HomeLearn Fast select Course
Show Unit/ChapterChange UnitChange SubjectLast Menu
NEET-XII-Chemistry

Previous Year Paper year:2018

with Solutions -
Qstn# 46 Prvs-QstnNext-Qstn
  • #46
    A mixture of 2.3 g formic acid and 4.5 g
    oxalic acid is treated with conc. ``H_2SO_4``.
    The evolved gaseous mixture is passed through
    KOH pellets. Weight (in g) of the remaining
    product at STP will be
    (1) 1.4
    (2) 3.0
    (3) 2.8
    (4) 4.4
    digAnsr:   3
    Ans : (3)
    Sol.
    HCOOH 2 4H SO
    Dehydrating Agent
     CO + H2O
    H O abosrbed
    by H SO
    2
    2 4
    (moles)i =
    2.3 1
    46 20
    = 0 0
    (moles)f 0
    1
    20
    1
    20
    H2C2O4 2 4H SO CO + CO2 + H2O
    [H2O absorbed by H2SO4]
    (moles)i
    4.5 1
    90 20
    = 0 0 0
    (moles)f 0
    1
    20
    1
    20
    1
    20
    CO2 is absorbed by KOH.
    So the remaning product is only CO.
    moles of CO formed from both reactions
    =
    1 1
    20 20
    + =
    1
    10
    Left mass of CO = moles × molar mass
    =
    1
    28
    10
    ï‚´
    = 2.8 g Ans.