'A' Level H2 Chemistry Qn
1.15 g of a metallic element reacts with 300 cm3 of oxygen at
298 K and 1 atm pressure, to form an oxide which contains
dinegative O2– anions. What could be the identity of the
metal?
MCQ options : A calcium - B magnesium - C potassium - D
sodium.
My BedokFunland JC Solution :
Let the molar mass of the metal be y grams
No of moles of metal present = 1.15 / y
Using PV=nRT, molar volume of gas = 24.45 dm3.
Hence moles of gas present = 12.27 x 10^-2 mol.
By stoichiometry, moles of dinegative O2- anions generated = 24.54
x 10^-2 mol
Assuming the metal cation has a unipositive charge :
moles of metal cation = 2 x moles of oxide anion
Hence 1.15 / y = 49.08 x 10^-2, and y = 23.43 g
Assuming the metal cation has a dipositive charge :
moles of metal cation = moles of oxide anion
Hence 1.15 / y = 24.54 x 10^-2, and y = 46.86 g.
Conclusion : Comparing against the molar masses and ionic charges
of the 4 metals listed in the options, the closest match and hence
the best answer, is Na (sodium).
Commentary : if we had used 24dm3 as the molar volume of gas, as
might have been intended by the question author, we would obtain
molar masses of 23.0g and 46.0g for unipositive and dipositive
ionic charges respectively, making Na (sodium) a more obvious
answer. However, 24dm3 is actually the molar volume of a gas to 3
sig fig when room temperature is taken to refer to 20 deg Celsius,
rather than 25 deg Celsius (both may be described as "room
temperature", itself an ambiguous term). Alternatively,
recalcitrant folks might argue that 24dm3 is still correct for 25
deg Celsius, if rounded off to 2 sig figs. Yeah well, 2 sig fig is
just bad math, particularly when the question itself gave the
sample mass of the metal to 3 sig figs. This is commonly
misunderstood (even by school teachers), but will usually not be
much of a problem in the actual A levels, as Cambridge will usually
specify along the lines of"You may assume the molar volume of a
gas under these conditions to be 24dm3" *or* explicitly require
you to use PV=nRT.