IB Chemistry SL S2.4 From models to materials questions

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Question 1

Multiple choiceMedium1 mark

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

Batch	Mass of Cu / g	Mass of Zn / g
I	6.40	6.50
II	12.8	6.50

Show answer and explanation

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

[1] Batch I gives $6.40/64.0 : 6.50/65.0 = 0.100:0.100 = 1:1$
[1] Batch II gives $12.8/64.0 : 6.50/65.0 = 0.200:0.100 = 2:1$
[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.

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