At $T(K),100 ext{ }g$ of $98\% H_{2}{SO}_{4}(w/w)$ aqueous solution is mixed with 100 g of $49\% H_

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At $T(K),100	ext{ }g$ 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)

Accepted Answer

Total weight of $H_{2}{SO}_{4}$$$= \left( 100 	imes \frac{98}{100} ight) + \left( 100 	imes \frac{49}{100} ight) = 147gm $$Total weight of $H_{2}O = 200 - 147 = 53gm$Mole fraction of $H_{2}{SO}_{4} = \frac{\frac{147}{98}}{\left( \frac{147}{98} + \frac{53}{18} ight)} = 0.337$

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