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Equivalent Mass

The mass of 9...

Question

The mass of 98% pure sulphuric acid required for the neutralization of 2 mol of $N a O H$ is:
(Molar Mass of S = 32 $g m o l^{- 1}$ , O = 16 $g m o l^{- 1}$ and Na = 23 $g m o l^{- 1}$ )

100 g

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98 g

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200 g

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196 g

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Solution

The correct option is A 100 g
$H_{2} S O_{4} + 2 N a O H \to N a_{2} S O_{4} + H_{2} O$

According to the stoichiometry, 2 moles of NaOH required 1 mole of sulphuric acid.
Mass of sulphuric acid required = 98 g
Mass of 98% sulphuric acid $\times \frac{98}{100} = 98$
$∴$ Mass of 98 % pure sulphuric acid required for the titration = 100 g

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N a O H

g m o l^{- 1}

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N a O H

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70 % H_{2} S O_{4}

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N a = 23 g m o l^{- 1}

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