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The weight of pure NaOH required to prep...

The weight of pure `NaOH` required to prepare `250 cm^(3)` of `0.1N` solution is

A

4g

B

1g

C

2g

D

5g

Text Solution

Verified by Experts

The correct Answer is:
B
Eq. mass of NaOH
`= ("Mol mass")/(1) = 40`
Normality = `(w_(B)xx1000)/(GEM xx V(cm^(3)))`
`0.1 = (w_(B)xx1000)/(40xx250)`
`w_(B)=1g`
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