2019 H2 Chem P1 Q11 — Mole Solution

The answer is A (0.82).

In 100 g of solder glass there are 16 g of \(\text{B}_2\text{O}_3\) and 84 g of PbO.
\(M_r\) of \(\text{B}_2\text{O}_3 = 2(10.8) + 3(16.0) = 69.6\); \(M_r\) of PbO = 207.2 + 16.0 = 223.2
Amount of \(\text{B}_2\text{O}_3 = \dfrac{16}{69.6} = 0.2299\) mol, so amount of B \(= 2 \times 0.2299 = 0.4598\) mol (each \(\text{B}_2\text{O}_3\) contains 2 B atoms)
Amount of PbO \(= \dfrac{84}{223.2} = 0.3763\) mol, so amount of Pb = 0.3763 mol
Pb/B molar ratio \(= \dfrac{0.3763}{0.4598} = 0.818 = 0.82\)

• B is wrong: 0.91 does not follow from the data; with the correct molar masses and 2 mol of B per mol of \(\text{B}_2\text{O}_3\), the ratio is 0.818.
• C is wrong: 1.22 is the ratio the other way up, B/Pb \(= \dfrac{0.4598}{0.3763}\).
• D is wrong: 1.64 is Pb/\(\text{B}_2\text{O}_3\) \(= \dfrac{0.3763}{0.2299}\); it forgets that each \(\text{B}_2\text{O}_3\) contains 2 B atoms.