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Molarity

What will be ...

Question

What will be density ( in $g {m L}^{- 1}$ ) of 3.60 molar sulphuric acid having 29% by mass?
$(M o l a r m a s s = 98 g {m L}^{- 1})$

1.88

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1.22

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1.45

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1.64

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Solution

The correct option is B 1.88
$29$ % by mass means $29 g$ $H_{2} S O_{4}$ in $100 g$ solution.
Number of moles of $H_{2} S O_{4}$ = $\frac{29}{98} = 0.295$ mol
Molarity of $H_{2} S O_{4} = \frac{n o . o f m o l e s o f H_{2} S O_{4}}{V o l u m e (m l)} \times 1000$
Or, $3.6 = \frac{0.295}{\frac{100}{ρ}} \times 1000$
$ρ = \frac{3.6 \times 100}{0.295 \times 1000} = 1.22 g / m l$
$∴$ Answer is option B.

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Similar questions

m l^{- 1}

3.60

29 %

H_{2} S O_{4}

= 98

m o l^{- 1}

g m L^{- 1}

3.60 M

H_{2} S O_{4}

98 g m o l^{- 1}

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1.139 {g mL}^{- 1}

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20 % H_{2} S O_{4}

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P b + S O_{4}^{2 -} \to P b S O_{4} + 2 e^{-}

P b O_{2} + 4 H^{+} + S O_{4}^{2 -} + 2 e^{-} \to P b S O_{4} + 2 H_{2} O

98 % H_{2} S O_{4} (w / v)

1.80 {g m L}^{- 1}

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