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Standard XII

Chemistry

Percentage Composition

What amount o...

Question

What amount of water is added in 40 ml of NaOH (0.1N) which is neutralised by 50 ml of HCl (0.2N)?

80 ml

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60 ml

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40 ml

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90 ml

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Solution

The correct option is B 60 ml
Given $N_{1} = 0.1 V_{1} = ?$
volume of NaOH in solution = 40 ml
$N_{2} = 0.2, V_{2} = 50 m l$
$N_{1} V_{1} = N_{2} V_{2}$
or $V_{1} = \frac{0.2 \times 50}{0.1} = 100 m l$
Amount of water to be added
$= V_{1}$ -Volume of NaOH in solution
= 100-40 =60 ml

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What amount of water is added in 40 ml of NaOH (0.1N) which is neutralised by 50 ml of HCl (0.2N)?

The correct option is B 60 mlGiven N1=0.1V1=?volume of NaOH in solution = 40 mlN2=0.2,V2=50mlN1V1=N2V2or V1=0.2×500.1=100mlAmount of water to be added=V1 -Volume of NaOH in solution= 100-40 =60 ml

The correct option is B 60 ml
Given $N_{1} = 0.1 V_{1} = ?$
volume of NaOH in solution = 40 ml
$N_{2} = 0.2, V_{2} = 50 m l$
$N_{1} V_{1} = N_{2} V_{2}$
or $V_{1} = \frac{0.2 \times 50}{0.1} = 100 m l$
Amount of water to be added
$= V_{1}$ -Volume of NaOH in solution
= 100-40 =60 ml