Question

A 37.0 mL aliquot of a 0.600M stock solution must be diluted to 0.100 M. Assuming the volumes are additive, how much water should be added? volume of water 222 mL

Answer 1

Solution:- Given that in Volume of the stock Solution (M) = 3760 MI Molarity of the Stock Solution (S.) 0.600 ml Erolarity of the diluted Sdution (S2) = 0.100 ml Volume of the diluted solution = VAL Favating the Moler of initial and final Solution : v X.piloo -0.600 x 3710 v s (0.600 x 3710) Oileo = 22.2 0.100 202]mi Hence. An The Amant The Stock of water should be added Solution - ( 232 – 34.0)

chlorine = 5.30 x 100 g. Solution = 1.20 ppm. The simple conversion is 1 ppm al mg/l density 21.00 9/1 5.30 x 106 g of solution 5:20x106ml = 5130x102 litey of Solution, you required 1.20 ppm: 3 l. 20 X5.30 x 10 a 6.360 chlorine gas (6.360] Assuming 100 % dissociation : Nag sou -0.020 mel. In 1.00 kg Water moles of Mag sen (n) = 0.620 mol Mass of water (wo). 1.00 kg . molarity m 0.820 Tico 10.8am

boileng Point (Jo) alooc Freezing point (To) = 0°c . Boiling point In Дть = }}ь уп Vanthaff factor - i-3 for Nassou (Nag sou Qreat +son 2) Key = 0.511°c m for water, t = 1.86* c m fer water, .To-To = 3x0:518cm x 0.820 TH -|ooc - (25° i Tb = 10011.05 ) freezing point : To If ik mi ok - = 3 x 8G cm x 600 con O*If 1054°c ... If =-4.57°C freezing point If = -4.576