Weight of Urea required 200ml of 2 M solution will be

Options

A. 12 gm
B. 24 gm
C. 20 gm
D. 60 gm

Explanation

To find the weight of urea required to prepare a 200 ml of a 2 M solution, we can follow these steps:

### Step 1: Understand the formula for molarity
Molarity (M) is defined as the number of moles of solute divided by the volume of solution in liters.

\[ M = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \]

### Step 2: Convert volume from ml to liters
Given that the volume of the solution is 200 ml, we need to convert this to liters:

\[ \text{Volume in liters} = \frac{200 \text{ ml}}{1000} = 0.2 \text{ L} \]

### Step 3: Calculate the number of moles of urea
We know the molarity (2 M) and the volume (0.2 L). We can rearrange the molarity formula to find the number of moles:

\[ \text{Number of moles} = M \times \text{Volume in liters} \]

Substituting the values:

\[ \text{Number of moles} = 2 \, \text{mol/L} \times 0.2 \, \text{L} = 0.4 \, \text{moles} \]

### Step 4: Calculate the weight of urea
The weight of urea can be calculated using the formula:

\[ \text{Weight of urea} = \text{Number of moles} \times \text{Molecular weight} \]

The molecular weight of urea (NH₂CONH₂) is 60 g/mol. Now substituting the values:

\[ \text{Weight of urea} = 0.4 \, \text{moles} \times 60 \, \text{g/mol} = 24 \, \text{g} \]

### Final Answer
The weight of urea required for 200 ml of a 2 M solution is **24 grams**.

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