How do you calculate the mass of one mole of such hydrogen giving your answer to atoms four decimal places? (The Avogadro constant, L= $6.0225 \times 10^{23} m o l^{- 1}$ $6.0225 xx 10^23 mol^-1$ )?

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How do you calculate the mass of one mole of such hydrogen giving your answer to atoms four decimal places? (The Avogadro constant, L= $6.0225 \times 10^{23} m o l^{- 1}$ $6.0225 xx 10^23 mol^-1$ )?

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Answer 1 · 2017-01-07T15:28:39

$Mass of 1 mole of hydrogen atoms$ $"Mass of 1 mole of hydrogen atoms"$ $=$ $=$ $Mass of 1 hydrogen atom \times N_{A}$ $"Mass of 1 hydrogen atom"xx"N"_A$ , where $N_{A} = L = 6.0225 \times 10^{23} \cdot m o l^{- 1}$ $"N"_A=L=6.0225xx10^23*mol^-1$ .

Explanation:

$Mass of 1 hydrogen atom$ $"Mass of 1 hydrogen atom"$ $=$ $=$ $1.6727 \times 10^{- 27} \cdot kg$ $1.6727xx10^{-27}*"kg"$ (from this site).

And thus, $mass of 1 mole of hydrogen atoms$ $"mass of 1 mole of hydrogen atoms"$ $=$ $=$

$1.6727 \times 10^{- 27} \cdot kg \times 6.0225 \times 10^{23} \cdot m o l^{- 1} \times 10^{3} \cdot g \cdot k g^{- 1}$ $1.6727xx10^(-27)*"kg"xx6.0225xx10^23*mol^-1xx10^(3)*"g"*kg^-1$

$= 1.007 \cdot g \cdot {mol}^{- 1}$ $=1.007*"g"*"mol"^-1$

Clearly, you would not be expected to know these masses. You would be expected to be able to do such a calculation if provided with the masses of the nucleon, and with $Avogadro's number$ $"Avogadro's number"$ .

What is the mass of one mole of hydrogen molecules?

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How do you calculate the mass of one mole of such hydrogen giving your answer to atoms four decimal places? (The Avogadro constant, L= $6.0225 \times 10^{23} m o l^{- 1}$ $6.0225 xx 10^23 mol^-1$ )?

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Explanation:

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How do you calculate the mass of one mole of such hydrogen giving your answer to atoms four decimal places? (The Avogadro constant, L= 6.0225 xx 10^23 mol^-16.0225×1023mol−16.0225 xx 10^23 mol^-1)?

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Explanation:

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How do you calculate the mass of one mole of such hydrogen giving your answer to atoms four decimal places? (The Avogadro constant, L= $6.0225 \times 10^{23} m o l^{- 1}$ $6.0225 xx 10^23 mol^-1$ )?