The question requires us to calculate the amount of water needed to prepare a KCl solution that is 0.500 m when 5.00 g of KCl are used.
Since the question provides the solution concentration in terms of molality (m) and provides the mass of solute, we need to calculate the number of moles of solute (in this case, KCl) and use it to calculate the amount of solvent (water) that is required.
To solve this question, we'll use the following definition of molality:
First, we need to calculate the number of moles of KCl that exist in 5.00 g of this substance. To do that, we'll need the molar mass of KCl, which can be calculated from the atomic mass of potassium (K, 39.10 u) and chlorine (Cl, 35.45 u):
molar mass (KCl) = (1 * 39.10) + (1 * 35.45) = 74.55 g/mol
Now, we can calculate how many moles of KCl are there in 5.00 g of this compound:
74.55 g of KCl --------------- 1 mol of KCl
5.00 g of KCl ----------------- x
Solving for x, we have that there are 0.0671 moles of KCl.
At this point, we have the following information:
number of moles of KCl = 0.0671 mol
molality of solution = 0.500 m
With the information above and the molality equation written previoulsy, we can calculate the amount of water (the solvent), in kg, required to prepare the solution:
Therefore, we would need 0.134 kg of water (or 134 g of water) to prepare a 0.500 m KCl solution from 5.00 g of this solute. (since the density of water is approximately 1 g/mL, we could also say that 134 mL of water are required)