How many helium atoms are there in a helium blimp containing 535 kg of helium?

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How many helium atoms are there in a helium blimp containing 535 kg of helium?

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Answer 1 · 2017-09-15T18:38:03

Approx..... $8 \times 10^{28} \cdot helium atoms$ $8xx10^28*"helium atoms"$

Explanation:

We work out (i) the molar quantity of helium gas....

$Moles of helium = \frac{535 \cdot k g \times 10^{3} \cdot g \cdot k g^{- 1}}{4.0 \cdot g \cdot m o l^{- 1}} = 133750 \cdot m o l .$ $"Moles of helium"=(535*kgxx10^3*g*kg^-1)/(4.0*g*mol^-1)=133750*mol.$

And then (ii) we mulitply this molar quantity by Avogadro's number.... $N_{A} = 6.022 \times 10^{23} \cdot m o l^{- 1}$ $N_A=6.022xx10^23*mol^-1$ to get the number of helium atoms.....

$6.022 \times 10^{23} \cdot m o l^{- 1} \times 133750 \cdot m o l = an unfeasibly large number$ $6.022xx10^23*mol^-1xx133750*mol="an unfeasibly large number"$ .

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How many helium atoms are there in a helium blimp containing 535 kg of helium?

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