For a solution containing 0.03659 g/mL of HC | vedclass

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Question

(A) Normality $(N)$ is calculated as: $N = \frac{	ext{Mass concentration (g/L)}}{	ext{Equivalent mass}}$.For $HCl$: Equivalent mass = $36.5 \ g/mol$

Vedclass · Accepted Answer

$N_{HCl}$ is more

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(A) Normality $(N)$ is calculated as: $N = \frac{	ext{Mass concentration (g/L)}}{	ext{Equivalent mass}}$.For $HCl$: Equivalent mass = $36.5 \ g/mol$. Concentration = $0.03659 \ g/mL = 36.59 \ g/L$.$N_{HCl} = \frac{36.59}{36.5} \approx 1.002 \ N$.For $CH_3COOH$: Equivalent mass = $60 \ g/mol$. Concentration = $0.04509 \ g/mL = 45.09 \ g/L$.$N_{CH_3COOH} = \frac{45.09}{60} \approx 0.7515 \ N$.Comparing the two,$1.002 > 0.7515$,therefore $N_{HCl}$ is more.

For a solution containing $0.03659 \ g/mL$ of $HCl$ and a solution containing $0.04509 \ g/mL$ of acetic acid $(CH_3COOH)$,which of the following is true?

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