Question

5- Find the mass of Kr needed for an ideal gas in a 1200 ml container having a pressure of 3850 torr and a temperature of -173 C 6- A 2 M solution is prepared by adding 3.8 gram of KOH to water, find the volume of water. 7- How can we prepare a 0.4 M solution from 50 mL of 6 M solution? 8-10 gram of KOH in water is diluted with 50 mL of water to give a final volume of 120 mL. What is the Molarity of Final Solution?

Answer 1

V = 1200m1 = 12000 =1-2 l P = 3850 torr- 3850 atm = 5.06 atos 760 T= - 173°C = - 173+273k = 100k R = 0.082 l-ato/mol-k From PV = nRT n= PL = 5:06 X 1.2 RT - 0.082 x 100 - n = 0.74 moles of kr m = 0.74 =w = 0.74883.8 = 62.0129 Let's say V vol of water required It we ignore role of added kol then. Vol. of solo = ve WKOH = 3.89 inkoH = 3.8 moles molarity = - 56v > 2= 38 = v= 30 t = 0.0339 vot of water required=0-0339 x rood = 33.9 ml We can get 0. 4m sols from on solution by adding solvent, water. Let's say v vol. of water required no. of milimoles = molarity X Vol(mi) upon adding water, moles of solute remains same. PROH

som om Add um HO (50+vml. Now mitimoles of solute = 50x6 - 0.4m =(50+v x0.4 v = 300 - --50 – 300-20 0.4 0.4 = v= 700 ml Hence by adding 700ml of Hoto som of om solo, we can get o.4 m soln- 8 Let's say V Initial role of soln in ml 80 V+50 = 120 = V = 70ml Now AkoH = 10/56 moles solo = 120ml = 0.12 l Hence final molarity = 1 KOH +680 12 = 1.488 m