1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# The mass of 3.01×1023 molecules of a gas iss 32.0 g. What volume does 8.00 g of the gas occupy at STP?

A
2.80 L
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
8.00 L
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
11.2 L
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
22.4 L
No worries! We‘ve got your back. Try BYJU‘S free classes today!
E
64.0 L
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A 2.80 LMass of 3.01 × 1023 molecules is 32 grams.Mass of 1 molecule is 323.01 × 11023 grams.Thus, mass of 6.023 × 1023 molecules is 323.01 × 11023 × 6.023 × 1023 grams, which is almost equal to 64 grams.Thus, 64 grams is the molecular weight of the gas.One mole of any gas occupy 22.4 L at STP.64 grams occupy 22.4 L at STP.1 gram occupy 22.464 L at STP.Thus, 8 grams will occupy 22.464 × 8 L at STP, which is equal to 2.80 L.Thus, option A is the correct answer.

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
CHEMISTRY
Watch in App
Explore more
Join BYJU'S Learning Program