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NEET-XII-Chemistry

Previous Year Paper year:2018

with Solutions -
Qstn# 57 Prvs-QstnNext-Qstn
  • #57
    In which case is the number of molecules of
    water maximum?
    (1) 18 mL of water
    (2) 0.18 g of water
    (3) 0.00224 L of water vapours at 1 atm and 273 K
    (4) ``10^{-3}`` mol of water
    digAnsr:   1
    Ans : (1)
    Sol. (1) 18 mL water
    As dH2O = 1 g/mL So WH2O = 18g
    nH2O =
    18
    1
    18
    =
    molecules = 1 × NA
    (2) 0.18 g of water
    nH2O =
    0.18
    18
    = 0.01
    (molecules)H2O = 0.01 × NA
    (3) (VH2O(g))STP = 0.00224 L
    nH2O =
    V 0.00224
    22.4 22.4
    = = 0.0001
    molecules = 0.0001 × NA
    (4) nH2O = 10
    -3
    (molecules)H2O = 10
    -3 × NA