Question

Match the following aqueous solutions with the appropriate letter from the column on the right. 1.0.16 m Fe(CH3COO)2 2. 0.15 m K2SO4 3. 0.18 m CaBr2 14.0.52 m Sucrose(nonelectrolyte) A. Lowest freezing point B. Second lowest freezing point C. Third lowest freezing point D. Highest freezing point

Answer 1

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Anewer. howe This quetion goes by it woes Basic Theory Motality (m) . It is defined as the number of moks of the solure that is present in 1 kg (kilogram) of the Bohusen. Solvent. Example 4 mol/kg 4modal or 4m Vant - Hoff factor (i) & It is defined as the measure of the solute or on vartous collegative properties such as osmolte pressure , borang potnt elevation, sem depression in relative towering of vapour pressure, & freezing pornt. etc &s known as / termed as Vant Hoff factor. So, here in this questton, The freezing points of given solutions are lowered by the particles present in solution . The first z compounds breaks anto 3 different tons in their solution form. so we multiply the motalities by 3. 9 te the non electrolyte does not dissociate, the molalite for it remains same

So, we obtain the results as ir 0.16 mx 3 = 0.48m. Fe (Cth 600) Rg 304 Cabry 0 • 15 mx 3 = 0.45m. 0.18mx 3 = 0.54m. 0.52mx 1 = 0.52m. Sucrose - (non electroly ke) VA Now, according to the above results obtained, the lowest freezing point es possessed by the solution with the - highest molality Ames of the Vant Hoff factor So, according to that 4X Lowest freezing point by second lowest freezing point c) Thard towest freezing point > Highest freezing point 3 048 m Cabra 87 0.52m Sucrose. i) D. 16 m Fe (C12600), 2) Or15m K250g.