Please explain how to solve and show all work! I\'m extremely lost. Part 1: A 18
ID: 886227 • Letter: P
Question
Please explain how to solve and show all work! I'm extremely lost.
Part 1:
A 180,000 gal. swimming pool requires treatment with 8.85 L of 7.00% "by mass" chlorinating solution to give chlorine level of 1.00 ppm (1.00 g chlorine per 10^3 kg of pool water). Based on these data, calculate d (in g/ml) for the chlorinating solution nothing that it is close to 1 g/mL. * "% by mass" means "mass of solute/100g solution". thus, this chlorinating solution contains 7.00 g chlorine/100 g chlorinating solution.
Part 2:
Take pool water as pure and, for the poo, calculate the H2O/Cl2 molecular ration after treatment with the 8.85 L of the chlorinating soln.
Part 3:
If density, d (pool water) = 1.00 g/mL, and d (chlorinating solution) = 1.10 g/mL, calculate volume, V, of chlorinating solution to add to the 180,000 gal pool water to up d of resulting pool wter to 1.01 g/mL.
Explanation / Answer
1 gallon =3.78 liters
1,80,000 gallons =3.78* 1,80,000 = 6,80,400 liters =6,80,400*103 ml
Denisty =1 gm/ml, hence gms of water = 6,80,400 X103 gm
Final chlorine water contains 1gm of chlorine per 1000 kg water or 1gm chlrorine per 106 gm of water = 1gm per 106 gm of water
Total chlorine = 6,80,400*103/106 = 680.4 gms
7 gms of chlorine corresponds to 100 gm of chlorinating solution
680.4 gms of chlrine corresponds to 680.4*100/7= 9720gms of chlrorinating solution
its volume is given as 8.85 liters =8.85*1000 ml
Density of chlorinating solution = 9720/ (8850)=1.09 gm/ml
b) Water in the pool = 6,80,000*103 gms
Chlorine added = 680.4 gms , mass of chlrorinating solution =9720, mass of water= 9720-680.4 =9039.6
Total water after treatment = 6,80,000*103 + 9039.6 =689439600gms
H20 moles = 689439600/18= 38302200, moles of chlrine = 680.4/71=9.583 moles
Molar ratio of water to chlorine = 38302200/9.583=3996890.3 moles of water/ mole of chlorine
part 3 :
mass of water = 6,80,400X103 gms
density of chlorinating solution =1.1gm/ml, let the volume of chlorinating solution =V, mass of chlorinating solution = V*1.1 gms
total mass = 6,80,400*103+ V*1.1 gms
Final volume = 6,80,400*103+V
Density of the pool after chlorination =1.01gm/ml, mass of pool after chlorination =(6,80,400*103+V)*1.01 gm
mass in =mass out , 6,80,400X103 +V*1.1 = (6,80,400*103+V)*1.01
V*(1.1-1.01) = 6,80,400*103(1.01-1)
V*0.09 = 6,80,400*103*0.01 , V =6,80,400*103*0.01/0.09=7,56,000 ml =756 liters