How many grams of Al(NO3)3 are needed to prepare 25.0 mL of a 0.750 M aqueous solution.
How many moles do you want? That is M x L = ??
moles = grams/molar mass.
Solve for grams.
To answer this question, we need to use the formula for molarity:
Molarity (M) = moles of solute / volume of solution (in liters)
First, let's convert the given volume from milliliters to liters:
25.0 mL = 25.0 / 1000 = 0.025 L
Now we can use the formula to find the moles of solute:
Molarity = moles of solute / volume of solution
0.750 M = moles of solute / 0.025 L
We can rearrange the formula to solve for the moles of solute:
moles of solute = Molarity x volume of solution
moles of solute = 0.750 M x 0.025 L
Now let's calculate the moles of Al(NO3)3 needed:
moles of Al(NO3)3 = 0.750 M x 0.025 L
The last step is to convert moles to grams using the molar mass of Al(NO3)3:
Molar mass of Al(NO3)3 = (1 x atomic mass of Al) + (3 x atomic mass of N) + (9 x atomic mass of O)
Molar mass of Al(NO3)3 = (1 x 26.98 g/mol) + (3 x 14.01 g/mol) + (9 x 16.00 g/mol)
Now we can calculate the grams of Al(NO3)3 needed:
grams of Al(NO3)3 = moles of Al(NO3)3 x molar mass of Al(NO3)3
Plug in the values and calculate the result.