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A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If your car's cooling system holds 4.40 gallons, what is the boiling point of the solution?
Make the following assumptions in your calculation: at normal filling conditions, the densities of engine coolant and water are 1.11 g/mL and 0.998 g/mL respectively. Assume that the engine coolant is pure ethylene glycol (HOCH2CH2OH), which is non-ionizing and non-volatile, and that the pressure remains constant at 1.00 atm. Also, you'll need to look up the boiling-point elevation constant for water.

Respuesta :

Answer:

109.09°C

Explanation:

Given data:

The capacity of the car cooling system =  4.40 gallons

Ratio of water and solute = 50/50

thus,

Volume of solute = Volume of water = 4.40/ 2 gallons = 2.20 gallons

thus, the Mass of solute, M₁ = Density × Volume = 1.1 g/mL × [tex]2.20\times\frac{3785.41\ \textup{mL}}{1\ \textup{gallon}}[/tex] = 9160.69 grams

Mass of water = 0.998 g/mL × [tex]2.20\times\frac{3785.41\ \textup{mL}}{1\ \textup{gallon}}[/tex] = 8310.346 grams

molality , m = [tex]\frac{mass\ of \ solute\ \times 1000}{molar\ mass\ of\ solute\times\ mass\ of\ water}[/tex]

on substituting the values, we get

m = [tex]\frac{9160.69 \times 1000}{62.07\times8310.346}[/tex] =  17.759 m

Now, the elevation in boiling point is given as:

[tex]\Delta T_b = k_bm[/tex]

where,

[tex]k_b[/tex] is a constant = 0.512 °C/m

on substituting the values we get

Boiling point = 100°C + 17.759 × 0.512= 109.09°C

annyksl

The boiling point of the solution will be equal to 109.09°C.

We can arrive at this answer as follows:

First, we must take into account that the water and solute ratio is 50/50, therefore, we must divide the capacity of the car's cooling system by 2, to know the volume of solute and solvent, which will be equal per account of the reason between them.

  • Therefore, we can calculate:

[tex]\frac{4.40}{2} = 2.20[/tex] gallons

  • After that, we can calculate the mass of the solute with the following formula:

[tex]Density * Volume\\1.1*(2.20*3785.41)\\9160.69 g[/tex]

  • It will also be necessary to calculate the mass of water and for that, we will use the same formula. Therefore, the result will be:

[tex]0.998*(2.20*3585.41)\\8310.346 g[/tex]

  • Now that we know the value of the masses, we can calculate the molality. In this calculation, let's consider that the mass of the solute will be represented by the symbol "[tex]M_S[/tex]" while the mass of water will be represented by the symbol "[tex]M_W[/tex]". With that, we can follow the formula:

[tex]Molality = \frac{M_s*1000}{molar M_S*M_W}\\\\Molality = \frac{9160.69*1000}{62.07*8310.34} \\\\Molality= 17.759 m[/tex]

  • Now, we can calculate the boiling point elevation using the formula:

[tex]\Delta T= K_b*molality\\[/tex]

We must take into account that [tex]K_b[/tex] is a constant and its value is always 0.512 °C/m.

  • With that, we can calculate:

[tex]\Delta T= 0.512*(100C + 17.759)\\\Delta T= 109.09C[/tex]

More information:

https://brainly.com/question/20659868?referrer=searchResults

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