The number of oxygen atoms in 20 grams of sulphur trioxide is the same as those present in x grams of ozone. The atomic mass of sulphur and oxygen are 32 u and 16 u respectively. Find the value of x.
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20 g of SO3
molecular weight
= 32+(16×3)
= 32+48
= 80 u
Number of moles
= 20/80
= 1/4
= 0.25 moles
No. of oxygen atoms
= 0.25 × 6.022 × 10^23
Ozone = O3
molecular weight
= 16×3
= 48 u
x grams weight of Ozone has no. of atoms
= 0.25×6.022×10^23
x/48 = 0.25
x = 48×0.25
x = 12 g
---------------------------------
Hence the value of x is 12 grams.
molecular weight
= 32+(16×3)
= 32+48
= 80 u
Number of moles
= 20/80
= 1/4
= 0.25 moles
No. of oxygen atoms
= 0.25 × 6.022 × 10^23
Ozone = O3
molecular weight
= 16×3
= 48 u
x grams weight of Ozone has no. of atoms
= 0.25×6.022×10^23
x/48 = 0.25
x = 48×0.25
x = 12 g
---------------------------------
Hence the value of x is 12 grams.
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