Sulphuric acid is $95.8$ % by mass. Calculate mole fraction of $H_{2} S O_{4}$ of density $1.91 c m^{- 3} .$ $(H = 1, S = 32, O = 16$ ).

0.895

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0.2982

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0.2056

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0.8075

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Solution

The correct option is D 0.8075
For mole Fraction:-
No. of moles of $H_{2} S O_{4} = \frac{95.8}{98} = 0.9776$
No. of Moles of $H_{2} O = \frac{4.2}{18} = 0.2333$
$∴$ Mole fraction of $H_{2} S O_{4} = \frac{m o l e s o f H_{2} S O_{4}}{T o t a l m o l e s} = \frac{0.9776}{0.9776 + 0.2333} = 0.8073$

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(H = 1, S = 32, O = 16)

S = 32, H = 1, O = 16

4.22 M

H_{2} S O_{4}

1.21 g c m^{- 3}

H_{2} S O_{4}

m = 5.298 m o l k g^{- 1}

C + 2 H_{2} {S O}_{4} \to {C O}_{2} + {2 H}_{2} O + 2 {S O}_{2}

H_{2} S O_{4}

H = 1, S = 32, O = 16

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