A flask conntains 3.2g of sulphur dioxide calculate the following
1.the moles of sulphur dioxide present inthe flask
2.the number of molecules of sulphurdioxide present in the flask
3 the volume occupied by 3.2g of sulphur dioxide at s.t.p[s=32,o=16]

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Solution

1. Number of moles = given mass/ molar mass = 3.2 g/64 gmol =0.05 moles
2.Number of molecules = number of moles $\times$ Avogadro number = 0.05 x 6.022 $\times$ 10²³ =3.01 $\times$ 10²²
3. At STP , volume of one mole of any ideal gas =22.4 L
Now, volume occupied by 0.05 moles of SO₂ = 22.4 $\times$ 0.05 = 1.12 L

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(a) A flask contains 3.2 g of sulphur dioxide. Calculate the following :
(i) The moles of sulphur dioxide present in the flask.
(ii) the number of molecules of sulphur dioxide present in the flask.
(iii) the volume occupied by 3.2 g of sulphur dioxide at STP.
(S = 32, O = 16)
(b) A n experiment showed that in a lead chloride solution 6.21 g of lead is combined with 4.26 g of chlorine. What is the empirical formula of this chloride?
(Pb = 2107; Cl = 35.5).

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)

(a) A flask contains 3.2 g of sulphur dioxide. Calculate the following:
(i) The moles of sulphur dioxide present in the flask.
(ii) The number of molecules of sulphur dioxide present in the flask.
(iii) The volume occupied by 3.2 g of sulphur dioxide at STP. (S = 32, O = 16)

(b) The reaction of potassium permanganate (VII) with acidified iron (II) sulphate is given below:
2KMnO₄ + 10FeSO₄ + 8H₂SO₄ $\overset{}{\to}$ K₂SO₄ + 2MnSO₄ + 5Fe₂(SO₄)₃ + 8H₂O
If 15.8 g of potassium permanganate (VII) was used in the reaction, calculate the mass of iron (II) sulphate used in the above reaction. (K = 39, Mn = 55, Fe = 56, S = 32, O = 16)

Q. If

1

gram of Sulphur dioxide contains x molecules, how many molecules will be present in

1

gram of oxygen? (S =

32

u ; O =

16

Calculate the number of molecules in 12.8 g of sulphur dioxide gas. Take Avogadro's number as 6 × 10²³.
[S = 32, O = 16]

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A flask conntains 3.2g of sulphur dioxide calculate the following1.the moles of sulphur dioxide present inthe flask2.the number of molecules of sulphurdioxide present in the flask3 the volume occupied by 3.2g of sulphur dioxide at s.t.p[s=32,o=16]

1. Number of moles = given mass/ molar mass = 3.2 g/64 gmol =0.05 moles 2.Number of molecules = number of moles ×Avogadro number = 0.05 x 6.022 ×1023 =3.01 ×1022 3. At STP , volume of one mole of any ideal gas =22.4 L Now, volume occupied by 0.05 moles of SO2 = 22.4 × 0.05 = 1.12 L

A flask conntains 3.2g of sulphur dioxide calculate the following
1.the moles of sulphur dioxide present inthe flask
2.the number of molecules of sulphurdioxide present in the flask
3 the volume occupied by 3.2g of sulphur dioxide at s.t.p[s=32,o=16]