Question

1g of oleum sample is diluted with water. The solution required 222.45mL of 0.1M NaOH solution. % of free SO3 in this oleum sample is

Answer 1

Answer:

40%

Explanation:

Given 1g of oleum sample is diluted with water. The solution required 222.45mL of 0.1M NaOH solution. % of free SO3 in this oleum sample is

We know that  

To completely neutralize milli equivalent of oleum must be equal to milli equivalent of sodium hydroxide.

So equivalent of h2so4 + equivalent of So3 = equivalent of NaOH

So in oleum we have h2so4 and So3

So let so3 be x (1g) and h2so4 be 1 – x

So molecular weight of h2so4 is 98

So we can write as

H2so4 + So3

(1 – x) / 98 / 2 + x / 80 /2 (because n factor = 2)

(1 – x) / 49 + x / 40 = 222.45 x 0.1 x 10^-3 (convert to litre)

40 + 9x = 43.6002

9x = 3.6002 / 9

So x = 0.40 x 100

Or x = 40 %

Answer 2

your answer is

40%

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