Question

calculate the mass of 120cc nitrogen present at STP. how many number of molecules are present in it?​

Answer 1

`0.15008\ \text{g}`

`3.23* 10^(21)`

Step-by-step explanation:

1 mol of nitrogen at STP = 22.4 L = 22400 cc

n = Mol of
`N_2` =
`(120)/(22400)=0.00536\ \text{mol}`

M = Molar mass of
`N_2` =
`28\ \text{g/mol}`

`N_A` = Avogadro's number =
`6.022* 10^(23)\ \text{mol}^(-1)`

Mass of
`N_2` is

`m=nM\\\Rightarrow m=0.00536* 28\\\Rightarrow m=0.15008\ \text{g}`

Mass of the nitrogen is
`0.15008\ \text{g}`

Number of molecules is given by

`nN_A=0.00536* 6.022* 10^(23)=3.23* 10^(21)\ \text{molecules}`

The number of molecules present in it are
`3.23* 10^(21)`