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lbvjy [14]
3 years ago
6

What is the boiling point of a solution that contains 3 moles of KBr in 2000 g of water? (Kb = 0.512 C/m)

Chemistry
1 answer:
pochemuha3 years ago
4 0

Answer:

The answer to your question is   Tb = 100.768°C

Explanation:

Data

Tb = ?

moles of KBr = 3

mass of water = 2000 g

Kb = 0.512 C/m

Process

1.- Calculate the molality of the solution

molality = moles of solute/mass of solvent (kg)

-Substitution

molality = 3/2

-Result

molality = 1.5

2.- Calculate ΔT

    ΔTb = Kbm

-Substitution

    ΔTb = (0.512)(1.5)

-Result

    ΔTb = 0.768

3.- Calculate the boiling point, the boiliing point of pure water is 100°C

   ΔTb = Tb - Tb(solvent)

-Solve for Tb

   Tb = Tb(solvent) + ΔTb

-Substitution

   Tb = 100 + 0.768

-Result

  Tb = 100.768°C

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Answer:

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B) 13-karat

Explanation:

A) From the problem we have

<em>1)</em> Mg + Ms = 9.40 g

<em>2)</em> Vg + Vs = 0.675 cm³

Where M stands for mass, V stands for volume, and g and s stand for gold and silver respectively.

We can rewrite the first equation using the density values:

<em>3)</em> Vg * 19.3 g/cm³ + Vs * 10.5 g/cm³ = 9.40

So now we have<em> a system of two equations</em> (2 and 3) <em>with two unknowns</em>:

We <u>express Vg in terms of Vs</u>:

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  • Vg = 0.675 - Vs

We <u>replace the value of Vg in equation 3</u>:

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  • (0.675-Vs) * 19.3 + Vs * 10.5 = 9.40
  • 13.0275 - 19.3Vs + 10.5Vs = 9.40
  • -8.8 Vs + 13.0275 = 9.40
  • <u>Vs = 0.412 cm³</u>

Now we <u>calculate Vg</u>:

  • Vg + Vs = 0.675 cm³
  • Vg + 0.412 cm³ = 0.675 cm³
  • Vg = 0.263 cm³

We <u>calculate Mg from Vg</u>:

  • 0.263 cm³ * 19.3 g/cm³ = 5.08 g

We calculate the mass percentage of gold:

  • 5.08 / 9.40 * 100% = 54.04%

B)

We multiply 24 by the percentage fraction:

  • 24 * 54.04/100 = 12.97-karat ≅ 13-karat
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