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OpenStudy (anonymous):
how many molecules of dinitrogen pentoxide are present if there are 0.0543 g of oxygen atoms in the sample
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OpenStudy (anonymous):
@ribhu
OpenStudy (ribhu):
u know dinitrogen pentoxide is N2O5, mol wt =108
OpenStudy (anonymous):
yes
OpenStudy (ribhu):
1 mol n2o5 contains 5 mol oxygen
OpenStudy (anonymous):
yes
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OpenStudy (ribhu):
108 gm n2o5 contains 80 gm oxygen
OpenStudy (anonymous):
so there is 28 grams of nitrogen
OpenStudy (ribhu):
0.05423 mol of oxygen since gm atoms = mol
OpenStudy (ribhu):
for this 1 mol oxygen would be contained in 1/5 mol n205
so .05423 mol ox we need to have 1/5 *.05423 mol n205
OpenStudy (ribhu):
so the molecules of n2o5 are
1/5 *.05423 * 6.023*10^23 molecules
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OpenStudy (ribhu):
6.53*10^21 molecules of n2o5 @ryankean12 this would be the answer
OpenStudy (anonymous):
i kinda get it just not the "for this 1 mol oxygen would be contained in 1/5 mol n205
so .05423 mol ox we need to have 1/5 *.05423 mol n205"
OpenStudy (anonymous):
how did you get the 1/5 *.05423 mol N2O5?
OpenStudy (ribhu):
see gram of atoms are same as moles
OpenStudy (ribhu):
and we know 1 mol n2o5 has 5 mol oxygen
so 1 mol oxygen would be contained in 1/5 mol n2o5
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OpenStudy (anonymous):
i get it now
OpenStudy (ribhu):
and hence from the question .05423 gram of oxygen atoms or mol of oxygen.
OpenStudy (ribhu):
i hope i am clear now.
OpenStudy (ribhu):
@ryankean12
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