Answer: 0.03798 kilograms
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given volume}}{\text {Molar Volume}}=\frac{12L}{22.4L}=0.535moles[/tex]
1 mole of chlorine gas [tex](Cl_2)[/tex] weighs = 71 g
Thus 0.535 moles of chlorine gas [tex](Cl_2)[/tex] will weigh = [tex]\frac{71}{1}\times 0.535=37.98g=0.03798kg[/tex] (1kg=1000g)
Thus there are 0.03798 kilograms in 12 L of chlorine gas.
