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Calculate the weight of 6.022 xx 10^(23)molecules of CaCO_(3) |
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Answer» Solution :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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