JEE Mains · Chemistry · STD 11 - 1. Some basic concept of chemistry
At \(T(K)\) 100 g of 98% \(H_{2}SO_{4}\) \((w/w)\) aqueous solution is mixed with 100 g of 49% \(H_{2}SO_{4}\) \((w/w)\) aqueous solution. What is the mole fraction of \(H_{2}SO_{4}\) in the resultant solution?
(Given: Atomic mass \(H=1u; S=32u; O=16u\))
(Assume that temperature after mixing remains constant)
- A 0.9
- B 0.1
- C 0.337
- D 0.663
Answer & Solution
Correct Answer
(C) 0.337
Step-by-step Solution
Detailed explanation
Total weight of \(H_{2}SO_{4}\) \(= (100 \times \frac{98}{100}) + (100 \times \frac{49}{100}) = 147~gm\) Total weight of \(H_{2}O = 200 - 147 = 53~gm\) Mole fraction of \(H_{2}SO_{4} = \frac{\frac{147}{98}}{(\frac{147}{98} + \frac{53}{18})} = 0.337\)
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