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IRISSAK [1]
3 years ago
14

A student dissolves 7.9 g of stilbene (C14H12) in 475. mL of a solvent with a density of 1.03 g/mL. The student notices that the

volume of the solvent does not change when the stilbene dissolves in it. Calculate the molarity and morality of the students solution. Round both of your answers to 2 significant digits.
Chemistry
1 answer:
andrey2020 [161]3 years ago
6 0

Answer:

Molarity: 0.092M

Molality: 0.090m

Explanation:

Molarity, M, is defined as the moles of solute (In this case, C14H12 -Molar mass: 180.25g/mol-) in 1L of solution.

The molality, m, are moles of solute per kg of solvent.

<em>Molarity:</em>

<em>Moles solute: </em>

7.9g * (1mol/180.25g) = 0.04383 moles

<em>Liters solution:</em>

475mL = 0.475L

Molarity: 0.04383 moles / 0.475L = 0.092M

<em />

<em>Molality:</em>

<em>kg solvent:</em>

475mL * (1.03g/mL) = 489.25g = 0.48925kg

Molality:

0.04383 moles / 0.48925kg = 0.090m

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2 years ago
And with solution...
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Answer:

The answer to your question is: V = 6.93 L

Explanation:

Data

N₂ = 5.6 g

Volume of NH₃ = ?

                              14 g of N   ----------------  1 mol

                              5.6 g -----------------------   x

                             x = (5.6 x 1) / 14 = 0.4 mol of N

Reaction

                                N₂    +     3H₂    ⇒    2NH₃

                                1 mol of N₂   ----------------  2 moles of NH₃

                                0.4 mol of N₂ --------------   x

                               x = (0.4 x 2) / 1

                               x = 0.8 mol of NH₃

Formula

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Substitution

                     V = (0.8)(0.082)(723) / 6.84

                     V = 6.93 L

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3 years ago
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Answer:

Insoluble in water:

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