Density of the sample"=2.72\\frac{g}{cm^3}"
Edge length"=2.205in=2.205\\times2.54=5.6007cm"
So the volume of the cubic sample"=5.6007\\times5.6007\\times5.6007=175.68cm^3"
Hence the mass of the sample"=2.72\\times175.68=477.85g"
Molar mass of "CaCO_3" "=100\\frac{g}{mol}"
Therefore number of moles of "CaCO_3" in the given sample"\\frac{477.85}{100}=4.7785mol"
In "1mol" "CaCO_3", number of "O" atoms"=3\\times N_A"
So in "4.7785mol" "CaCO_3", number of "O" atoms"=4.7785\\times3\\times N_A=8.634\\times10^{24}atoms"
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