how much of a 0.225 M KCl solution contains 55.8g of KCl
Molarity*volume*molmass=mass
.225*Volumeinliters*molmassKCl= mass in grams.
3.33L
To determine the amount of a 0.225 M KCl solution that contains 55.8g of KCl, you need to use the molar mass of KCl and the formula for calculating the amount of a solute in a solution.
1. Find the molar mass of KCl:
The molar mass of KCl is the sum of the atomic masses of potassium (K) and chlorine (Cl):
M(K) = 39.10 g/mol
M(Cl) = 35.45 g/mol
M(KCl) = M(K) + M(Cl) = 39.10 g/mol + 35.45 g/mol = 74.55 g/mol
2. Use the molar mass to convert the given mass of KCl (55.8g) to moles:
moles of KCl = mass of KCl / molar mass of KCl
moles of KCl = 55.8g / 74.55 g/mol ≈ 0.748 mol
3. Use the given molarity (0.225 M) to calculate the volume of the solution containing the given amount of KCl:
Molarity (M) = moles of solute / volume of solution (in liters)
Volume of solution = moles of solute / Molarity
Volume of solution = 0.748 mol / 0.225 mol/L ≈ 3.32 L
Therefore, approximately 3.32 liters of the 0.225 M KCl solution contains 55.8g of KCl.
To find the amount of a specific compound in a solution, you need to use the formula:
Amount = Concentration × Volume
In this case, we are given the concentration of the KCl solution as 0.225 M, and we want to find the amount of KCl in the solution that corresponds to 55.8 g.
To calculate the amount of KCl, we need to convert the given mass to moles by dividing it by the molar mass of KCl. The molar mass of KCl is 74.55 g/mol (potassium has a molar mass of 39.10 g/mol and chlorine has a molar mass of 35.45 g/mol).
Amount of KCl (in moles) = mass of KCl / molar mass of KCl
Amount of KCl (in moles) = 55.8g / 74.55 g/mol ≈ 0.748 mol
Now that we have the amount of KCl in moles, we can use the concentration of the solution to find the volume of the solution.
Concentration (in mol/L) = Amount / Volume
0.225 M = 0.748 mol / Volume
Rearranging the equation to solve for Volume:
Volume = Amount / Concentration
Volume = 0.748 mol / 0.225 mol/L
Volume ≈ 3.32 L
Therefore, approximately 3.32 liters of the 0.225 M KCl solution contains 55.8 g of KCl.