Home
Class 11
CHEMISTRY
How much volume of 0.40 M Na(2)S(2)O(3) ...

How much volume of 0.40 M` Na_(2)S_(2)O_(3)` would be required to react with the `I_(2)` liberated by adding excess of KI of 50 mLof 0.20 M `M CuSO_(4)`

A

12.5 mL

B

25 mL

C

50 mL

D

2.5 mL

Text Solution

Verified by Experts

The correct Answer is:
B
m.eq of `Na_(2)S_(2)O_(3)` = m eq of `CuSO_(4)`
`0.4xxVxx1=50xx0.2xx1 " V=25ml"`
Promotional Banner

Similar Questions

Explore conceptually related problems

The weight of iodine liberated when excess of KI reacts with 500 ml of 1 M H_2O_2 is

The volume of 0.05N Na_2CO_3 solution required to neutralise 200ml of 0.02M H_2SO_4 solution is

How many grams of Al_2(SO_4)_3 , will be required to prepare 4 litres of 0.025M solution ?

What volume of 0.01 M K_(2)Cr_(2)O_(7) would be required to oxidize Fe(II) in 50 ml of 0.03 M solution of ferrous ammonium sulphate in acidic medium?

The reaction of I, with Na_(2)S_(2)O_(3) gives Nal and

There are two types of iodine titrations (a) Iodometric & (b) Iodimetric, Iodometric method is indirect method of I_(2) estimation. Any oxidant which liberates I_(2) from KI solution, the liberated iodine is estimated by titrating it with Na_(2)S_(2)O_(3) solution as : I_(2)+2Na_(2)S_(2)O_(3)rarr2NaI+Na_(2)S_(4)O_(6) 100 mL of .x. M K_(2)Cr_(2)O_(7) solution is added to excess of KI solution in acidic medium. The liberated iodine required 50 mL of 0.1N Na_(2)S_(2)O_(3) solution. The value of x is

Calculate the volume of 0.1 MKMnO_(4) required to react with 100 ml. of 0.1 M H_(2)C_(2)O_(4) . 2H_(2)O solution in the presence of H_(2)SO_(4) .

50 ml solution of H_(2)O_(2) was treated with excess KI (s) and the solution was acidified with acetic acid. The liberated I_(2) required 40 ml of 0.5 M Na_(2)S_(2)O_(3) solution for the end point using starch is indicator. Find the molairty and volume strength of the H_(2)O_(2) solution.