The molarity of 6.3g of `H_2C_2O_(4).2H_2O` in 250ml of water is `Xxx10^(-2)`. Find X?

Answer

To find the molarity of 6.3 g of \( H_2C_2O_4 \cdot 2H_2O \) in 250 ml of water, we can follow these steps:

### Step 1: Calculate the molar mass of \( H_2C_2O_4 \cdot 2H_2O \)

The molar mass of oxalic acid dihydrate can be calculated as follows:

- Molar mass of \( H_2C_2O_4 \):
- \( H: 1 \, \text{g/mol} \times 2 = 2 \, \text{g/mol} \)
- \( C: 12 \, \text{g/mol} \times 2 = 24 \, \text{g/mol} \)
- \( O: 16 \, \text{g/mol} \times 4 = 64 \, \text{g/mol} \)

Total for \( H_2C_2O_4 = 2 + 24 + 64 = 90 \, \text{g/mol} \)

- Molar mass of \( 2H_2O \):
- \( H: 1 \, \text{g/mol} \times 4 = 4 \, \text{g/mol} \)
- \( O: 16 \, \text{g/mol} \times 2 = 32 \, \text{g/mol} \)

Total for \( 2H_2O = 4 + 32 = 36 \, \text{g/mol} \)

- Total molar mass of \( H_2C_2O_4 \cdot 2H_2O = 90 + 36 = 126 \, \text{g/mol} \)

### Step 2: Calculate the number of moles of \( H_2C_2O_4 \cdot 2H_2O \)

Using the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles} = \frac{6.3 \, \text{g}}{126 \, \text{g/mol}} = 0.05 \, \text{mol} \]

### Step 3: Convert the volume from ml to L

Since molarity is defined as moles of solute per liter of solution, we need to convert the volume from milliliters to liters:
\[ 250 \, \text{ml} = 0.250 \, \text{L} \]

### Step 4: Calculate the molarity

Using the formula for molarity:
\[ \text{Molarity (M)} = \frac{\text{number of moles}}{\text{volume (L)}} \]
Substituting the values:
\[ \text{Molarity} = \frac{0.05 \, \text{mol}}{0.250 \, \text{L}} = 0.20 \, \text{M} \]

### Step 5: Express the molarity in the form \( X \times 10^{-2} \)

We can express 0.20 M as:
\[ 0.20 = 20 \times 10^{-2} \]

### Conclusion

Thus, the value of \( X \) is \( 20 \).

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