This question is from an old test that I took and I gotwrong. The answer is -7.8

Question

This question is from an old test that I took and I gotwrong. The answer is -7.86 degrees C not-2.62 degrees C. But please show me how I can get to thatanswer. Thank you! If freezing point depression constant (Kfp)of water is -1.86 C/m, at what temperature will a solution preparedby dissolving 200. g of Na2SO4 (molar mass142.05 g/mol) into 1.00 kg of water freeze (assuming no ion pairingin the solution)? This question is from an old test that I took and I gotwrong. The answer is -7.86 degrees C not-2.62 degrees C. But please show me how I can get to thatanswer. Thank you! If freezing point depression constant (Kfp)of water is -1.86 C/m, at what temperature will a solution preparedby dissolving 200. g of Na2SO4 (molar mass142.05 g/mol) into 1.00 kg of water freeze (assuming no ion pairingin the solution)?

Explanation / Answer

Molality of sodium sulfate =( mass / molar mass ) * 1 kg
                                       =( 200 g / 142.05 g/mol) * 1 kg                                        =1.4079 m 1 mole of Sodium sulfate ionizes to give 2 mole of sodium ionsand one mole of sulfate ions. So depression in freezing point = 3 * Kf. m                                                =3 * 1.86 C/m * 1.4079 m                                                 =7.86 C So freezing point = 0 - 7.86 C = - 7.86 C

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