# If 60.0L of nitrogen is collected over water at 40.0 degrees celcius when the atmospheric pressure is 760 torr, how many moles of the nitrogen are present? Vapour pressure of water at 40ºc=55.3torr

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#### Explanation:

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Oct 22, 2017

$2.17 \text{mol} {N}_{2} \left(g\right)$

#### Explanation:

Since this volume of gas was collected over water, we would use Dalton's Law of Partial pressure of gases to determine the pressure contributed only by the ${N}_{2} \left(g\right)$

${P}_{T} = {P}_{\text{nitrogen"+P_"water at 40C}}$

We are told that the atmospheric pressure was $760 m m H g$ and hence it is the total pressure or ${P}_{T}$ in our equation.

Therefore:
${P}_{T} - {P}_{\text{water at 40C"=P_"nitrogen}}$
Giving
$760 - 55.3 m m H g = 704.7 m m H g$=${P}_{\text{nitrogen}}$

Since we now know the ${P}_{\text{nitrogen}}$, we can now use the ideal gas law to determine the number of moles that was actually collected.

$P \cdot V = n \cdot R \cdot T$ hence $n = \frac{P \cdot V}{R \cdot T}$

Before implementing the use of the formula, each quantity must be in their appropriate units hence both $P$ and $T$ are to be converted.

That is
${P}_{\text{atm}} = \left(704.7 \cancel{m m H g}\right) \cdot \left[\frac{1 a t m}{760 \cancel{m m H g}}\right] = 0.927236842 a t m$

${T}_{\text{Kelvin}} = 40.0 + 273.15 K = 313.15 K$

Therefore

$n = \frac{0.927236842 \cancel{a t m} \cdot 60.0 \cancel{L}}{\left(\frac{0.08206 \cancel{a t m \cdot L}}{\cancel{K} \cdot m o l}\right) \cdot 313.15 \cancel{K}}$=$2.1650005 m o l$

Adjusting this figure to 3 sf as given in the values is your question , we obtain $2.17 \text{mol} {N}_{2} \left(g\right)$

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