Question

the normality of 10% (weight/volume) acetic acid is.....​

Answer 1

Answer;

Thus, 100 g of acetic acid would be present in 1000 ml of water. 1 mole of acetic acid has a mass of 60 g. So 100 g of acetic acid means 100/60 =1.67 moles of acetic acid. Thus normality of 10% acetic acid is 1.67

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Answer 2

Answer:

Acetic acid Ch3COOH

molar mass 60g

Let volume be 1L

then Mass of acetic acid present is 100g

Then no. of moles=100/60=1.67 moles

Now Molarity is 1.67M

then Normality=Molarity×n factor

N factor of acetic acid 1

Then Normality is 1.67N

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