2 g of 109 So, gm eq. of H_2SO_4= 2.18/98× 2 Similarly, gm eq. of Ca(OH)_2=44.49 × 10^ 3× M × 2 Since, gm eq. of H_2SO_4=gm eq. of Ca(OH)_2 Therefore, 2.18/98× ...

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Introduction to Oxidation and Reduction

2 g of 109

Question

2 g of 109% oleum is diluted with water. To neutralize this solution completely, 44.49 ml of $C a (O H)_{2}$ solution is required. What is the molarity of $C a (O H)_{2}$ solution?

1.0 M

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0.1 M

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0.05 M

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0.5 M

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Solution

The correct option is D 0.5 M
2 g of 109% oleum $= 2 \times 1.09 = 2.18$ g of $H_{2} S O_{4}$ in dilute solution.
So, gm eq. of $H_{2} S O_{4} = \frac{2.18}{98} \times 2$
Similarly, gm eq. of $C a (O H)_{2} = 44.49 \times 10^{- 3} \times M \times 2$
Since,
$g m e q . o f H_{2} S O_{4} = g m e q . o f C a (O H)_{2}$
Therefore,
$\frac{2.18}{98} \times 2 = 44.49 \times 10^{- 3} \times M \times 2$
$\Rightarrow M = 0.5$

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2 g of 109% oleum is diluted with water. To neutralize this solution completely, 44.49 ml of Ca(OH)2 solution is required. What is the molarity of Ca(OH)2 solution?

The correct option is D 0.5 M2 g of 109% oleum =2×1.09=2.18 g of H2SO4 in dilute solution. So, gm eq. of H2SO4=2.1898×2 Similarly, gm eq. of Ca(OH)2=44.49×10−3×M×2 Since, gm eq. of H2SO4=gm eq. of Ca(OH)2 Therefore, 2.1898×2=44.49×10−3×M×2 ⇒M=0.5

2 g of 109% oleum is diluted with water. To neutralize this solution completely, 44.49 ml of $C a (O H)_{2}$ solution is required. What is the molarity of $C a (O H)_{2}$ solution?