Question

The molecules present in 5.6 L of sulphur dioxide at STP
a) 1.5 x 1023
b) 1.5 x 10-23
C) 4x 1023
d)15 x 1023​

Answer 1

Answer:- (a) 1.5× 10²³

Explanation:- At STP 22.4 L of volume is occupied by 1 mole of gas.

So 5.6 L is occupied by 5.6/22.4 = 0.25 moles.

1 mole= 6.022 X 10^23 molecules

=> 0.25 moles= 1.5055 X 10^23 molecules.

Answer 2

Answer:

22.4 litre contains 6.022 X 10^23

therefore, 5.6 litres would contain

X = (6.022 X 10^23) X ( 5.6) / 22.4

X = 1.424 X 10^24 molecules

so your answer is option(a)