What is the volume occupied by 1 oxygen molecule at STP?

3.71 \times 10^{- 23} l

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

7.42 \times 10^{- 23} l

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.855 \times 10^{- 23} l

No worries! We‘ve got your back. Try BYJU‘S free classes today!

14.84 \times 10^{- 23} l

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $3.71 \times 10^{- 23} l$
At STP 1 mole of oxygen occupies $22.4 l$ .
Thus $N_{a}$ molecules occupie $22.4 l$ .
Where, $N_{a}$ = Avogadro's number= $6.022 \times 10^{23}$
$∴$ volume occupied by 1 molecule is: $22.4 \div N a = 3.71 \times 10^{- 23}$

Suggest Corrections

Similar questions

8

C H_{4}

6.02 \times 10^{20}

C H_{4}

C H_{4}

S T P

H_{2} O

1 g / m l

1 m o l \to 18 g = 6 \times 10^{+ 23}

4.045 \times 10^{23}

27^{\circ} C

700 t o r r

A flask contains 3.2 g of sulphur dioxide. Calculate the followings:

i) The number of molecules of sulphur dioxide at STP.

ii) The volume occupied by 3.2 g of sulphur dioxide at STP.

(S = 32, O = 16)

Join BYJU'S Learning Program