Answered

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If 75 g. of Potassium Chloride (ionic compound) is dissolved in 250 grams of
water, what will be the freezing point of the solution? Kf = 1.86

Sagot :

sebassandin

Answer:

[tex]T_{f,sol}=-15.0\°C[/tex]

Explanation:

Hello there!

In this case, for these problem about the colligative property of freezing point depression, it is possible set up the following equation:

[tex]T_{f,sol}-T_{f,water}=-i*m*Kf[/tex]

Whereas the van't Hoff's factor, i, is 2 since KCl is ionized in two ions (K+ and Cl-); and the molality (m) of the solution is computed by:

[tex]m=\frac{75g*\frac{1mol}{74.55g} }{250g*\frac{1kg}{1000g} } \\\\m=4.02mol/kg[/tex]

Thus, since the freezing point of water (ice) is 0°C, we obtain the following freezing point of the solution by plugging in:

[tex]T_{f,sol}=-(2)(4.02mol/kg)(1.86\°C/mol*kg)\\\\T_{f,sol}=-15.0\°C[/tex]

Best regards!

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