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Arrhenius Acid

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Question

A solution of a mixture of KCl and KOH was neutralised with 120 mL of 0.12 N HCl. Calculate the amount of KOH in the mixture.

A

0.205 g

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B

2.14 g

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C

0.806 g

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D

None of these

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Solution

The correct option is C 0.806 g
$K O H + H C l ⟶ K C l + H_{2} O$
$56 g$ $36.5 g$
Volume of $H C l = 120 m L$
Concentration of $H C l = 0.12 N = 0.12 M$
$M = [(w e i g h t o f K O H) / (M o l e c u l a r w e i g h t o f K O H)] \times [100 / 12]$
$0.12 \times 0.12 = W / 56$
$0.144 \times 56 = W$
$W = 0.8064 g$

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