A mixture of CO and `CO_(2)` when treated with `I_(2)O_(5)` gives `I_(2)` vapours accoeding to the following equation:
`5CO+I_(2)O_(5)to5CO_(2)+I_(2)`
`I_(2)` vapour was separated and treated with `HClO_(4)` and the resultant `HIO_(4)` required 0.001 " mol of "hlycerol for complete oxidation. After treatment with `I_(2)O_(5)` and removal of `I_(2)` ,t he mixture was treated with excess of 0.1 N NaOH solution and finally this solution required 20 " mL of " 1 N HCl to reach end point using phenolphthalein as indicator, followed by methyl orange as indicator after the first end point, 10 " mL of " further HCl was consumed.
Q. The total volume of NaOH used in the problem is

Options

A. 30 Ml
B. 300 mL
C. 60 mL
D. 600 mL

Answer

B

Explanation

In the beginning mixture of CO and `CO_(2)` is given and 5 moles of `CO_(2)` is formed by the reaction of `5CO` with `I_(2)O_(5)`.
Let the number of m" mol of "`CO_(2)` initially`=n`)
`therefore` Number of m " mol of "`CO_(2)=(n+5)`
`CO_(2)+2NaOHtoNa_(2)CO_(3)+H_(2)O`
Let the mmoles of `NaOH=b`
mmoles of `NaOH` left`=b-2(n+5)`
When phenolphthalein is used
`b-2(n+5)+(n+5)=20`
`b-(n+5)=10` ..(i)
When methyl orange is used
`(n+5)=10` ..(ii)
From equation (i) and (ii) we get
`n=5impliesm" mol of "CO_(2)=5`
`thereforeb=30impliesmmopl of NaOH=30`
Thus, m" mol of "`NaOH` taken `=30` Volume of NaOH:
`30=0.1xxV`
`V=300mL`