Respuesta :
Answer:
The volume of a chlorine solution a homeowner should add to her swimming pool is [tex]4.480\times 10^3 mL[/tex].
Explanation:
The solution contains 5.50 percent chlorine by mass.
Mass of the chlorine required to added in the pool= x
Mass of the water in the pool = y
Volume of the water ,V= [tex]6.52\times 10^4 gal=2.464\times 10^5 L[/tex]
(1 gal = 3.78 L)
V = [tex]2.464\times 10^5 L=2.464\times 10^8 mL[/tex]
( 1 L = 1000 mL)
Density of the water ,d= 1.00 g/mL
[tex]y=d\times V=1.00 g/mL\times 2.464\times 10^8 mL=2.464\times 10^8 g[/tex]
y = [tex]2.464\times 10^8 g[/tex]
1.00 g of chlorine per million grams of water .
Then for [tex]2.464\times 10^8 g[/tex] of water, we required x amount of chlorine:
[tex]z=\frac{2.464\times 10^8 g}{10^6 g}=246.4 g[/tex]
x = 246.4 grams of chlorine
Mass of the chlorine solution = M'
[tex](w/w)\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100[/tex]
[tex]5.5\%=\frac{z}{M'}\times 100[/tex]
[tex]M'=\frac{246.4}{5.5}\times 100=4,480 g[/tex]
Density of the chlorine solution = D = 1 g/mL
Volume of the chlorine solution to be added in the pool : v
[tex]v=D\times M'=1 g/mL\times 4,480 g=4480 mL[/tex]
[tex]v=4.480\times 10^3 mL[/tex]
The volume of a chlorine solution a homeowner should add to her swimming pool is [tex]4.480\times 10^3 mL[/tex].
Answer:
4.45 L
Explanation:
The concentration of chlorine in the pool must be 1.00g/10⁶g, and the solution has 5.50g/100g, so it will be diluted in the pool.
The product of the concentration by the volume is always constant in a solution, thus, we can use the equation:
C1V1 = C2V2
Where C is the concentration, V is the volume, 1 represents the solution, and 2 the pool. So, C1 = 5.50g/100g, C2 = 1.00g/10⁶g, V2 = 6.52x10⁴ gal = 247108 L. Thus,
(5.50/100)*V1 = (1.00/10⁶)*247108
0.055V1 = 0.247108
V1 = 4.5 L
