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Question: The number of carbon atoms present in of acetone (density \(1.2\dfrac{g}{{c{m^3}}}\)) at 300 K is: \...

The number of carbon atoms present in of acetone (density 1.2gcm31.2\dfrac{g}{{c{m^3}}}) at 300 K is: 2.0cm32.0c{m^3}
A.0.083 N
B.0.042 N
C.0.124N
D.0.166 N

Explanation

Solution

Acetone is a ketone and its molecular formula is CH3COCH3C{H_3}COC{H_3}. It means that in one molecule of acetone, the number of carbon atoms present are 3. The formula, Density=MassVolumeDensity = \dfrac{{Mass}}{{Volume}} will be used to calculate the mass of the acetone and the mass will be converted to moles.

Complete step by step answer:
Given,
Volume of acetone =2.0cm3 = 2.0c{m^3}
Density of acetone =1.2gcm3 = 1.2\dfrac{g}{{c{m^3}}}
We know
Density=MassVolumeDensity = \dfrac{{Mass}}{{Volume}}
Therefore, Mass=Density×VolumeMass = Density \times Volume
Mass of acetone =2×1.2=2.4g = 2 \times 1.2 = 2.4g
We know that
Number of moles =WeightMolermass = \dfrac{{Weight}}{{Moler\,mass}}
Molar mass of acetone =CH3COCH3 = C{H_3}COC{H_3}
=12+(1×3)+12+16+12+(1×3)=58g= 12 + \left( {1 \times 3} \right) + 12 + 16 + 12 + \left( {1 \times 3} \right) = 58g
Number of moles of acetone =2.458=4.14×102 = \dfrac{{2.4}}{{58}} = 4.14 \times {10^{ - 2}}
Number of moles of Carbon atoms =3×4.14×102=0.124 = 3 \times 4.14 \times {10^{ - 2}} = 0.124
So, the number of carbon atoms present in 2.0cm32.0c{m^3} of acetone is 0.124.

Therefore, the correct answer is option (C).

Note: Acetone also known as dimethyl ketone. Its chemical formula is CH3COCH3C{H_3}COC{H_3}. It is a member of the ketone family. It is a colourless, volatile and flammable liquid and has a fruity and sweetish odour. Acetone may also be present in urine and blood in small quantities. It is used in the nail polish remover. It is also used to make drugs, plastics. Fibres etc. It is a manufactured chemical but also occurs naturally in plants, trees, forest fires. Its melting point is 95.4C - {95.4^\circ }C and boiling point is 56.53C{56.53^\circ }C. Flash point of acetone is 0F{0^\circ }F.