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.

Answer 1

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*10^3*0.01/0.09=7,56,000 ml =756 liters