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erastovalidia [21]
3 years ago
10

A liquid solvent is added to a flask containing an insoluble solid. The total volume of the solid and liquid together is 80.0 mL

. The liquid solvent has a mass of 21.0 g and a density of 0.865 g/mL. Determine the mass of the solid given its density is 2.00 g/mL.
Chemistry
1 answer:
Lyrx [107]3 years ago
8 0

Answer:

111.44 g

Explanation:

We'll begin by calculating the volume of the solvent. This can be obtained as follow:

Mass of solvent = 21 g

Density of solvent = 0.865 g/mL.

Volume of solvent =?

Density = mass /volume

0.865 = 21 /volume of solvent

Cross multiply

0.865 × volume of solvent = 21

Divide both side by 0.865

Volume of solvent = 21 / 0.865

Volume of solvent = 24.28 mL

Next, we shall determine the volume of the solid. This can be obtained as follow:

Volume of solvent + solid = 80.0 mL.

Volume of solvent = 24.28 mL

Volume of solid =?

Volume of solid = (Volume of solvent + solid) – (Volume of solvent)

Volume of solid = 80 – 24.28

Volume of solid = 55.72 mL

Finally, we shall determine the mass of the solid. This can be obtained as follow:

Density of solid = 2.00 g/mL.

Volume of solid = 55.72 mL

Mass of solid =.?

Density = mass / volume

2 = mass of solid / 55.72

Cross multiply

Mass of solid = 2 × 55.72

Mass of solid = 111.44 g

Therefore, the mass of the solid is 111.44 g

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At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+

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M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M

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n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-

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M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M

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