Question

What weight of 98.5% KHC8H4O4 is equivalent to 35 ml of 0.01 N NaOH?

The answer is 0.07257g

But what I want to know is the solution. How to get the answer above. I need to write the solution.

Thanks. Please respect.​

Answer 1

Explanation:

Hence the Answer is 0.07246 gm.

Answer 2

What weight of 98.5 % KHC8H4O4 is equivalent to 35 ml of 0.01 N NaOH ?

solution : volume of NaOH = 35 ml = 0.035 L

normality of NaOH solution = 0.01 N

so no of equivalents = 0.035 × 0.01

n factor of NaOH = 1

so no of moles of NaOH = no of equivalents = 0.035 × 0.01 mol ....(1)

let mass of KHC8H4O4 is m

molar mass of KHC8H4O4 = 204.22 g/mol

so no of moles of KHC8H4O4 = (98.5 % of m)/(molar mass of KHC8H4O4)

= 98.5m/100 × 204.22 ....(2)

no of moles of NaOH = no of moles of KHC8H4O4

⇒0.035 × 0.01 = 98.5m/100 × 204.22

⇒0.035 × 0.01 × 204.22 × 100 = 98.5 m

⇒m = 0.0725654822 ≈ 0.07257 g

Therefore the mass of KHC8H4O4 is 0.07257 g