At s.t.p volume of a gas in a dischargetube is 1.12×10.is the power of _4ml.number of molecules present in the gas are
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At S.T.P 1mole=6.023×10^23 molecules which occupy a volume of 22400ml.Let the number of molecules present in 1.12×10^-7ml be `X`.So, 6.023×10^23 ×1.12*10^-7= X × 22400 X=6.023×10^23 ×1.12*10^-7 ÷ 22400 Number of molecules (X)=3.01 ×10^12 molecules
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