Answer to Question #144829 in General Chemistry for Jarrett Smith

Question

What volume of gaseous chlorine at STP would you need to obtain a 269.5-g sample of gas?

Answer 1

M(Cl₂) = 70.9 g/mol

$n = \frac{m}{M}$

n(Cl₂) $= \frac{269.5}{70.9} = 3.8 \;mol$

1 mole of any gas occupies 22.4 L at STP.

V(Cl₂) $= 3.8 \times 22.4 = 85.12\; L$