1) 0.084 M

2) 5 mL

3) 15.0 grams

4) 6.75 grams of sodium chloride

5) 1.1*10^-2 ppm

6) 24.8 ppm

7) We should the volume dilute to 6.7 mL ( add 1.7 mL)

8) The new molarity = 2.25 M

Explanation:

1. Calculate the molarity when 5.0 g KBr is dissolved to make 0.500 L solution.

Step 1: Calculate moles of KBr

Moles KBr = mass KBr / molar mass KBr

Moles KBr = 5.0 grams / 119 g/mol

Moles KBr = 0.042 moles

Step 2: Calculate molarity

Molarity = moles / volume

Molarity = 0.042 moles / 0.500 L

Molarity = 0.084 M

2. How many ml of isopropyl alcohol are needed to prepare 125 ml of 4.0% isopropyl alcohol solution (by volume/volume)?

4.0 % means the ratio v/v isopropyl alcohol / solution = 0.04

0.04 = volume isopropyl alcohol / 125 mL

volume isopropyl alcohol = 125 mL * 0.04 = 5 mL

3.How many grams of glucose are needed to prepare 300 ml of a 5.0% glucose solution (by mass/volume) ?

5.0 % means means there is 5 grams / 100 mL

For a volume of 300 mL the mass should be 5.0 grams *3 = 15.0 grams

4. How many grams of sodium chloride are needed to prepare 750 g of a 0.90% sodium chloride solution (by mass/mass)?

0.90 % means there is 0.90 grams of sodium chloride per 100 grams of solution

To get 0.90 (m/m)% for 750 grams of solution we need 0.90 * 7.5 = 6.75 grams of sodium chloride

5. A teaspoon of salt has a mass of approximately 8.0 g. If this quantity is added to a swimming pool that is 25 meters long and 15 meters wide, with a average of 2.0 meters deep, how many parts per million of salt have been added to the contents of the pool? Assume that the density of water in the pool is 1.00 g/mL.

Step 1: Calculate volume of the pool

volume = 15m * 25m * 2m = 750 m³ = 750 * 10³ L = 7.5 *10^8 mL

Step 2: Calculate mass of the water

Density = 1g/mL

Mass = volume * density

Mass = 7.5*10^8 mL * 1g/mL = 7.5 *10^8 grams

Step 3: Calculate ppm salt in the pool

We have 8.0 grams per 7.5 * 10^8 grams of water= 1.067 *10^-6 (m/m%)

1.067 *10^-6 m/m = 0.01067 ppm = 1.1*10^-2 ppm

6. A car is tested for tail pipe emissions. A 8-Kg sample of exhaust was collected. The sample contained 8 grams of carbon monoxide and 198 mg of unburned hydrocarbons. How many parts per million of unburned hydrocarbons were found in the exhaust.

Total mass of the exhaust = 8 kg = 8000 grams

We have 0.198 grams of unburned hydrocarbons in 8000 grams

That's 0.00248 grams per 100 gram ( 0.00248 m/m)

This is 24.8 ppm

7. To what volume should 5.0 mL of an 2.0 M citric acid solution be diluted in order to obtain a final solution that is 1.50 M? Round to two significant digits.

2.0 M means 2.0 moles per liter

For 5.0 mL this means we have 0.01 mol / 5 mL to have a molarity of 2.0 M

Since the number of moles will not change, the volume will be

0.01/1.5 = 0.0067 L = 6.7 mL

We should the volume dilute to 6.7 mL ( add 1.7 mL)

8. What is the morality of a solution of sodium chloride prepared by diluting 750.0 mL of a 6.0 M sodium chloride solution to 2.0 L?

A 6.0 M solution means it has 6.0 moles per 1000mL

For 750 mL we need 4.5 moles to have a molarity of 6.0 M

When we dilute this 6.0 M 750 mL solution to a 2.0 L solution, the number of moles will not change, only the volume

We have 4.5 moles / 2L

The new molarity = 2.25 M