Sign in
Please select an account to continue using cracku.in
↓ →
Prep with like-minded aspirants; Get access to free daily tests and study material.
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=1 u ; s=32 u ; 0 = 16 u).
(Assume that temperature after mixing remains constant)
100g of 98% H₂SO₄: 98g H₂SO₄ + 2g H₂O. 100g of 49%: 49g + 51g. Total: 147g H₂SO₄ + 53g H₂O.
Moles H₂SO₄ = 147/98 = 1.5. Moles H₂O = 53/18 = 2.944.
Mole fraction = 1.5/(1.5+2.944) = 1.5/4.444 = 0.337.
The answer is Option 4: 0.337.
Create a FREE account and get:
Predict your JEE Main percentile, rank & performance in seconds
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.