Question

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

Answer 1

Answer:

$0.15008\ \text{g}$

$3.23\times 10^{21}$

Explanation:

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

n = Mol of $N_2$ = $\dfrac{120}{22400}=0.00536\ \text{mol}$

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

$N_A$ = Avogadro's number = $6.022\times 10^{23}\ \text{mol}^{-1}$

Mass of $N_2$ is

$m=nM\\\Rightarrow m=0.00536\times 28\\\Rightarrow m=0.15008\ \text{g}$

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

Number of molecules is given by

$nN_A=0.00536\times 6.022\times 10^{23}=3.23\times 10^{21}\ \text{molecules}$

The number of molecules present in it are $3.23\times 10^{21}$