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# Express mass of $6.022×{10}^{23}$ molecules of ${\mathrm{CO}}_{2}$ in grams.

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## Step 1: Formula usedA mole is the quantity of any material which contains one Avogadro number ($6.022×{10}^{23}$) of atoms, molecules, or ions.$1\mathrm{mol}$ of ${\mathrm{CO}}_{2}$$=6.022×{10}^{23}$ molecules of ${\mathrm{CO}}_{2}$$\text{Numberofmoles=}\frac{\mathrm{Given}\mathrm{mass}}{\mathrm{Molar}\mathrm{mass}}$Step 2: Expressing the mass in gramsThe molar mass of ${\mathrm{CO}}_{2}$$=\mathrm{Atomic}\mathrm{mass}\mathrm{of}\mathrm{C}+2×\mathrm{Atomic}\mathrm{mass}\mathrm{of}\mathrm{O}\phantom{\rule{0ex}{0ex}}=12+2×16\phantom{\rule{0ex}{0ex}}=12+32\phantom{\rule{0ex}{0ex}}=44\mathrm{u}$$\therefore$Mass of ${\mathrm{CO}}_{2}$ in terms of grams is$=1\mathrm{mol}×44\mathrm{u}\phantom{\rule{0ex}{0ex}}=44\mathrm{grams}$Therefore, The mass of $6.022×{10}^{23}$ molecules of ${\mathrm{CO}}_{2}$ is $44$ grams.  Suggest Corrections  4      Similar questions  Related Videos   Atomic mass and atomic mass unit_tackle
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