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   ocr: CHAPTER 13 Answers to Solutions Properties of Solutions 13.48 AT = K(m) first calculate the molality of the solute particles. (a) 0.15 m (b) 20.0 0.350 9 kg C12H26 CCl X 170.3 1mol g C12H28 C12H26 = 0.336 m (c) 2.75 0.0800 9 kg MgClz H20 X 1r 95.21 mol gi MgClz MgClz X 1 3 mol moi ions MgCi2 = 1.08 m Then, f.p. = T,- K(m) : bp.=T,+h Kp(m) : Tin°C m T, -K,(m) f.p. T. +K,(m) b.p. (a) 0.15 -114.6 1.99(0.15) = -0.30 -114.9 78.4 1.22(0.15) = 0.18 78.6 (b) 0.336 -22.3 29.8(0.336) = -10.0 -32.3 76.8 5.02(0.336) == 1.69 78.5 (c) 1.08 0.0 1.86(1.08) == -2.01 -2.0 100.0 0.52(1.08) == 0.56 100.0 13.49 0. ...