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      • Molar mass can be used as a conversion factor because it provides a direct link between the mass of a substance (in grams) and the number of moles, which corresponds to the number of particles present in the sample.
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  2. Aug 10, 2022 · In such a conversion, we use the molar mass of a substance as a conversion factor to convert mole units into mass units (or, conversely, mass units into mole units). We also established that 1 mol of Al has a mass of 26.98 g (Example). Stated mathematically, 1 mol Al = 26.98 g Al.

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      Solution concentrations expressed in molarity are the...

  3. Molar mass can be used as a conversion factor because it provides a direct link between the mass of a substance (in grams) and the number of moles, which corresponds to the number of particles present in the sample.

  4. The first conversion factor comes from the balanced chemical equation, the second conversion factor is the molar mass of H 2 O 2, and the third conversion factor comes from the definition of percentage concentration by mass.

    • David W. Ball, Jessie A. Key
    • 2014
  5. May 28, 2024 · Apply a molarity conversion factor to convert between a molar quantity of solute and the volume of a solution. Apply multiple conversion factors to solve problems that involve complex molar relationships.

  6. The first conversion factor comes from the balanced chemical equation, the second conversion factor is the molar mass of H 2 O 2, and the third conversion factor comes from the definition of percentage concentration by mass.

  7. Sep 13, 2024 · The simplest type of manipulation using molar mass as a conversion factor is a mole-mass conversion (or its reverse, a mass-mole conversion). In such a conversion, we use the molar mass of a substance as a conversion factor to convert mole units into mass units (or, conversely, mass units into mole units).

  8. Start by using the balanced chemical equation to convert to moles of another substance and then use its molar mass to determine the mass of the final substance. In two steps, we have: [latex]33.9\cancel{\text{ mol }\ce{H2}}\times \dfrac{2\text{ mol }\ce{NH3}}{3\cancel{\text{ mol }\ce{H2}}}=22.6\text{ mol }\ce{NH3}[/latex]