Answer:
The added mole will be "11.388 moles".
Explanation:
The given values are:
Initial volume,
V₁ = 4.11 L
Final volume,
V₂ = 11.3 L
Number of moles,
n₁ = 6.51
On applying Avogadro's law,
⇒ [tex]\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}[/tex]
On putting the estimated values, we get
⇒ [tex]\frac{4.11}{6.51} =\frac{11.3}{n_{2}}[/tex]
On applying cross-multiplication, we get
⇒ [tex]4.11 \ n_{2}=11.3\times 6.51[/tex]
⇒ [tex]4.11 \ n_{2}=73.563[/tex]
⇒ [tex]n_{2}=\frac{73.563}{4.11}[/tex]
⇒ [tex]n_{2}=17.898 \ moles[/tex]
So that the number of moles at added gas will be:
[tex]=17.898-6.51[/tex]
[tex]=11.388 \ moles[/tex]