1.

`0.1 g KIO_(3)` and excess KI when treated with HCl, the iodine is liberated. The liberated iodine required 45 mL solution thiosulphate for titration . The molarity of sodium thisoulphate will be equivalent to :A. 0.252 MB. 0.126 MC. 0.0313 MD. 0.0623 M

Answer» Correct Answer - D
The reaction involved are :
`IO_(3)^(-)+5I^(-)+6H^(+) to 3I_(2)+H_(2)O`
`2Na_(2)S_(2)O_(3)+I_(2) to 2 NaI+Na_(2)S_(4)O_(6)`
Number of moles of `I_(2)=3xx " Number of moles of " KIO_(3)`
`=(3xx0.1)/(214)`
Number of moles of `Na_(2)S_(2)O_(3)=2xx " Number of moles of " I_(2)`
`(Mxx45)/(1000)=(2xx3xx0.1)/(241)`
M=0.623


Discussion

No Comment Found

Related InterviewSolutions