Chemistry Stoichiometry MDCAT 2019
PMDC Verified Question 76 of 105
How many moles of calcium carbonate are present in 1.75 kg of calcium carbonate? (Ar of Ca = 40, Ar of C = 12, Ar of O=16)
A
0.0175 mol
B
1.75 mol
C
17.5 mol
D
1750 mol
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: 17.5 mol
Concept:

Mole calculations require the mass to be inputted in grams. Therefore, kilogram values must be converted to grams before division by molar mass.

Formula:

$$ n = \frac{m \text{ (in grams)}}{M} $$

Solution:

  • First, determine the molar mass of Calcium Carbonate (\( \text{CaCO}_3 \)):


  • \( M = 40 + 12 + (3 \times 16) = 40 + 12 + 48 = 100 \text{ g/mol} \).


  • Convert the given mass to grams: \( 1.75 \text{ kg} = 1.75 \times 1000 = 1750 \text{ grams} \).


  • Calculate moles: \( n = 1750 \text{ g} / 100 \text{ g/mol} = 17.5 \text{ moles} \).


Why other options are incorrect:

  • 0.0175 mol: Results from erroneously dividing the kilogram mass directly by molar mass (1.75 / 100).


  • 1.75 mol: Results from a simple magnitude assumption error.


  • 1750 mol: Forgets to divide by the molar mass (100) after converting to grams.

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