Question

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.)

• A. \(0.337\)
• B. \(0.1\)
• C. \(0.9\)
• D. \(0.663\)

Hint:

For w/w solutions, always convert percentage into actual mass first before calculating mole fraction.

Answer 1

The Correct Option is A

Step 1: Calculate masses of \(H_2SO_4\) and water.
From \(100\,\text{g}\) of \(98%\) solution: \[ \text{Mass of } H_2SO_4 = 98\,\text{g}, \quad \text{Mass of } H_2O = 2\,\text{g}. \] From \(100\,\text{g}\) of \(49%\) solution: \[ \text{Mass of } H_2SO_4 = 49\,\text{g}, \quad \text{Mass of } H_2O = 51\,\text{g}. \]
Step 2: Total masses after mixing.
\[ \text{Total } H_2SO_4 = 98 + 49 = 147\,\text{g}, \] \[ \text{Total } H_2O = 2 + 51 = 53\,\text{g}. \]
Step 3: Convert masses into moles.
Molar mass of \(H_2SO_4 = 98\,\text{g mol}^{-1}\), \[ n(H_2SO_4) = \frac{147}{98} = 1.5\ \text{mol}. \] Molar mass of \(H_2O = 18\,\text{g mol}^{-1}\), \[ n(H_2O) = \frac{53}{18} \approx 2.94\ \text{mol}. \]
Step 4: Calculate mole fraction of \(H_2SO_4\).
\[ X_{H_2SO_4} = \frac{1.5}{1.5 + 2.94} = \frac{1.5}{4.44} \approx 0.337. \]
Final Answer: \[ \boxed{0.337} \]