To find out how many grams of KCl must be dissolved to produce a 1.50 M solution in 600.0 mL, we can use the formula for molarity:
\[ M = \frac{n}{V} \]
Where:
- \( M \) is the molarity (in moles per liter, M),
- \( n \) is the number of moles of solute,
- \( V \) is the volume of solution in liters.
First, we need to convert the volume from milliliters to liters:
\[ V = 600.0 , \text{mL} = 0.6000 , \text{L} \]
Next, we can rearrange the molarity formula to solve for \( n \) (the number of moles):
\[ n = M \times V \]
Substituting in the values we have:
\[ n = 1.50 , \text{M} \times 0.6000 , \text{L} = 0.9000 , \text{moles of KCl} \]
Now we need to convert moles of KCl to grams. The molar mass of KCl can be calculated as follows:
- Potassium (K) = 39.10 g/mol
- Chlorine (Cl) = 35.45 g/mol
Thus, the molar mass of KCl is:
\[ \text{Molar mass of KCl} = 39.10 , \text{g/mol} + 35.45 , \text{g/mol} = 74.55 , \text{g/mol} \]
Now we can calculate the mass of KCl required:
\[ \text{mass} = n \times \text{molar mass} \] \[ \text{mass} = 0.9000 , \text{moles} \times 74.55 , \text{g/mol} = 66.195 , \text{grams} \]
Therefore, to produce a solution of 1.50 M of KCl in 600.0 mL, you need to dissolve approximately 66.20 grams of KCl (rounded to two decimal places).