0.5g sample of a sulphite salt was dissolved in 200 ml solution and 20

1.	0.5g sample of a sulphite salt was dissolved in 200 ml solution and 20 ml of this solution required10 ml of 0.02 M acidified permanganate solutionsalt is:(A) 20%(C) 60%Hence, the % by mass of sulphite in the sulphite(B) 40%(D) 80%
Answer» The answer coming is 8%. See the solution given below. Lets say the salt is MSO3, M is any metal The n factor of this salt will be obviously 2 . Now its given 20 ml of this requires 10 ml of 0.02M acidified KMno4 ( whose n factor will be 5 as its acidic medium). So we can write 10* 0.02 5 = 0.51000/Ewhere E is equivalent weight,E is coming 500 g, so the molar wt. of the salt is coming as 1000 g. % SO3 will be = (80/1000) *100 = 8 %

0.5g sample of a sulphite salt was dissolved in 200 ml solution and 20 ml of this solution required10 ml of 0.02 M acidified permanganate solutionsalt is:(A) 20%(C) 60%Hence, the % by mass of sulphite in the sulphite(B) 40%(D) 80%

Answer»

The answer coming is 8%. See the solution given below.

Lets say the salt is MSO3, M is any metal

The n factor of this salt will be obviously 2 .

Now its given 20 ml of this requires 10 ml of 0.02M acidified KMno4 ( whose n factor will be 5 as its acidic medium).

So we can write 10* 0.02 *5 = 0.5*1000/Ewhere E is equivalent weight,E is coming 500 g, so the molar wt. of the salt is coming as 1000 g.

% SO3 will be = (80/1000) *100 = 8 %