Question

How many molecules are present in 1.28 g of SO2 gas, if the atomic mass of sulphur and oxygen is 32 u and 16 u respectively ?

Answer 1

Molecular weight of $SO_2$ = Molecular weight of $S$ + 2*Molecular weight of $O_2$

M$SO_2$ = 32 + 2*16
M$SO_2$ = 64

By this we understand

64 gm $SO_2$ = $6.022*10^{23}$ atoms
Or, 1 gm $SO_2$ = $\frac{6.022*10^{23}}{64}$ atoms
Or, 1.28 gm $SO_2$ = $\frac{6.022*10^{23}}{64}* 1.28 = 1.2044*10^{22}$ atoms

Answer 2

Number of molecules present in 1.28 gm of SO2 = 1.2044 * 10^22

Explanation:

Molecular weight of   = Atomic weight of Sulphur  + 2 (atomic weight of oxygen)

= 32 + 2*16

= 64 gm

Avogadro's Constant = 6.022 × 1023 mol-1

So, number of molecules in 1 gm of SO2 =  Avogadro’s constant

So, number of molecules present in 1.28 gm of SO2 = Avogadro’s constant / molecular weight of SO2 * 1.28

= 6.022 * 1023 / 64 * 1.28

= 1.2044 * 10^22