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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% Hâ‚‚SOâ‚„: 98g Hâ‚‚SOâ‚„ + 2g Hâ‚‚O. 100g of 49%: 49g + 51g. Total: 147g Hâ‚‚SOâ‚„ + 53g Hâ‚‚O.
Moles Hâ‚‚SOâ‚„ = 147/98 = 1.5. Moles Hâ‚‚O = 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.
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