A solution of lead(II) chloride (M_r = 278.2) contains 1.08 g of PbCl₂ in 100 cm³ of solution - AQA - A-Level Chemistry - Question 26 - 2017 - Paper 3
Question 26
A solution of lead(II) chloride (M_r = 278.2) contains 1.08 g of PbCl₂ in 100 cm³ of solution. In this solution, the lead(II) chloride is fully dissociated into ions... show full transcript
Worked Solution & Example Answer:A solution of lead(II) chloride (M_r = 278.2) contains 1.08 g of PbCl₂ in 100 cm³ of solution - AQA - A-Level Chemistry - Question 26 - 2017 - Paper 3
Step 1
Calculate the number of moles of PbCl₂
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Answer
To find the number of moles of PbCl₂, use the formula: Number of moles=molar massmass
Given that the molar mass of PbCl₂ is 278.2 g/mol and the mass is 1.08 g: Number of moles of PbCl₂=278.21.08=0.00388 mol
Step 2
Determine the number of moles of chloride ions produced
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Answer
Since one mole of PbCl₂ produces two moles of chloride ions (Cl⁻): Number of moles of Cl−=2×Number of moles of PbCl₂=2×0.00388=0.00776 mol
Step 3
Calculate the concentration of chloride ions
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Answer
Concentration is defined as the number of moles per unit volume. Here, we have the total volume as 100 cm³, which is equivalent to 0.1 dm³. Therefore, the concentration of chloride ions is: Concentration (mol dm−3)=volume (dm3)Number of moles of Cl−=0.10.00776=0.0776 mol dm−3
This can be expressed as: 7.76×10−2 mol dm−3
