Question

Volume of 2.8gm of nitrogen at STP in cc

Answer 1

1 Mole of any gas at STP has a volume = 22.4L

This means 1 mole of N2 = 22.4L

28g of NE =22.4L. 【1 mole NE=28g】

So ,

1g=22.4÷28 L

2.8g = (22.4÷28)×2.8

=2.24 L

The retire 2.8g of NE at STP =2.24L

=2240 cc

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