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Secondary IV • 1yr.

Good evening,

I have a problem with the following question:

"In increasing order of concentration, order the following solutions:

Sol. A: 0.1 g / L of CaCO3 in water

Sol. B: 5% m / V solution of ethanol in water

Sol. C: solution of 3 mol of CO2 in 60 mol of air

Sol. D: solution of 560 ppm "

Thank you for your help

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Explanations (1)

  • Explanation from Alloprof

    Explanation from Alloprof

    This Explanation was submitted by a member of the Alloprof team.

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    Team Alloprof • 1yr.

    Thank you for your question!


    To answer it, we must express all of the given concentrations the same way. In this case, switching everything to ppm is probably the easiest since the concentration of solution D is already expressed in ppm, and that solution C is easily transformable into ppm.


    First, let's start by converting the concentration of solution A to ppm. We know that there is 0.1 g of CaCO3 per 1000 mL of water. Since, under normal conditions, 1 mL of water ≈ 1 g (Note: this approximation is only valid for water!), We can consider that there is 0.1 g of CaCO3 per 1000 g of water. Thus, we can find the concentration of solution A in ppm:

    C = msolute / msolution • 1000000

    C = 0.1g / 1000g • 1000000 = 100ppm


    Then, we can find the concentration of solution B in ppm. We know that there is 5% ethanol in terms of mass/volume in an aqueous solution. Using the reasonable approximation of 1 mL of water ≈ 1 g, all it remains to do is to convert the concentration in its decimal form and multiply it by 1 million to obtain the concentration in ppm:

    é

    C = msolute / msolution • 1000000

    C = (5% / 100) / 1000g • 1000000 = 50ppm


    Finally, we can find the concentration of solution C in ppm. We know that there is 3 mol of carbon dioxide (CO2) for 60 mol of air. (I believe here that air means "dinitrogen" (N2). I remember seeing this number somewhere, and otherwise, the question wouldn't make sense!).

    Knowing that the molar mass of carbon dioxide is 44.01 g/mol, we conclude that


    3molCO2 = 3 • 44.01g/mol = 132.03g

    Then, knowing that the molar mass of dinitrogen is 28.0134 g/mol, we conclude that

    60molN2 = 60 • 28,0134g/mol = 1680,804g

    To find the total mass of the solution (we must include the mass of the solute because it is not negligible compared to the one of the solution), we add the respective masses of carbon and nitrogen:

    msolution = 132.03g + 1680.804g = 1812.834g

    Then, it only remains to apply the ppm formula to find the concentration of the solution:

    é

    C = msolute / msolution • 1000000

    C = (132.03g / 1812 , 834g) • 1000000≈72800ppm


    In short, the following concentrations are observed:

    [SolutionA] = 100ppm