Question

The following solutions of ethanol (C₂H₆O, M = 46.0 g mol^-1 ; d = 0.790 g mL^-1 ) were mixed:

• 50.0 mL of concentration 8.50 % (v/v)

• 75.0 mL of concentration 72.0 g L-1

• 1.35 L of concentration 950 ppm

• 8.00 x 104 μL of concentration 25.0 mg mL-1

• 125 mL of concentration 1.25 M

Assuming that mixing has no effect on the total volume, what is the concentration of ethanol in the final solution, expressed as % (m/v)?

Answer 1

Solution 1:-

50.0 mL solution having 8.50% v/v ethanol.

V₁ = 50.0 mL

So, 8.50% of 50 mL = (8.50/100)(50 mL) = 4.25 mL

Volume of ethanol = 4.25 mL

Density of ethanol = 0.790 g mL^-1

Mass of ethanol, m₁ = Volume x Density = (4.25 mL)(0.790 g mL^-1) = 3.3575 g

So, m₁ = 3.3575 g

Solution 2:-

75.0 mL of concentration 72.0 g L^-1

Volume of solution, V₂ = 75.0 mL

Now, 1 L = 1000 mL

Volume of solution = (75.0 mL)(1 L/1000 mL) = 75.0/1000 L = 0.0750 L

So, amount of ethanol in the solution of the given concentration = (72.0 g L^-1)(0.0750 L) = (72.0 x 0.0750) g = 5.4 g

or m₂ = 5.4 g

Solution 3:-

1.35 L of concentration 950 ppm

950 ppm = 950 parts per million

which means in 1 million mL of solution, there is 950 mL of ethanol present.

or 1 million mL = 10⁶ mL has 950 mL of ethanol in it.

So, 1 mL of solution will have 950/10⁶ mL of ethanol.

Therefore, "V mL" of solution will have (950/10⁶)V mL of ethanol.

Also, 1 L = 1000 mL = 10³ mL

Given Volume of Solution, V₃ = 1.35 L = 1.35(10³ mL) = 1350 mL

Here, V = V₃

So, Volume of ethanol in it = (950/10⁶)V₃ mL = (950/10⁶)1350 mL = 1.2825 mL

Therefore, Mass of ethanol = Volume x Density = (1.2825 mL)(0.790 g mL^-1) = 1.013175 g

or m₃ = 1.013175 g

Solution 4:-

8.00 x 10⁴ μL of concentration 25.0 mg mL^-1

1 μL = 10^-6 L

Also, 1 L = 1000 mL

So, 1 μL = 10^-6(1000 mL) = 10^-3 mL

So, Volume of solution, V₄ = 8.00 x 10⁴ μL = (8.00 x 10⁴ μL)(10^-3 mL/1 μL) =  8.00 x 10⁴ x 10^-3 mL = 8.00 x 10^-1 mL = 0.800 mL

Mass of ethanol in solution, m₄ = (concentration)(volume) = (25.0 mg mL^-1)(0.800 mL) = 20 mg

Now, 1 g = 1000 mg

So, m₄ = (20 mg)(1 g/1000 mg) = 20/1000 g = 0.020 g

Solution 5:-

125 mL of concentration 1.25 M

Volume of solution, V₅ = 125 mL

Concentration = 1.25 M = 1.25 mol L^-1

Now, 1 L = 1000 mL

So, V₅ = (125 mL)(1 L/1000 mL) = 125/1000 L = 0.125 L

So, Number of moles of ethanol, n = Concentration x Volume = (1.25 mol L^-1)(0.125 L) = (1.25 x 0.125) mol = 0.15625 mol

Molar Mass of ethanol, M = 46.0 g mol^-1

So, Mass of ethanol present in solution, m₅ = n.M = (0.15625 mol)(46.0 g mol^-1) = 7.1875 g

or m₅ = 7.1875 g

Now, when these 5 solutions are mixed, the final:-

Volume of Solution, V = V₁ + V₂ + V₃ + V₄ + V₅

Mass of ethanol in the solution, m = m₁ + m₂ + m₃ + m₄ + m₅

So, V = (50.0 mL) + (75.0 mL) + (1350 mL) + (0.8 mL) + (125 mL) = 1600.8 mL

And, m = (3.3575 g) + (5.4 g) + (1.013175 g) + (0.020 g) + (7.1875 g) = 16.978175 g

Concentration of solution, C = m/V = (16.978175 g)/(1600.8 mL) = 0.0106 (m/v)

or C = 0.0106 x 100 %(m/v)

or

C = 1.06 %(m/v)

Therefore, the concentration of ethanol in the final solution is 1.06 %(m/v).