Question 1

Multiple choiceMedium1 mark

Two batches of the same copper–zinc metallic material have the compositions shown. Use Ar(Cu)=64.0A_\mathrm{r}(\mathrm{Cu})=64.0 and Ar(Zn)=65.0A_\mathrm{r}(\mathrm{Zn})=65.0 Which option gives the amount ratios n(Cu):n(Zn)n(\mathrm{Cu}):n(\mathrm{Zn}) and the correct classification?

BatchMass of Cu / gMass of Zn / g
I6.406.50
II12.86.50
Show answer and explanation

Answer: C. For an element, MM its molar mass in g mol−1\mathrm{g\,mol^{-1}} has the same numerical value as ArA_\mathrm{r} use n=m/Mn=m/M

[1] Batch I gives 6.40/64.0:6.50/65.0=0.100:0.100=1:16.40/64.0 : 6.50/65.0 = 0.100:0.100 = 1:1
[1] Batch II gives 12.8/64.0:6.50/65.0=0.200:0.100=2:112.8/64.0 : 6.50/65.0 = 0.200:0.100 = 2:1
[1] The differing ratios are consistent with a mixture, which does not have a fixed ratio of elements.

Total: [3]

A: calculates both ratios correctly but classifies the material as a compound, which has a fixed element ratio.
B: reverses the Cu:Zn ratio for batch II.
D: swaps the ratios for batches I and II.

Remember: alloys are mixtures, not compounds with a fixed element ratio.

More S2.4 From models to materials questions