Question

Molecular mass of ( mathrm{N}_{2} mathrm{O}=44 )
weight of ( mathrm{N}_{2} mathrm{O}=20 mathrm{Omg}=0.2 mathrm{g} )
moles of ( mathrm{N}_{2} mathrm{O} ) present ( =0.2 / 44=4.55 times 10^{wedge}(-3) )
Let moles of ( mathrm{N}_{2} mathrm{O} ) removed ( =mathrm{y} ) moles of ( mathrm{N}_{2} mathrm{O} ) remained ( =2.89 times 10^{wedge}(-3) )
Thus ( 4.55 times 10^{wedge}(-3)-y=2.89 times 10^{wedge}(-3) )
( Rightarrow-y=2.89 times 10^{wedge}(-3)-4.55 times 10^{wedge}(-3) )
( Rightarrow y=4.55 times 10^{wedge}(-3)-2.89 times 10^{wedge}(-3) )
( Rightarrow mathrm{y}=1.66 times 10^{wedge}(-3) ) moles
No. of molecules in 1 mole ( =6.022 times 10^{wedge}(23) ) No. of molecules in y moles ( =y times 6.022 times 10^{wedge}(23) ) ( =1.66 times 10^{wedge}(-3) times 6.022 times 10^{wedge}(23) )
( =9.97 times 10^{wedge}(20) )
Thus ( x=9.97 times 10^{wedge}(20) ) molecules

# 45. From 200 mg of Co, when x molecules are removed, 2.89 x 10-3 moles of Co, are left. x will be (1) 1020 molecules (2) 1010 molecules (3) 21 molecules (4) 1021 molecules

Solution