At \(T\) K, \(100\,\text{g}\) of \(98%\) \(H_2SO_4\) (w/w) aqueous solution is mixed with \(100\,\text{g}\) of \(49%\) \(H_2SO_4\) (w/w) aqueous solution. What is the mole fraction of \(H_2SO_4\) in the resultant solution? (Given: Atomic mass \(H = 1\,u,\; S = 32\,u,\; O = 16\,u\). Assume that temperature after mixing remains constant.)

Question

Which one of the following is not correct for an ideal solution?

Answer 1

To find the mole fraction of \( H_2SO_4 \) in the resultant solution, we follow these steps:

First, calculate the moles of \( H_2SO_4 \) and water in each solution separately.
Determine the amount of \( H_2SO_4 \) in the first solution:
- Weight of \( H_2SO_4 \) = \( 98\% \) of \( 100\,\text{g} = 98\,\text{g} \)
- Molecular weight of \( H_2SO_4 = (2 \times 1) + 32 + (4 \times 16) = 98 \, \text{u} \)
- Moles of \( H_2SO_4 = \dfrac{98\,\text{g}}{98\,\text{g/mol}} = 1 \, \text{mol} \)
- Moles of water in the first solution = \(\dfrac{2\,\text{g}}{18\,\text{g/mol}} \approx 0.111 \, \text{mol} \)
Determine the amount of \( H_2SO_4 \) in the second solution:
- Weight of \( H_2SO_4 = 49\% \) of \( 100\,\text{g} = 49\,\text{g} \)
- Moles of \( H_2SO_4 = \dfrac{49\,\text{g}}{98\,\text{g/mol}} = 0.5 \, \text{mol} \)
- Moles of water = \(\dfrac{51\,\text{g}}{18\,\text{g/mol}} \approx 2.833 \, \text{mol} \)
Add the moles of \( H_2SO_4 \) from both solutions:
- Total moles of \( H_2SO_4 = 1 + 0.5 = 1.5 \, \text{mol} \)
Add the moles of water from both solutions:
- Total moles of water = \( 0.111 + 2.833 = 2.944 \, \text{mol} \)
Calculate the total moles in the resultant solution:
- Total moles = \( 1.5 + 2.944 = 4.444 \, \text{mol} \)
Calculate the mole fraction of \( H_2SO_4 \):
- Mole fraction of \( H_2SO_4 = \dfrac{\text{moles of } H_2SO_4}{\text{total moles}} = \dfrac{1.5}{4.444} \approx 0.337 \)

Thus, the mole fraction of \( H_2SO_4 \) in the resultant solution is \(0.337\), which matches the correct answer option provided.

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