What is the freezing point of radiator fluid that is 50% antifreeze by mass? k f for water is 1.86 ∘ c/m?

Question

What is the freezing point of radiator fluid that is 50% antifreeze by mass? k f for water is 1.86 ∘ c/m?

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2022-04-08T17:43:49+00:00 0 Answers 0

Answers ( 2 )

  1. Allison
    0
    2022-04-08T17:44:53+00:00

    Answer:

    -30.02 u00baC

    Explanation:

    Assuming the antifreeze to be ethylene glycol (Cu2082Hu2086Ou2082) which is popular antifreeze.

    Molar mass of ethylene glycol (Cu2082Hu2086Ou2082) = 62g/mol

    Step 1: calculate the freezing point depression of the solution

    u0394T = -Kf*M

    where,

    u0394T= depression in the freezing point.

    M = the molarity of the solution (mol solute / Kg solvent)

    Kf = molar freezing point constant of water = 1.86u00b0C/m

    To determine depression in the freezing point (u0394T), first we need to calculate;

    • molarity of solute (ethylene glycol) in mol
    • mass of solvent (water) in kg
    • molarity of the solution (water +ethylene glycol)

    Step 2: calculate the molarity of the solute (ethylene glycol)

    Molar mass ethylene glycol = 62 g/mol

    molarity of ethylene glycol in mol = 50 g / 62g/mol = 0.807 mol

    Step 3: calculate mass of solvent in kg

    There is 1kg of ethylene glycol which is present in 1kg of water

    mass of solvent (water) in kg= 50 g/ 1000 g/ Kg = 0.050 Kg

    Step 4: calculate the molarity of the solution (M)

    M = 0.807 mol / 0.050 Kg = 16.14 m

    Step 5: calculate the freezing point depression of the solution (u0394T)

    u0394T = – Kf*M = -1.86 u00baC/m x 16.14 m

    = -30.02 u00baC

  2. Myna
    0
    2022-04-08T17:45:32+00:00

    Ethylene glycol is termed as the primary ingredients in antifreeze.
    The ethylene glycol molecular formula is Cu2082Hu2086Ou2082.
    Molar mass of Cu2082Hu2086Ou2082 is = (2×12) +(6×1) + (216) = 62g/mol
    Now that antifreeze by mass is 50%, then there is 1kg of ethylene glycol which is present in 1kg of water.
    u0394Tf = Kfxm
    u0394Tf = depression in the freezing point.
    = freezing point of water freezing point of the solution
    = Ou00b0c – Tf
    = -Tf
    Kf = depression in freezing constant of water = 1.86u00b0C/m
    M is the molarity of the solution.
    =(mass/molar mass) mass of solvent in kg
    =1000g/62 (g/mol) /1kg
    =16.13m
    If we plug the value we get
    -Tf = 1.86x 16.13 = 30
    Tf = -30u00b0c

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