1. A sample of air was found to contain 21% oxygen by volume. how many ml of oxygen are present in 2.0 L of air?
2. A 25.0g solution containing salt and water is 2.2% salt by mass. what is the mass (in g) of salt in the solution?
please show me how to solve :)
1. To solve this problem, we need to calculate the volume of oxygen in 2.0 L of air.
Step 1: Convert the percentage to a decimal
The percentage of oxygen is given as 21%. To convert this to a decimal, divide it by 100:
21 ÷ 100 = 0.21
Step 2: Calculate the volume of oxygen
Multiply the decimal value by the volume of air to get the volume of oxygen:
0.21 × 2.0 = 0.42 L
Step 3: Convert the volume to milliliters
Since 1 liter (L) is equivalent to 1000 milliliters (mL), we can convert the volume of oxygen from liters to milliliters by multiplying by 1000:
0.42 × 1000 = 420 mL
Therefore, there are 420 mL of oxygen present in 2.0 L of air.
2. To find the mass of salt in the solution, we need to calculate it based on the given mass percentage.
Step 1: Convert the percentage to a decimal
The percentage of salt is given as 2.2%. To convert this to a decimal, divide it by 100:
2.2 ÷ 100 = 0.022
Step 2: Calculate the mass of salt
Multiply the decimal value by the mass of the solution:
0.022 × 25.0 = 0.55 g
Therefore, the mass of salt in the solution is 0.55 grams.