Suppose that you used 0.75g of iron. What is the minimum volume of 1.5 M_ CuSO4 solution required
Using eqn. (3-1)? and eqn. (3-2)?
To find the minimum volume of the 1.5 M CuSO4 solution required, we need to use Equation (3-1) and Equation (3-2).
Equation (3-1): n = m / M
Equation (3-2): n = V * C
Where:
n is the number of moles
m is the mass in grams
M is the molar mass in grams/mole
V is the volume in liters
C is the concentration in moles/liter
First, using Equation (3-1), we need to find the number of moles of iron.
Given:
Mass of iron (m) = 0.75g
Molar mass of iron (M) = 55.845 g/mol (you can find this from a periodic table)
Substituting the values into Equation (3-1):
n = m / M
n = 0.75g / 55.845 g/mol
Now we have the number of moles of iron.
Next, using Equation (3-2), we need to find the volume of the CuSO4 solution required.
Given:
Concentration of CuSO4 (C) = 1.5 M
Substituting the values into Equation (3-2):
n = V * C
V = n / C
Now we can substitute the value of n (which we found from Equation (3-1)) and the value of C into Equation (3-2) to find the volume:
V = (0.75g / 55.845 g/mol) / (1.5 mol/L)
Calculating this, we get the desired volume of the CuSO4 solution in liters.