Can you show me step by step how to do the following problems?
How many grams of KI are in 25.0 mL of a 3.0 % (m/v) KI solution?
How many milliliters of a 2.5 M MgCl2 solution contain 17.5 g MgCl2?
Of course! I'd be happy to help you solve those problems step by step. Let's start with the first one:
Problem 1: How many grams of KI are in 25.0 mL of a 3.0 % (m/v) KI solution?
Step 1: Convert the percentage to a decimal.
To do this, divide the given percentage (3.0%) by 100.
3.0 / 100 = 0.03
Step 2: Calculate the mass of the KI in the solution.
To find the mass, multiply the volume of the solution (25.0 mL) by the density of the solution.
Density is given by: density = mass / volume
Since the density is not given, we can assume it to be equal to water's density, which is approximately 1 g/mL.
mass = volume * density
mass = 25.0 mL * 1 g/mL
mass = 25.0 g
Step 3: Calculate the mass of KI in the solution.
Multiply the mass of the solution (25.0 g) by the decimal representation of the percentage (0.03).
mass of KI = 25.0 g * 0.03
mass of KI = 0.75 g
Therefore, there are 0.75 grams of KI in 25.0 mL of a 3.0% (m/v) KI solution.
Now, let's move on to the second problem:
Problem 2: How many milliliters of a 2.5 M MgCl2 solution contain 17.5 g MgCl2?
Step 1: Calculate the molar mass of MgCl2.
The molar mass is the sum of the atomic masses of all the atoms in the compound. The atomic mass of Mg is 24.31 g/mol, and the atomic mass of Cl is 35.45 g/mol (x2 because there are two Cl atoms).
molar mass of MgCl2 = (24.31 g/mol) + 2 * (35.45 g/mol)
molar mass of MgCl2 = 95.21 g/mol
Step 2: Convert the mass of MgCl2 to moles.
To do this, divide the given mass (17.5 g) by the molar mass (95.21 g/mol).
moles of MgCl2 = 17.5 g / 95.21 g/mol
moles of MgCl2 = 0.1837 mol
Step 3: Calculate the volume of the solution.
Use the equation: moles = volume (in liters) * molarity
Since we want to find the volume in milliliters, we need to convert liters to milliliters (1 L = 1000 mL).
volume (in liters) = moles / molarity
volume (in liters) = 0.1837 mol / 2.5 mol/L
volume (in liters) = 0.0735 L
volume (in milliliters) = volume (in liters) * 1000 mL/L
volume (in milliliters) = 0.0735 L * 1000 mL/L
volume (in milliliters) = 73.5 mL
Therefore, 17.5 g of MgCl2 will require 73.5 mL of a 2.5 M MgCl2 solution.
I hope this helps! Let me know if you have any other questions.