Question

Normality of 10% acetic acid

Answer 1

Answer:

Explaining the normality of 10 percent Acetic Acid.

Explanation:

To answer and calculate this value, let us take into consideration that w/v be the given percentage.

So, let percentage be = w/v.

Now, 10 percent certainly does be equivalent to 10 gram of Acid that is mixed in water (100 ml).

This would also make 100 g of Acetic Acid = 1000 ml of water.

Now focusing on mole concept, we know that

1 mole Acetic acid = 60 g by mass.

100 g Acetic acid = 100 / 60 = 1.67 moles.

So in 1 l of water it is certain that 1.67 moles of Acetic acid is present.

Now for acetic acid we have 1 Normality = 1 Mole

So 10 percent Normality = 1.67 .

Answer 2

Normality of 10% acetic acid is 1.67 N.

Explanation:

Normality (N) is defined as number of gram equivalents of solute present in one litre of solution.

Normality=$\frac{\text { Number of gram equivalents of solute }}{\text { Volume of solution }}$

From the given , 10% w/v acetic acid means 10 grams of acetic acid is dissolved in 100 ml of water.

Molar mass of acetic acid = 60 .05 g/mol

Per 1000 ml of solution has 100 g of solute = $\frac{1000}{60.05} = 1.665 \mathrm{M} \approx 1.67 \mathrm{M} So,10%  acetic acid molarity=1.67 M$

Normality of acetic acid = Molarity of acetic acid = 1.67 N