One mole of an ideal gas will occupy a volume of 22.4 liters at STP (Standard Temperature and Pressure, 0°C and one atmosphere pressure).
Proportion:
1 mole – 22400 ml
n moles – 112 ml
n = 0.005 moles of NO2
m=n×M
M(NO2) = 46 g/mol
m(NO2) = 0.005×46=0.23 g
ρ(NO2) =1.15 g/ml
V = m/ρ
V = 0.23/1.15 = 0.2 ml
One mole of NO2 molecules contains 6.02×1023 NO2 molecules
Number of molecules = n×6.02×1023 = 0.005×6.02×1023 = 3.01×1021 molecules
Answer: 0.2 ml and 3.01×1021 molecules
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