volume occupied by 2.8g of no2 in litres
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No of moles of NO2 =MASS /MOLAR MASS
THEREFORE , MOLES =2.8/32+14=2.8/46=0.06
Now 1 mole has volume 22.4L
Therefore 0.06 moles will have volume =22.4×0.06=1.34L
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Mass of NO2 =2.8g
Moles of NO2 = mass /molar mass =2.8 / 14+32
Moles of NO2 = 0.0608
Moles of NO2 = volume in litre / volume at STP
Volume of NO2 = moles × volume at STP
Volume of NO2 = 0.0608×22.4
Volume of NO2 = 1.36litre
Moles of NO2 = mass /molar mass =2.8 / 14+32
Moles of NO2 = 0.0608
Moles of NO2 = volume in litre / volume at STP
Volume of NO2 = moles × volume at STP
Volume of NO2 = 0.0608×22.4
Volume of NO2 = 1.36litre
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