Question

1. A solution of breach contains 3.62% by mass of NaOCl is in 2.50 kg of bleach solution?

2. An ore that is mined for it’s silver content is found to contain 2.86g sliver per ton of ore. Determine the parts per million (ppm) of silver in the ore. (1 ton = 2000 lbs, 1lb = 454 grams)

3. Rubbing alcohol is a 70% (v/v) solution of isopropyl alcohol in water. If a reaction required 3.45 grams of isopropyl alcohol, how many mL of rubbing alcohol solution would you need ? (the density of pure isopropyl alcohol is .7785 g/mol)

4. A solution of acetaminophen is 40% (m/v) in water. What volume of this solution would you need to administer to a 120 kg patient whose dosage is 15 mg/kg?

5. A solution is prepared by mixing ethanol with water. 25 mL of ethanol (d=0.785 g/mL) is added to 175g of water (d= 1.0 g/mL).

a) What is the v/v% of this solution?

b) what is the m/m% of this solution?

c) what is the m/v% of this solution?

Answer 1

1. A commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCl.

that means for 100 g of the solution the mass of NaOCl is 3.62 g

OR

1000 g of the solution contains 3.62 *10 = 36.2 g of NaOCl

1 Kg of solution contains 36.2 g of NaOCl

2.50 Kg of solution contains X g of NaOCl

Thus,

X = ( 36.2 *2.50) / 1

= 90.5 g

2. Mass of silver in a ton of ore= 2.86 g

Mass of ore= 1 ton×2000× 454/ 1 ton

= 908000 g

Therefore, concentration of silver in ppm=( mass of silver/ mass of ore)× 10⁶

= (2.86/908000)× 10⁶

= 3.14 ppm.

3. Given, (v/v) % = 70

I.e volume of isopropyl alcohol= 70 ml

Volume of solution= 100 ml

Density of isopropyl alcohol= 0.7785 g/ml

Therefore, mass of isopropyl alcohol= volume of isopropyl alcohol× density of isopropyl alcohol.

= 70 ml× 0.7785 g/ml

= 54.495 g.

Now, for 70 ml volume of isopropyl alcohol we have 54.495 g of isopropyl alcohol mass.

Therefore, for 3.45 g, volume of isopropyl alcohol= (70/54.495)× 3.45

= 4.4315 ml

4. Given, (m/v)%= 40

Therefore, mass of acetaminophen= 40 g

Volume of solution= 100 ml

Given, 120 kg patient dosage is 15 mg/kg.

So, total dosage= 120× 15= 1800 mg= 1.8 g

Thus, for 1.8 g acetaminophen, volume of solution needed= (100/40)× 1.8

= 4.5 ml.

5. Given,

Volume of ethanol= 25 ml

Density of ethanol= 0.785 g/ml

Therefore, mass of ethanol= 25× 0.785 g

= 19.625 g.

Also, mass of water= 175 g

Density of water= 1 g/ml

Therefore, volume of water= 175 ml.

Now, v/v % = (volume of ethanol/ volume of water+ volume of ethanol)× 100

=( 25/200)× 100= 12.50 %

Also, m/m% = (mass of ethanol/ mass of water+ mass of ethanol)× 100

= (19.625/ 194.625)× 100

= 10.08 %

Again, m/v% = (mass of ethanol/ volume of solution)× 100

= (19.625/200)× 100

= 9.812 %

Hope that it helps, if yes.. do give a thumbs up