1.76moles
The number of moles of gas "n" is directly proportional to its volume "v". This can be expressed as:
[tex]\begin{gathered} n\alpha v \\ n=kv \\ k=\frac{n_1}{v_1}=\frac{n_2}{v_2} \end{gathered}[/tex]where:
n1 and n2 are initial and final moles respectively
v1 and v2 are the initial and final volumes
Given the following parameters
[tex]\begin{gathered} n_1=0.49mole \\ v_1=1.20L \\ v_2=4.30L \end{gathered}[/tex]Required
Final number of moles "n2"
Substitute the given parameters into the formula
[tex]\begin{gathered} n_2=\frac{n_1v_2}{v_1} \\ n_2=\frac{0.49\times4.30}{1.20} \\ n_2=\frac{2.107}{1.20} \\ n_2=1.76moles \end{gathered}[/tex]Hence the moles of gas added to increase the volume to 4.30L is 1.76moles
