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Counting Particles by Mass: The Mole · Paper 1 — Free Chemistry HL Practice Questions

1FoundationMCQConversions between moles, mass, and number of particles1 markPaper 1~2 min
A student dissolves 12.485 g12.485 \text{ g} of hydrated copper(II) sulfate, CuSO4⋅5H2O\text{CuSO}_4 \cdot 5\text{H}_2\text{O} (molar mass =249.7 g mol−1= 249.7 \text{ g mol}^{-1}), in water. Excess sodium hydroxide solution is added, precipitating all copper(II) ions according to: Cu2+(aq)+2OH−(aq)→Cu(OH)2(s)\text{Cu}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Cu(OH)}_2(\text{s}) What is the maximum number of moles of Cu(OH)2\text{Cu(OH)}_2 produced?
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2FoundationMCQConversions between moles, mass, and number of particles1 markPaper 1~2 min
A technician obtains 3.024 g3.024 \text{ g} of pure paracetamol (C8H9NO2\text{C}_8\text{H}_9\text{NO}_2, molar mass =151.2 g mol−1= 151.2 \text{ g mol}^{-1}) after recrystallisation. How many molecules of paracetamol are present in this sample? (NA=6.02×1023 mol−1N_A = 6.02 \times 10^{23} \text{ mol}^{-1})
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3FoundationMCQConversions between moles, mass, and number of particles1 markPaper 1~2 min
A sample of dry air contains 3.01×10203.01 \times 10^{20} molecules of carbon dioxide. What is the mass of CO2\text{CO}_2 in this sample? (Molar mass of CO2=44.01 g mol−1\text{CO}_2 = 44.01 \text{ g mol}^{-1}; NA=6.02×1023 mol−1N_A = 6.02 \times 10^{23} \text{ mol}^{-1})
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4MasteryMCQConversions between moles, mass, and number of particles1 markPaper 1~2 min
A student heats a 2.50 g sample of hydrated copper(II) sulfate, CuSO4⋅xH2O\text{CuSO}_4 \cdot x\text{H}_2\text{O}, to constant mass and obtains 1.60 g of anhydrous CuSO4\text{CuSO}_4. Using M(CuSO4)=159.6 g mol−1M(\text{CuSO}_4) = 159.6 \text{ g mol}^{-1} and M(H2O)=18.0 g mol−1M(\text{H}_2\text{O}) = 18.0 \text{ g mol}^{-1}, what is the value of xx?
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