Question

volume of 56 g of nitrogen at stp ​

Answer 1

Explanation:

➡️1 mole of N2 = 28 grams = 22.4 L volume is occupied by it

28 g = 1 mole

so,

56 g = 56/28 = 2 moles.

1 mole occupies volume = 22.4 L

so,

2 moles will occupy = 22.4 x 2 = 44.8 L.✔️

Answer 2

CALCULATION OF VOLUME  >>>>>>>

∵  28g OF  $N_{2}$ OCCUPIES = 22.4 litre AT STP

∵ 56 g of   $N_{2}$OCCUPIES =$\frac{22.4}{28}*56 = 44.8 litres$