Calculate the mass of 6.022 × 1023 molecule of Calcium carbonate (CaCO3)
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6.022∗1023 molecules are equal to 1Mole
1Mole of CaCO_{3}CaCO3 has mass of 100g (mass of Calcium + mass of Carbon + 3 times the mass of Oxygen)
Therefore, 6.022*10^{23}6.022∗1023 molecules of CaCO_{3}CaCO3has mass 100g
6.022∗1023 molecules are equal to 1Mole
1Mole of CaCO_{3}CaCO3 has mass of 100g (mass of Calcium + mass of Carbon + 3 times the mass of Oxygen)
Therefore, 6.022*10^{23}6.022∗1023 molecules of CaCO_{3}CaCO3has mass 100g
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✨Molar mass of CaCO3 = 40+12+3×16 = 100 g
✨No. of moles of CaCO3
➡ No. of molecules/Avogadro constant
➡ 6.022 × 1023/ 6.022 × 1023
➡1 mole
Mass of CaCO3
=➡No. of moles × molar mass
➡ 1 × 100 g = 100 g.
✨No. of moles of CaCO3
➡ No. of molecules/Avogadro constant
➡ 6.022 × 1023/ 6.022 × 1023
➡1 mole
Mass of CaCO3
=➡No. of moles × molar mass
➡ 1 × 100 g = 100 g.
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