Answer:
The correct answer from the given options is option 1.2 L.
Explanation:
Volume of the gases when 0.65 moles of gas are present = [tex]V_1=750 mL=0.750 L[/tex]
Volume of the gas when there 0.35 moles of gas are added= [tex]V_2[/tex]
[tex]n_1=0.65 moles,n_2=0.65+0.35 moles=0.10 moles[/tex]
According Avogadro law of gases:
[tex]\frac{V_1}{n_1}=\frac{V_2}{n_2}[/tex]
[tex]V_2=\frac{V_1\times n_2}{n_1}=1.15 L[/tex]
The closest answer from the given options is option 1.2 L.
From the explanation, we are given that:
The volume of gas it occupies when [tex]0.35moles[/tex] = [tex]{V_2}[/tex] (unknown)
Therefore we have:
[tex]{n_2}[/tex] = [tex]0.65[/tex] + [tex]0.35[/tex] = [tex]0.10moles[/tex] , [tex]{n_1}[/tex] = [tex]0.65moles[/tex]
To solve this question, we have to consider the Avogadro’s law, which state that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules”
The law can be expressed as:
[tex]\frac{V_1}{n_1}=\frac{V_2}{n_2}[/tex]
[tex]V_2=\frac{V_1\times n_2}{n_1}[/tex]
[tex]V_2=\frac{0.750\times 0.10}{0.65}[/tex] = [tex]1.15L[/tex]
[tex]1.2L[/tex]
Therefore, the correct answer is 1.2L
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