Answer
$1.30\text { grams/Liter}$
Work Step by Step
We assume that we are dealing with only one mole of air.
1 mole of gas at STP = 22.42 Liters
$0.78\text{ mol $N_2$}*\frac{28.0134 \text{ grams $N_2$}}{1 \text{ mol $N_2$}}=22\text{ grams $N_2$}$
$0.21\text{ mol $O_2$}*\frac{31.9988 \text{ grams $O_2$}}{1 \text{ mol $O_2$}}=6.7\text{ grams $O_2$}$
$0.010\text{ mol Ar}*\frac{39.948\text{ grams Ar}}{1 \text{ mol Ar}}=0.40\text{ grams Ar}$
Total mass = $22+6.7+0.40=29.1\text{ grams}$
Volume = 22.42 L
Density = $\frac{\text{Mass}}{\text{ Volume}}=\frac{29.1}{22.42}=1.30\text { grams/Liter}$
