Answer
See the answer below.
Work Step by Step
1) $d=m/V=PM/RT$
$\bar{M}=0.195\ g×0.082×308\ K/((725/760×1\ atm)(0.125\ L))$
$\bar{M}=41.33\ g/mol$
2) Let x be the mol fraction of $NO_2$:
$x×46.00+(1-x)×32.00=41.33\ g/mol$
$x=0.666=66.6\%$
$O_2:0.334=33.4\%$
3) Number of moles:
$NO_2: 0.195\ g/41.33\ g/mol×0.666=3.142×10^{-3}\ mol$
$O_2: 0.195\ g/41.33\ g/mol×0.334=1.576×10^{-3}\ mol$
4)
The ratio of the number of moles of the two is: $2.00$
Stoichiometry gives:
$Cu(NO_3)_2\rightarrow CuO+1/2\ O_2+2\ NO_2$
Stoichiometric ratio: 4.0
Since $N_2O_4$ is liquid at this Temperature and Pressure, the difference in ratios could be explained by the formation of the tetroxide.
