Chemistry, asked by HilariousMonkey, 3 days ago

How many molecules are there in 45 x 10^-21 g of calcium carbonate

Answers

Answered by ruchikatdl2003

0

Explanation:

Molar mass of CaCO3 = 40+12+ 48

= 100gm

No. Of moles = w/M

= 45×10^-21/100

= 45×10^-23

No. Of molecules = 6.023 x10²³ x n

= 6.023 x 10²³ x 45×10^-23

= 271.035 molecules

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