25 grams of a compound with a formula weight of 110g/mol was dissolved in 500mL

Question

25 grams of a compound with a formula weight of 110g/mol was dissolved in 500mL of water, and the freezing point of the aqueous solution decreased to -1.68 degrees C. what does the formula of the compound look like? the freezing point depression constant of water is 1.86degreeesC/m. assume that the density of water is 1.00g/mL. please show all work. 25 grams of a compound with a formula weight of 110g/mol was dissolved in 500mL of water, and the freezing point of the aqueous solution decreased to -1.68 degrees C. what does the formula of the compound look like? the freezing point depression constant of water is 1.86degreeesC/m. assume that the density of water is 1.00g/mL. please show all work. 25 grams of a compound with a formula weight of 110g/mol was dissolved in 500mL of water, and the freezing point of the aqueous solution decreased to -1.68 degrees C. what does the formula of the compound look like? the freezing point depression constant of water is 1.86degreeesC/m. assume that the density of water is 1.00g/mL. please show all work.

Explanation / Answer

Molality = (25/110)*(1000/500) = 0.45

Depression in freezing point = i*kf*molality = 1.68

=> i*1.86*0.45 = 1.68

=> i = 2 (appx)

It implies that the compound splits into two ions.

The formula looks like AB which in solution splits into A+ and B-

Formula of compound is AB

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