Mole ConceptMedium
Question
At $T(K),100\text{ }g$ 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)
Options
A.0.9
B.0.1
C.0.337
D.0.663
Solution
Total weight of $H_{2}{SO}_{4}$
$$= \left( 100 \times \frac{98}{100} \right) + \left( 100 \times \frac{49}{100} \right) = 147gm $$Total weight of $H_{2}O = 200 - 147 = 53gm$
Mole fraction of $H_{2}{SO}_{4} = \frac{\frac{147}{98}}{\left( \frac{147}{98} + \frac{53}{18} \right)} = 0.337$
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