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Raoult's Law

The density i...

Question

The density (in $g m L^{- 1}$ ) of a $3.60 M$ sulphuric acid solution $H_{2} S O_{4}$ (molar mass = $98 g m o l^{- 1}$ ) will be:

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Solution

Given, Molarity $= 3.60 M$
$\Rightarrow 3.6$ moles $H_{2} S O_{4}$ is present in $1 L$ solution

$\Rightarrow 3.6 \times 98 \times \frac{100}{29} = 1216.5 g$ $H_{2} S O_{4}$ is present in $1 L$ solution

$ρ = \frac{1216.5}{1000} = 1.2165 g / m L$

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Similar questions

g {m L}^{- 1}

(M o l a r m a s s = 98 g {m L}^{- 1})

m l^{- 1}

3.60

29 %

H_{2} S O_{4}

= 98

m o l^{- 1}

^{- 1}

H_{2} S O_{4}

H_{2} S O_{4}

^{- 1}

^{- 1}

1.294

1.139 {g mL}^{- 1}

1.294 {g mL}^{- 1}

39 % H_{2} S O_{4}

1.139 {g mL}^{- 1}

20 % H_{2} S O_{4}

3.5 L

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

H_{2} S O_{4}

1.80 {g mL}^{- 1}

0.1 M H_{2} S O_{4}

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