common denominator (which is 180):
\[ \frac{3x}{180} + \frac{2x}{180} = 0.01 \]
\[ \frac{5x}{180} = 0.01 \]
\[ 5x = 1.8 \]
\[ x = 0.36 \, \text{g} \]

### Step 11: Calculate \( y \)
Now substituting back to find \( y \):
\[ y = 2x = 2 \cdot 0.36 = 0.72 \, \text{g} \]

### Conclusion
Thus, the correct values are:
- \( x = 0.36 \, \text{g} \)
- \( y = 0.72 \, \text{g} \)