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16. How many molecules are present in 3.0 X 102 grams of calcium

carbonate? [Calcium - 40 g/m;Carbon - 12g/m. Oxygen - 16 g/m]
A) 6.02 X 1023
B 81.1 X 1023.
C) 1.81 X 1024

User Tkane
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1 Answer

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To solve this problem, we need to use the concept of Avogadro's number and molecular weight.

The molecular weight of calcium carbonate (CaCO3) can be calculated as follows:

  • Calcium (Ca) atomic weight = 40 g/mol
  • Carbon (C) atomic weight = 12 g/mol
  • Oxygen (O) atomic weight = 16 g/mol
  • Molecular weight of CaCO3 = (1 x 40) + (1 x 12) + (3 x 16) = 100 g/mol

Now, we can calculate the number of molecules in 3.0 x 10^2 grams of CaCO3 as follows:

  • Convert the mass of CaCO3 to moles using the formula:

moles = mass / molecular weight

moles = 3.0 x 10^2 g / 100 g/mol

moles = 3.0 x 10^0 mol

  • Use Avogadro's number to calculate the number of molecules:

number of molecules = moles x Avogadro's number

number of molecules = 3.0 x 10^0 mol x 6.02 x 10^23 molecules/mol

number of molecules = 1.806 x 10^24

Therefore, the answer is C) 1.81 x 10^24.

User Joelreeves
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