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djyliett [7]
1 year ago
6

If a 66.5 g sample of ammonium nitrate (NH4NO3) is dissolved in enough water to make 315 mL of solution, what will be the molari

ty (M)?
Chemistry
1 answer:
11111nata11111 [884]1 year ago
7 0

Answer:

2.64 M

Explanation:

To find the molarity, you need to (1) convert grams to moles (via molar mass), then (2) convert mL to L, and then (3) calculate the molarity (via molarity ratio). The final answer should have 3 sig figs to match the sigs figs of the given values.

(Step 1)

Molar Mass (NH₄NO₃): 2(14.007 g/mol) + 4(1.008 g/mol) + 3(15.998 g/mol)

Molar Mass (NH₄NO₃): 80.04 g/mol

66.5 grams NH₄NO₃                1 mole
---------------------------------  x  ----------------------  =  0.831 moles NH₄NO₃
                                             80.04 grams

(Step 2)

1,000 mL = 1 L

 315 mL                1 L
--------------  x  ------------------  =  0.315 L
                        1,000 mL

(Step 3)

Molarity = moles / volume

Molarity = 0.831 moles / 0.315 L

Molarity = 2.64 M

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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)
miss Akunina [59]

Answer:

(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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Polymeric MDI is a mixture of. monomeric MDI as well as larger molecular weight oligomers of MDI, and is a brownish. liquid at room temperature and may have a slight odor. Commercial MDI products are. often mixtures of monomeric and polymeric MDI and can contain other additives as well.

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The reaction described in Part A required 3.40 L of sodium chloride. What is the concentration of this sodium chloride solution
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