Question

d. From the graph determine the freezing point of Pure Water: Unknown Solution: e. Calculate the freezing point depression of the solution AT f. Calculate the molality of the solution (m) g. Calculate the molar mass of the unknown. 2. What are colligative properties? List three colligative properties. 3. What effect does the presence of a nonvolatile solute have upon the boiling point of a solution? 4. What mass of glucose (CH1,.) should be dissolved in 10.00 kg of water to obtain a solution with a freezing point of -4.2°C? 5. How many grams of NaNO, would you add to 500g of H 0 to prepare a solution of 0.500 m NaNO,?

Answer 1

@ freezing point of puse water (Tº) - oc freezing point of unknown son (If) = -1.11% - Freezing beint debsession OTE= Tfº-TF ATF = 213-271.2916 [off = lilk) 1 °°+273 K = 273K (-1.77€ + 2136 = 271.29K ② ATf is disectly proportional to the modality of soth ATfam OR ATf = Kfm 0 Where. Ky is the more freezing point depression constant of solvent (water) & m is the malality for water Kf= 1.86 K kg mot substitute the value of Kft.Atf in equo TIK= 186k kg moil x m e m= 111 k kg mol = 0.92 m (mockg) – 1.86 K. 6 Relationship b/w DTf 4 molar mass of unanrum is gien by ean STS = kx WB KORD When We Given weight MaxWA * Црът WAT weight of solvent - XWB 7000 Me molar mass of ATfXWA 41gknown weight of innown (WB) & weight of water (WA) is not given in the equ If you have these valus, just put in this eqr, you will get MB (Molar Man) -

(2) Colligative properties. The properties which depends only on the number of salute particles & not on the nature of salute particles are called Colligative properties Important colligative properties ax: 0 Elevation in Boiling point. (2) Depression in freezing poind. ③ Relative towering in vapour pressure | Osmotic presus. (3) In the presence of non volalite salute, the boiling point of the solution increases. } [ATE = Kf x WB X 1000/-o [ AIf = Kfm AMB X WA L m (molality) = WB x 1000 / plex wA. W here Kfa molal depression in freezing pint A reposents salvent for water Kf = 1.86 K kg moi o represente salute WB = weight of takna glucose (salute) WA = weight of water / sellend) = 10 kg = 1000g MB = molar man of wea Glucose Gatizo = 6*12+1246 x 16 - 1Bog Given If = - 4.2°C = -4:2+273 k = 268.8% we known freezing point of fox worker If = 0c = 213k ATF = TfO-Tf = 273-268.83 4.2 Put all these values in e o 1.16 x Wg X 1000 4.2 = 0 180 x 100 - 4.2x180x10 ( To converted into ka OR Wo= 4064.5 g [ divide by 1000 a 4.2 X 180X10 7.& = 4.06 kg .

6 - X love - no of males of salute (1) molality (m) = ? Wt of solvent in g(WA) weight of salute (wo) no of males (n) = molar mass (M8) m = Wo xlose 0 salrent Ho solute & Nano mo X WA m = 0.500 wg = ? А = 50°3 - Śq MB = Nanoq = 23+ 14+ 3x16 g Mos menos Na N o Put in ean WB x Der 0.500 = 85×500 -0.500x 85 85 = WBF 21.25 g 10 : 10 ""259 v
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