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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

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A solution of lead nitrate, Pb(NO₃)₂(aq) has a concentration of 66.24 g/dm³. The relative formula mass, Mₑ, 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 the molar mass of lead(II) nitrate

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Answer

The formula for lead(II) nitrate is Pb(NO₃)₂. The relative atomic masses are:

  • Pb = 207.2 g/mol
  • N = 14.0 g/mol
  • O = 16.0 g/mol

The molar mass, Mₑ, can be calculated as:

Me=207.2+2(14.0+3imes16.0)Mₑ = 207.2 + 2(14.0 + 3 imes 16.0)

Calculating this, we have:

Me=207.2+2(14.0+48.0)Mₑ = 207.2 + 2(14.0 + 48.0)
Me=207.2+2(62.0)Mₑ = 207.2 + 2(62.0)
Me=207.2+124.0=331.2extg/molMₑ = 207.2 + 124.0 = 331.2 ext{ g/mol}

Step 2

Determine the number of moles of lead(II) nitrate

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Answer

To find the concentration in mol/dm³, we need to determine the number of moles present in 66.24 g of lead(II) nitrate. This is given by the formula:

n=massmolar massn = \frac{mass}{molar\ mass}

Substituting in the values:

n=66.24extg331.2extg/mol0.2extmoln = \frac{66.24 ext{ g}}{331.2 ext{ g/mol}} ≈ 0.2 ext{ mol}

Step 3

Calculate the concentration in mol/dm³

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Answer

Given that the volume of the solution is 1 dm³, the concentration (C) can be calculated as:

C=nvolumeC = \frac{n}{volume}

Substituting the values we have:

C=0.2extmol1extdm3=0.2extmol/dm3C = \frac{0.2 ext{ mol}}{1 ext{ dm}³} = 0.2 ext{ mol/dm}³

This can be expressed in scientific notation as:

C=2.0×101 mol/dm3C = 2.0 \times 10^{-1} \text{ mol/dm}³

Step 4

Choose the correct answer from the options

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Answer

Based on the calculations, the concentration is 2.0 x 10⁻¹ mol/dm³, which corresponds to option C.

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