How many atoms of carbon are present in 18g of glucose?
Correct answer: D. 3.6 x 1 023
- A. 6.02 x 10^22
- B. 6.02 x 10^23
- C. 3.6 x 1 022
- D. 3.6 x 1 023
Explanation
The molecular formula of glucose is C6H12O6, which means that there are 6 carbon atoms in every molecule of glucose. The molar mass of glucose is 180 g/mol, so 18 g of glucose contains 18/180 = 0.1 moles of glucose.Since there are 6 carbon atoms in every mole of glucose, there are 0.1 X 6 = 0.6 moles of carbon atoms in 18 g of glucose.Avogadro's number is the number of atoms in a mole of a substance, and it is equal to 6.022 x 1023 atoms/mol. So, there are 0.6 X 6.022 x 1023= 3.613 x 1023 carbon atoms in 18 g of glucose.Therefore, there are 3.613 x 1023carbon atoms present in 18 g of glucose.
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Mole calculations connect mass, number of particles and Avogadro's number, while balanced equations provide the mole ratios used in stoichiometry. Questions cover limiting and excess reactants, theoretical yield and percentage yield, including identifying which reactant is consumed first and comparing the actual product with the maximum possible product.
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