calculate the no. of molecule present in 15.4g of carbon tetrachloride
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2
Answer
GMM OF CCl4 = 12+(4×35.5)
= 154g
154g of ccl4= 6.022×10²³ molecules
So, 15.4g of ccl4 contains 6.022×10²³×15.4
=______________ molecules
154
=6.022×10²² molecules. (Ans)
Answered by
1
Explanation:
molecular mass of CCl4 = 12 + 4*35.5
= 154
no. of moles = 15.4/154
=1/10
no.of molecule = 1/10 *6.02*10 to the power 23
= 6.02*10 to the power 24
which is the answer
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