Question

Question #2 -/1 points How many grams of calcium nitrate need to be dissolved in 75 mL of water to form a solution that has a freezing point of -3.2 deg C? grams of calcium nitrate Evaluate Question #3 -/1 points What is the freezing point in °C) of a solution of 8.0 g of carbon tetrachloride, CCl4, in 80.0 g of benzene, C6H6. For benzene, Tp = 5.5°C and Kf = 5.12 deg kg mole?? Freezing point of solution degrees C Fvaluate

Answer 1

(degree of dissociation) Selute is calcium Nitrate Ca(Noul to call + 2NO- į = No. of cons furnished by the electrolyte Van't Hoff factor i= 3 (assume o&=1) solvent in water (density of water) Vuciter = 75ml, Sro = Ig/me 9th = 75g so By freezing point Depression (collegative property) AT = TG - If = 1km cryolepic constant of freezing print freezing of pure water point of JHO 01 grams of Water. Water = 1.86 km solution

0-(-3.2) C = 3 x 1086 * naman Ca(NO) 09. (9) (ma motality = noio) moe, falute exlooo gramus of solrent , Mealno, x75 3.2 3 x1.86 Ca(ND) < 40 +14+ ?X16) X2 Molar man of = 164 g/mole Calcium nitrate 3,2 x 164 x 759 2X1,86 Tooo ICANNO) 7.0599 = Icaro, for cely call covalent compound) solute Carbon Tetrachlorid Q3) OTE= like m Solvent → Benzone Te-To - IX 5912 °C Kg x 8 x100$ 'f mole 154 x 8 Molar mass of ecly Le 5.5-le = 5.12 x 100 = 3.324 154 - 5.5-3.324 -1.675 (Water) - 0 - - 0,4 ) = 0. 16 = x(x m = {x 7+ x 6 = 26xIglme for freezing points of fruse water lovalents compound (i-1) [tho # 269] 0.46°C = 1X1.86 °C K2 * 0.767 x/000 - M X 26 IM = 119,28 g/mole) nolarmers.

dorlarut Plotic acid (Chcoon) (K4)CH COOH = ? Т: • с °C = 1.049 , Solution tоr, сон) да, « ... /ы Schute water - 89 = 940 Моңсем - 3o44, 1, сон - Vу, свон * Грсон = 36 ( x ). оч1 2 дисон” 320.11*, / freezing point of solution А, г. - = тіс 4. «С - (-22-4 с) = 39.2°C Freezing pruit of from acetic acid 39.2 = A - іх ах 2 хоор 18x320.944 34. 2 х 12 х 320.944 кка . (*) Covн #оор 28.312 к ка/me. - (+ слоя |