A liter of air containing 1% Ar is repeatedly passed over hot Cu and hot Mg till no reduction of volume takes place. The final volume of Ar will be?

The trick here is to recognize the fact that argon will not react with the hot copper and the hot magnesium, but the oxygen and the nitrogen gas present in the sample will.

Basically, the fact that argon is a noble gas tells you that you shouldn't expect to see a reaction take place when the sample of air is passed over the two hot metals.

On the other hand, oxygen gas will react with the hot copper to form copper(II) oxide, #"CuO"#, and nitrogen gas will react with the hot magnesium to produce magnesium nitride, #"Mg"_3"N"_2#.

So if you pass the sample over the two hot metals until no reduction in volume takes place, you can assume that all the oxygen gas and all the nitrogen gas have reacted.

This means that the sample will only contain argon. Since this gas makes up #1%# of the initial volume, you can say that the final volume of argon will be #"10 mL"#, since

#1 color(red)(cancel(color(black)("%"))) * overbrace("1000 mL"/(100color(red)(cancel(color(black)("%")))))^(color(blue)("1 L is 100% of the volume")) = "10 mL"#

The answer is rounded to one significant figure.