Calculate the freezing point of an aqueous solution formed by dissolving 4.22 g

Question

Calculate the freezing point of an aqueous solution formed by dissolving 4.22 g of KCl in 25.00 mL of water. Kf for water is -1.86 degrees C/m. Assume the density of water is 1 g/mL. (Hint: Use the idea van't Hoff factor, i.)

Explanation / Answer

KCL ---> K^+ + Cl^- therefore i=2............ mol solute = 4.22 / 74.5 = 0.05664 mole........... density of water = 1 g / ml......... so,mass of water = 25 ml x 1 g / ml = 25 g = 25 / 1000 kg = 1/40 kg......... molality of solution (m ) = mol solute / mass of water = 0.05664/(1/40 ) =2.2656.... delta(T) = i m Kf = 2 x 2.2656 x 1.86 =8.428 oC...... Freezing point (T ) = 0.00 - 8.428 = - 8.428 oC.......

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