What is the $normality$ $"normality"$ of a solution prepared by dissolving a $5 \cdot g$ $5g$ mass of sodium hydroxide in a $1 \cdot L$ $1L$ volume of water?

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What is the $normality$ $"normality"$ of a solution prepared by dissolving a $5 \cdot g$ $5g$ mass of sodium hydroxide in a $1 \cdot L$ $1L$ volume of water?

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Answer 1 · 2017-08-19T14:42:33

$Normality = 0.125 \cdot m o l \cdot L^{- 1}$ $"Normality"=0.125*mol*L^-1$ .

Explanation:

Here $molarity$ $"molarity"$ $\equiv$ $-=$ $normality$ $"normality"$ . Why? Because hydroxide anion does not speciate to any degree in aqueous solution.

And $molarity$ $"molarity"$ $\equiv$ $-=$ $\frac{moles of solute}{volume of solution}$ $"moles of solute"/"volume of solution"$

$= \frac{\frac{5 \cdot g}{40.0 \cdot g \cdot m o l^{- 1}}}{1 \cdot L} = \frac{0.125 \cdot m o l}{1 \cdot L} = 0.125 \cdot m o l \cdot L^{- 1}$ $=((5*g)/(40.0*g*mol^-1))/(1*L)=(0.125*mol)/(1*L)=0.125*mol*L^-1$ with respect to $N a O H (a q)$ $NaOH(aq)$ .

What is $p H$ $pH$ of this solution? We know that $p H + p O H = 14$ $pH+pOH=14$ under standard conditions.

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What is the $normality$ $"normality"$ of a solution prepared by dissolving a $5 \cdot g$ $5g$ mass of sodium hydroxide in a $1 \cdot L$ $1L$ volume of water?

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What is the normality"normality" of a solution prepared by dissolving a 5⋅g5*g mass of sodium hydroxide in a 1⋅L1*L volume of water?

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What is the $normality$ $"normality"$ of a solution prepared by dissolving a $5 \cdot g$ $5g$ mass of sodium hydroxide in a $1 \cdot L$ $1L$ volume of water?