Question

The molarity of the solution prepared by dissolving \(6.3\text{ g}\) of oxalic acid (\(\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O}\)) in \(250\text{ mL}\) of water in mol L\(^{-1}\) is \(x \times 10^{-2}\). The value of \(x\) is \(\dots\dots\dots\). (Nearest integer)
[Atomic mass : H : 1.0, C : 12.0, O : 16.0]

Hint:

Always include the water of crystallisation when calculating the molar mass of hydrated crystals like oxalic acid or Mohr's salt.

Answer 1

Correct Answer: 20

To determine the molarity of the solution, we must first find the molar mass of oxalic acid dihydrate (\(\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O}\)). The formula units and respective atomic masses are calculated as follows:
- Hydrogen: \(2(\text{H}_2) + 4(\text{H}_2\text{O}) = 8\) atoms, \(8 \times 1.0 = 8.0\text{ g/mol}\)
- Carbon: \(2\text{ C}\), \(2 \times 12.0 = 24.0\text{ g/mol}\)
- Oxygen: \(4(\text{H}_2\text{C}_2\text{O}_4) + 2(\text{H}_2\text{O}) = 6\) atoms, \(6 \times 16.0 = 96.0\text{ g/mol}\)
The molar mass of \(\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O}\) is \(8.0 + 24.0 + 96.0 = 128.0\text{ g/mol}\).
Next, calculate the moles of oxalic acid:
\(\text{Moles} = \frac{\text{Mass in grams}}{\text{Molar mass}} = \frac{6.3}{128.0} = 0.04921875\text{ mol}\).
The volume of the solution is \(250 \text{ mL} = 0.250 \text{ L}\).
Molarity is calculated as:
\(\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in L}} = \frac{0.04921875}{0.250} = 0.196875\text{ mol L}^{-1}\).
Given that molarity is expressed as \(x \times 10^{-2}\), we have \(0.196875 = x \times 10^{-2}\). Solving for \(x\), we find:
\(x = 19.6875\).
Rounding to the nearest integer, \(x = 20\).
This falls within the expected range of 20, confirming the validity of the solution.