At T(K), 100 g of 98% (w/w) aqueous solution is mixed with 100 g of 49% (w/w) aqueous solution. What is the mole fraction of in the resultant solution ? (Given : Atomic mass H = 1 u ; S = 32 u ; O = 16 u). (Assume that temperature after mixing remains

Question

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 ; O = 16 u).

(Assume that temperature after mixing remains constant) [2026]

(Given : Atomic mass H = 1 u ; S = 32 u ; O = 16 u).

(Assume that temperature after mixing remains constant) [2026]

Smart Achievers · Accepted Answer

(4)

$Total weight of H_{2} {SO}_{4}$

$= (100 \times \frac{98}{100}) + (100 \times \frac{49}{100}) = 147 gm$

$Total weight of H_{2} O = 200 - 147 = 53 gm$

$Mole fraction of H_{2} {SO}_{4} = \frac{\frac{147}{98}}{(\frac{147}{98} + \frac{53}{18})} = 0.337$

Solution

1 0.9

2 0.663

3 0.1

4 0.337

Ans.
(4)

$Total weight of H_{2} {SO}_{4}$

$= (100 \times \frac{98}{100}) + (100 \times \frac{49}{100}) = 147 gm$

$Total weight of H_{2} O = 200 - 147 = 53 gm$

$Mole fraction of H_{2} {SO}_{4} = \frac{\frac{147}{98}}{(\frac{147}{98} + \frac{53}{18})} = 0.337$

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