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Calculate the weight of 6.022 xx 10^(23)...

Calculate the weight of `6.022 xx 10^(23)` molecules of `CaCO_(3)`

Text Solution

Verified by Experts

No. Of moles of `CaCO_(3) = ("no.of molecules")/("Av.cons")`
`=(6.022 xx 10^(23))/(6.022 xx 10^(23))=1`
Weight of `CaCO_3` = no. of moles x molecular wt.
`=1 xx 100 = 100g`
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