The volume occupied at S.T.P by 6.4 g of Sulphur dioxide is?

Question

1)2240cm^3
2)224cm^3
3)22.4cm^3
4)2.24cm^3

Answer 1

Molecular wight of SO2= 32+2*16 (S=32,O=16)
=64

hence,6.4 gm of SO2=6.4/64 moles of SO2=0 .1 moles of SO2
at stp, 1 mole of a gas occupies 22.4 litre=22400 cm^3 volume.
hence, at S.T.P. 0.1 moles of SO2 will occupy (0.1/1)*22400 cm^3
= 2240 cm^3

hence, option 1) is correct

Answer 2

First of all, u have got to knw this: " 1 mole of any gas occupies 24,000cm^3 at s.t.p"

So, now we have got to find the volume occupied by 6.4g of sulphur dioxide at s.t.p. Now this means that we have to convert that 6.4g of sulphur dioxide into moles. In other words, we will have to first find out the number of moles in 6.4g of sulphur dioxide.
In order to do that -> Mass / Relative Formula Mass whereby the Mass is 6.4g and the Relative Formula Mass is 64g (32 + 2(16)).
So, when u divide that u get, 0.1 moles.

Now lets go back to the part where we said 1 mole of a gas occupies 24,000 cm^3 of volume. That means 0.1 mole of a gas will occupy : 0.1*24,000cm^3

And that is 2400cm^3.
Thats the answer. And I see its not in the options that u have give. But I am quite sure thats the answer for this question.

GoodLuck!!

Answer 3

Remember that a cubic centimeter is equal to one milliliter.

Now the molecular mass of SO2 is 64 g/mole, so 6.4 g is 0.1 moles. Since one mole of a gas occupies 22.4 Liters at STP then 6.4 grams would occupy 2.24 Liters.

Now back to the 1 cm^3 = 1 mL. 2.24 Liters will be the same as 2.34 L x 10^3 mL/L or 2240 mL which means 2240 cm^3.

The answer is 1).

Answer 4

6.4g =.1 mole
1 mole=22.4 L
.1 mole=2.24 L=2240mL
2240cm^3 choice 1

Answer 5

first find the amount of SO2 in moles (divide mass by molar mass)
molar mass = 1 sulfur + 2 oxygen = 1(32.1) + 2(16.0) = 64.1 g/mol
moles = mass / molar mass = 6.4 g / 64.1 g/mol = 0.1 mol
now, molar volume of any gas at STP in 22.4 L/mol - so for 0.1 mol it's 2.24 L
1 L = 1000 cm^3
therefore 2.24 = 2240 cm^3
so correct answer is 1)

Answer 6

PV=nrT

at stp, 1 atm and 20 c , r is constant

convert 6.4g S02 to moles first using molar mass(6.4g x 1 mol/64.07g mol= .0999moles so2) and convert T from c to K by adding 273.15.(20+273.15=293.15)

V= (.0999)(.0821)(293.15)/1
V=2.40 L

Answer 7

no of mols= mass/molecular weight
also
no of mols=vol/22400
mols=6.4/64=0.1

0.1=vol/22400
so vol=0.1*22400=2240 cm^3

1 is the answer.

Answer 8

2.24 cm3