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The density of ethanol is 1.09 g/cm³ The density of propylene is 0.97 g/cm³ 60 litres of ethanol are mixed with 128 litres of propylene to make 188 litres of antifreeze - Edexcel - GCSE Maths - Question 14 - 2019 - Paper 3

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Question 14

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The density of ethanol is 1.09 g/cm³ The density of propylene is 0.97 g/cm³ 60 litres of ethanol are mixed with 128 litres of propylene to make 188 litres of antifr... show full transcript

Worked Solution & Example Answer:The density of ethanol is 1.09 g/cm³ The density of propylene is 0.97 g/cm³ 60 litres of ethanol are mixed with 128 litres of propylene to make 188 litres of antifreeze - Edexcel - GCSE Maths - Question 14 - 2019 - Paper 3

Step 1

Calculate the mass of ethanol

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Answer

To find the mass of ethanol, use the formula:

Mass=Density×Volume\text{Mass} = \text{Density} \times \text{Volume}

For ethanol, we have:

Massethanol=1.09g/cm3×60litres=1.09×60×1000g=65400g\text{Mass}_{\text{ethanol}} = 1.09\,\text{g/cm}^3 \times 60\,\text{litres} = 1.09 \times 60 \times 1000\,\text{g} = 65400\,\text{g}

Step 2

Calculate the mass of propylene

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Answer

To find the mass of propylene, use the same formula:

Masspropylene=0.97g/cm3×128litres=0.97×128×1000g=12416g\text{Mass}_{\text{propylene}} = 0.97\,\text{g/cm}^3 \times 128\,\text{litres} = 0.97 \times 128 \times 1000\,\text{g} = 12416\,\text{g}

Step 3

Find the total mass of the mixture

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Answer

The total mass of the antifreeze mixture is given by:

Total Mass=Massethanol+Masspropylene\text{Total Mass} = \text{Mass}_{\text{ethanol}} + \text{Mass}_{\text{propylene}}

Calculating this gives:

Total Mass=65400g+12416g=77816g\text{Total Mass} = 65400\,\text{g} + 12416\,\text{g} = 77816\,\text{g}

Step 4

Calculate the density of the antifreeze

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Answer

The density of the antifreeze can now be calculated using the formula:

Density=Total MassTotal Volume\text{Density} = \frac{\text{Total Mass}}{\text{Total Volume}}

The total volume is 188 litres, which is equivalent to 188000 cm³:

Densityantifreeze=77816g188000cm30.4145g/cm3\text{Density}_{\text{antifreeze}} = \frac{77816\,\text{g}}{188000\,\text{cm}^3} \approx 0.4145\,\text{g/cm}^3

Rounding this to two decimal places gives:

Densityantifreeze=0.41g/cm3\text{Density}_{\text{antifreeze}} = 0.41\,\text{g/cm}^3

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