Question

A flask contains 3.2g of sulphur dioxide. Calculate the followings:
The moles of sulphur dioxide present in the flask.
(i) The number of molecules of sulphur dioxide at STP.
(i) The volume occupied by 3.2g of sulphur dioxide at STP.
(5=32,0 = 16)

Answer 1

Answer:

1) no of moles = given mass/ molar mass

so n= 3.2÷64=

0.05

2) 64g has 6.022×10^23 molecule

so 3.2g has 6.022×10^23 ×32÷64×10

=3.011 ×10^22

3)64g has 22.4 l volume

3.2g has 22.4×32÷64×10

2.24litre

Answer 2

Answer:

1)no of moles m=given mass/molar mass

so n=3.2÷64=0.05 moles

2)64g has 6×10^23

6×10^13×32÷64×10=3×10^22

3)64g has 22.4L Volume

3.2 has 22.4×32÷64=2.24 L

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