(A) The formula for normality is $N = \frac{w}{E \times V_{(L)}}$,where $w$ is the weight in grams,$E$ is the equivalent weight,and $V_{(L)}$ is the v

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(A) The formula for normality is $N = \frac{w}{E \times V_{(L)}}$,where $w$ is the weight in grams,$E$ is the equivalent weight,and $V_{(L)}$ is the volume in liters.For $NaOH$,the equivalent weight $E = 40 \ g/eq$.The volume $V = 250 \ cm^3 = 0.25 \ L$.Substituting the values: $0.1 = \frac{w}{40 \times 0.25}$.$0.1 = \frac{w}{10}$.$w = 0.1 \times 10 = 1.0 \ g$.

Class 11 Chemistry Some Basic Concepts of Chemistry Chemical stoichiometry

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The weight of pure $NaOH$ required to prepare $250 \ cm^3$ of $0.1 \ N$ solution is ......... $g$. (in $g$)

A
$1.0$
B
$0.1$
C
$2.0$
D
$0.25$

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The weight of pure $NaOH$ required to prepare $250 \ cm^3$ of $0.1 \ N$ solution is ......... $g$. (in $g$)

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