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
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
since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.