A solution used to chlorinate a home swimming pool contains 7.5% chlorine by mass. An ideal chlorine level for the pool is one part per million (1 ppm; treat this as exact). (Think of 1 ppm as being 1 g chlorine per million grams of water.) If you assume densities of 1.10 g/mL for the chlo- rine solution and 1.00 g/mL for the swimming pool water, what volume of the chlorine solution, in liters, is required to produce a chlorine level of 1 ppm in an 18,000-gallon swimming pool? (Hint: How much chlorine is required by the pool? How much solution is required to supply this much chlorine?)

0.8255L

Explanation:

The first we need to know is the mass of 18000 gallons of water in the pool (1 gallon = 3.7854L):

18000 gallons * (3.7854L / 1 gallon) = 68137.2L = 68137200mL = 68137200g of  water

Now, the concentration you want is 1g of chlorine per million of grams of water. The mass of chlorine in 68137200g of water:

68137200g of water * (1g chlorine / 1000000g of water) = 68.1g of chlorine are required.

The mass of the solution that contains 68.1g of chlorine is:

68.1g chlorine * (100g solution / 7.5g) = 908g of the solution are required.

In volume, using density of the solution (1.10g/mL):

908g of the solution * (1mL / 1.10g) = 825.5mL of the solution are required =

0.8255L

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm