A 29.300% by mass solution of Strontium Fluoride will freeze at what temperature? Kf water is 1.86 C/m.?

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  • 4 weeks ago

    We need the molality of the SrF2 solution. To do that, we first assume 100 g of the solution is present.

    29.300%(w/w) means 29.3 g of SrF2 in the 100 g of solution as well as 70.7 g of water.

    Since molality involves moles of solute, we do this:

    29.3 g / 125.62 g/mol = 0.233243 mol

    Now, for the molality:

    0.233243 mol / 0.0707 kg = 3.29905 m

    Now, the freezing point calculation:

    Δt = (3) (1.86) (3.29905) <--- the 3 is the van 't Hoff factor

    Δt = 18.408699

    The solution will freeze at -18.4 C

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