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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Oct 22, 2017

Answer:

#2.17"mol"N_2(g)#

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(g)#

#P_T=P_"nitrogen"+P_"water at 40C"#

We are told that the atmospheric pressure was #760mmHg# and hence it is the total pressure or #P_T# in our equation.

Therefore:
#P_T-P_"water at 40C"=P_"nitrogen"#
Giving
#760-55.3mmHg=704.7mmHg#=#P_"nitrogen"#

Since we now know the #P_"nitrogen"#, we can now use the ideal gas law to determine the number of moles that was actually collected.

#P*V=n*R*T# hence #n=(P*V)/(R*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_"atm"=(704.7cancel(mmHg))*[(1atm)/(760cancel(mmHg))]=0.927236842atm#

#T_"Kelvin"=40.0+273.15K=313.15K#

Therefore

#n=(0.927236842cancel(atm)*60.0cancel(L))/[((0.08206cancel(atm*L))/(cancel(K)*mol))*313.15cancel(K)]#=#2.1650005mol#

Adjusting this figure to 3 sf as given in the values is your question , we obtain #2.17"mol"N_2(g)#

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