Question

E GASES, LIQUIDS, AND SOLIDS Calculating and using the van't Hoff factor for electrolytes When 57.5 g of alanine (CH,NO) are dissolved in 1450. g of a certain mystery liquid X, the freezing point of the solution is 2.1 °C lower than the freezing point of pure X. On the other hand, when 575 g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is 29 °C lower than the freezing point of pure X. Calculate the van't Hoff factor for potassium bromide in X. Be sure your answer has a unit symbol, if necessary, and round your answer to 2 significant digits.

Answer 1

Molecular weight of G H NOR (Mm) - 8x12)+(1 * 7) + (1x 4) +(2016) or, M. – 89 gm.not weight of aniline (wg) = 57. 5 gm weigh of X (wi) = 1450 gm We know that, depression of freezing point (ATF) = ixkex W2 x 1000 M₂ x w, for aniline So, IT f i=vant - Hoff factor = 1 57.5.81000 = 1 x kg x 5 F* 89 X 1450 57500 or 201 = kg x 89 x 1450 Or os Kf = 57500 0 2.1 x 89 x 1450 - = 4.713 °C. motal! on the other hand, Molecular weight of KBr (M2) = 119 gen. most weight of KBr (W2) = 57.5 gen. X (W) = 1450 gm д; - іх к, х Ш х 10р M₂ xw, Maxw, 119 1450 W2 x 1000 4.H3 57.5*1880 = 1.846 ~ 1.85 so, i=1.85 Now 2.