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A solution of lead nitrate, Pb(NO₃)₂(aq) has a concentration of 66.24 g/dm³ - OCR Gateway - GCSE Chemistry: Combined Science - Question 7 - 2019 - Paper 9

Question 7

A solution of lead nitrate, Pb(NO₃)₂(aq) has a concentration of 66.24 g/dm³. The relative formula mass, Mₜ, of lead(II) nitrate is 331.2. What is the concentration, ... show full transcript

Worked Solution & Example Answer:A solution of lead nitrate, Pb(NO₃)₂(aq) has a concentration of 66.24 g/dm³ - OCR Gateway - GCSE Chemistry: Combined Science - Question 7 - 2019 - Paper 9

Step 1

Calculate moles of lead nitrate

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Answer

To find the concentration in mol/dm³, we first need to calculate the number of moles of lead nitrate present in the solution. This can be found using the formula:

$n = \frac{mass}{Mₜ}$

Where:

mass = 66.24 g
Mₜ (relative formula mass of Pb(NO₃)₂) = 331.2 g/mol

Substituting the values:

$n = \frac{66.24}{331.2} ≈ 0.200 mol$

Step 2

Calculate concentration in mol/dm³

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Answer

Since the volume of the solution is 1 dm³, the concentration (C) can be calculated using:

$C = \frac{n}{V}$

Where:

n = number of moles = 0.200 mol
V = volume = 1 dm³

Thus:

$C = \frac{0.200}{1} = 0.200 mol/dm³$

This can be expressed in scientific notation:

$C = 2.0 × 10^{-1} mol/dm³$

Therefore, the correct answer is option C: 2.0 × 10¹ mol/dm³.

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