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VikaD [51]
3 years ago
14

A certain molecular compound M has a solubility in acetone of 0.737g/mL at 10.°CCalculate the mass of M that's dissolved in 9.0L

of a saturated solution of M in acetone at this temperature.Be sure your answer has the correct unit symbol and number of significant digits.
Chemistry
1 answer:
Paha777 [63]3 years ago
5 0

Answer:

6633 grams of M are dissolved in 9L

Explanation:

If the compound M has a solubility in acetone of 0.737 g/mL, means that in 1 mLof solution, 0.737 grams of solute is dissolved.

Let's make a rule of three:

9 L = 9000 mL

In 1 mL ___ 0.737 g of M is dissolved

In 9000 mL ___ (9000 . 0.737) /1 = 6633 grams

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Which aqueous solution has the highest boiling point at standard pressure?(1) 1.0 M KC1(aq) (3) 2.0 M KCl(aq)(2) 1.0 M CaC12(aq)
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(4) 2.0 M CaCl₂(aq).

Explanation:

  • Adding solute to water elevates the boiling point.
  • The elevation in boiling point (ΔTb) can be calculated using the relation:

<em>ΔTb = i.Kb.m,</em>

where, ΔTb is the elevation in boiling point.

i is the van 't Hoff factor.

  • van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

Kb is the molal elevation constant of water.

m is the molality of the solution.

<u><em>(1) 1.0 M KCl(aq):</em></u>

i for KCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

suppose molarity = molality, m = 1.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (2)(Kb)(1.0 m) = 2(Kb).

<u><em>(2) 2.0 M KCl(aq):</em></u>

i for KCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

suppose molarity = molality, m = 2.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (2)(Kb)(2.0 m) = 4(Kb).

<u><em>(3) 1.0 M CaCl₂(aq):</em></u>

i for CaCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

suppose molarity = molality, m = 1.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (3)(Kb)(1.0 m) = 3(Kb).

<u><em>(4) 2.0 M CaCl₂(aq):</em></u>

i for CaCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

suppose molarity = molality, m = 2.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (3)(Kb)(2.0 m) = 6(Kb).

  • <em>So, the aqueous solution has the highest boiling point at standard pressure is: (4) 2.0 M CaCl₂(aq).</em>

<em></em>

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Explanation:

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In the open chain form, monosaaccharides have a carbonuyl group in one of their chains. If the carbonyl group is in the form of an aldehyde group, the monosaccharide is an aldose; if the carbonyl group is in the form of a ketone group, the monosaccharide is known as a ketose. glucose is an aldose while fructose is a ketose.

In D-glucose, there is an aldehyde functional group, and the carbonyl group is at carbon-1 when looking at the Fischer projection.

In D-fructose, there is a ketone functional group, and the carbonyl group is at carbon-2 when looking at the Fischer projection.

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