Question

A percent is a ratio expressed per 100 units. If you have 20 balls, and 1 is blue, you can say that 5% (or 5 out of 100 balls) is blue. Similarly, if a solution is 5% KI by mass, you can say that 5 g per 100 g solution is KI. When mass percent is used as a conversion factor, it can be expressed in the units required to solve the problem provided that the ratio remains constant. If the numerator is divided by 100, the denominator must also be divided by 100. Two examples are: express the grams of solute per 100 grams of solution, or express the centigrams (cg) of solute per 1 g of solution The latter example works because 100 cg is equivalent to 1 g. Complete the following statements regarding how many grams of KI are found in either 100 g or 1 kg of a 2.5% solution of KI.

a. A 2.5% (by mass) solution concentration signifies that there is ______ of solute in every 100 g
b. Therefore, when 2.5% is expressed as a ratio of solute mass over solution mass, that mass ratio would be ________
c. A solution mass of 1 kg is _______ times greater than 100 g, thus one kilogram (1 kg) of a 2.5% KI solution would contain ____________of

Answer 1

a. 2.5 g

b. 0.025

c. i. 10 times ii 25 g of KI

Step-by-step explanation:

a. A 2.5% (by mass) solution concentration signifies that there is ______ of solute in every 100 g

Since % by mass = mass of solute, m/mass of solvent,M × 100 %

2.5 % = m/M × 100%

since M = 100 g,

2.5 % = m/100 × 100 %

m/100 = 2.5/100

m = 2.5 g

b. Therefore, when 2.5% is expressed as a ratio of solute mass over solution mass, that mass ratio would be ________

The mass ratio = mass of solute/mass of solvent

= m/M

= 2.5 g/100 g

= 0.025

c A solution mass of 1 kg is _______ times greater than 100 g, thus one kilogram (1 kg) of a 2.5% KI solution would contain ____________of

i. A solution mass of 1 kg is _______ times greater than 100 g

Since 1 kg = 1000 g, 1 kg/100 g = 1000 g/100 g = 10

So,1 kg is 10 times greater than 100 g.

ii. Thus one kilogram (1 kg) of a 2.5% KI solution would contain ____________of KI

So, if M = 1 kg = 1000 g, we find m

Since % by mass = mass of solute, m/mass of solvent,M × 100 %

2.5 % = m/M × 100%

since M = 1000 g,

2.5 % = m/1000 × 100 %

m/1000 = 2.5/100

m = 2.5/100 × 1000 g

m = 2.5 × 10 g

m = 25 g