Question

UINCes. Calculate the molarity and molality of 268.3 mL of a solution containing 0.5142 mol of CsCl and 197.5 g of water.

Answer 1

Volume of solution = 268.3 ml = 0.2683 L no. of male of Coco = 0.5142 mal may of water = 197.5g = 0.1975 kg → molarity of solution = no. of mole of coce Volume of solution (ML) 0.5142 mol 0.2683 L -1.9165 m = 1.9165 mol/ solution = 1.9165m Hence, molarity of → molality of solution = no. of mode of call may of ₂0 (in kg). 0.5142 mol 0.1975 kg = 2.6 malling = 2.6 m Hence, molality of solution = 2.6m.