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torisob [31]
4 years ago
9

Calculate the freezing point (Kf = 1.86 °C/m) of a 2.1m aqueous solution of KCl.

Chemistry
1 answer:
Wittaler [7]4 years ago
7 0

Answer:

The freezing point of the solution is - 3.90 °C.

Explanation:

  • We can solve this problem using the relation:

<em>ΔTf = (Kf)(m),</em>

where, ΔTf is the depression in the freezing point.

Kf is the molal freezing point depression constant of water = - 1.86 °C/m,

m is the molality of the solution (m = 2.1 m).

<em>∴ ΔTf = (Kf)(m)</em> = (-1.86 °C/m)(2.1 m) =<em> - 3.90 °C.</em>

<em>∵ The freezing point if water is 0.0 °C and it is depressed by - 3.90 °C.</em>

<em>∴ The freezing point of the solution is - 3.90 °C.</em>

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5. Hydrogen &amp; oxygen react chemically to form water. How much water
stepladder [879]

39.25 g of water (H₂O)

Explanation:

We have the following chemical reaction:

2 H₂ + O₂ → 2 H₂O

Now we calculate the number of moles of each reactant:

number of moles = mass / molar weight

number of moles of H₂ = 14.8 / 2 = 7.4 moles

number of moles of O₂ = 34.8 / 32 = 1.09 moles

We see from the chemical reaction that 2 moles of H₂ will react with 1 mole of O₂ so 7.4 moles of H₂ will react with 3.7 moles of O₂ but we only have 1.09 moles of O₂ available. The O₂ will be the limiting reactant. Knowing this we devise the following reasoning:

if        1 moles of O₂ produces 2 moles of H₂O

then  1.09 moles of O₂ produces X moles of H₂O

X = (1.09 × 2) / 1 = 2.18 moles of H₂O

mass = number of moles × molar weight

mass of H₂O = 2.18 × 18 = 39.25 g

Learn more about:

limiting reactant

brainly.com/question/7144022

brainly.com/question/6820284

#learnwithBrainly

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