8. At STP 2 g of helium gas (molar mass = 4)occupies a volume of

1.	8. At STP 2 g of helium gas (molar mass = 4)occupies a volume of
Answer» Since 1 mole of gas occupies 22.4L at stp, Volume of He: 0.500mol x 22.4 = 11.2L 1 mole of any gas at STP occupies a volume of 22.4 litres. Mol.mass of He = 4. = 4g/mol 2g/4g/mol = 0.5 moles. = 0.5 x 22.4L = 11.2 L. 1 mol of gas occupies 22.4L at stopvolume of He=0.500mol×22.4L =11.2L

Answer»

Since 1 mole of gas occupies 22.4L at stp, Volume of He: 0.500mol x 22.4 = 11.2L

1 mole of any gas at STP occupies a volume of 22.4 litres. Mol.mass of He = 4. = 4g/mol 2g/4g/mol = 0.5 moles. = 0.5 x 22.4L = 11.2 L.

1 mol of gas occupies 22.4L at stopvolume of He=0.500mol×22.4L =11.2L

Discussion