What is the $normality$ $"normality"$ of a $4.0 \cdot g$ $4.0g$ mass of sodium hydroxide dissolved in a $1.0 \cdot L$ $1.0L$ volume of water?

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What is the $normality$ $"normality"$ of a $4.0 \cdot g$ $4.0g$ mass of sodium hydroxide dissolved in a $1.0 \cdot L$ $1.0L$ volume of water?

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Answer 1 · 2017-04-08T05:36:02

$Normality = 1.0 \cdot N$ $"Normality"=1.0*N$

Explanation:

$Normality$ $"Normality"$ is defined similarly to $molarity$ $"molarity"$ .

$Normality = \frac{Moles of H_{3} O^{+} or H O^{-}}{Volume of solution}$ $"Normality"=("Moles of "H_3O^+"or "HO^-)/"Volume of solution"$

Here, $normality = \frac{\frac{4.0 \cdot g}{40.0 \cdot g \cdot m o l^{- 1}}}{0.100 \cdot L} = 1.0 \cdot N$ $"normality"=((4.0*g)/(40.0*g*mol^-1))/(0.100*L)=1.0*N$ .

This will react quantitatively with one equiv of acid. One the other hand, a solution of diprotic $H_{2} S O_{4} (a q)$ $H_2SO_4(aq)$ that is $0.50 \cdot m o l \cdot L^{- 1}$ $0.50*mol*L^-1$ with respect to $H_{2} S O_{4}$ $H_2SO_4$ , but necessarily $1.0 \cdot m o l \cdot L^{- 1}$ $1.0*mol*L^-1$ with respect to $H_{3} O^{+}$ $H_3O^+$ (why?), would also be regarded as $1 \cdot N$ $1*N$ .

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What is the $normality$ $"normality"$ of a $4.0 \cdot g$ $4.0g$ mass of sodium hydroxide dissolved in a $1.0 \cdot L$ $1.0L$ volume of water?

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What is the normality"normality" of a 4.0⋅g4.0*g mass of sodium hydroxide dissolved in a 1.0⋅L1.0*L volume of water?

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What is the $normality$ $"normality"$ of a $4.0 \cdot g$ $4.0g$ mass of sodium hydroxide dissolved in a $1.0 \cdot L$ $1.0L$ volume of water?