Question

56 g of a gas occupies 280cm3 at n.t.p., then its molecular mass will be

Answer 1

HERE IS YOUR ANSWER____
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22.4 L is occupied by 1 mole of a gas at STP.
So, 280 cm3 (0.280 L) will be occupied by (1/22.4) x 0.280 = 0.0125 mole
Mass of 0.0125 mole of the given gas = 0.56 g
Therefore, mass of 1 mole of the given gas = (0.56/0.0125) x 1 = 44.8 g
 
Hence, the molecular mass of the gas is 44.8 g

@HOPE IT HELPS YOU!!

Answer 2

"Molecular mass of gas is 44.8 gm.

Given:

Mass of gas = 56 gm = 0.56 mg

Occupied volume of gas = $280 cm^{3}$ = 0.280 L

To Find:

Molecular mass of gas

Solution:

At NTP, 1 mole of gas will occupies 22.4 L volume.

1 mole gas -----22.4 L

$0.280 L ----- \frac {1}{22.4} \times 0.280 L$ = 0.0125 moles

0.0125 mole have 0.56 mg

Let’s calculate mass of one moles of gas =$\frac {0.56}{0.0125} \times {1} = 44.8 gm$"