Answer:
You will need 77.5 grams of NaNO₃.
Step-by-step explanation:
Molarity is a unit of concentration defined in terms of moles of solute per liter of solution.
Because we are given the molarity and volume of the solution, we can work backwards to compute how many moles of solute (NaNO₃) we need, from which we can determine the corresponding mass.
Mathematically, the molarity of a solution is represented by M = n/V, where M is the molarity of the solution, n is the number of moles of solute, and V is the volume of the solution (in liters). We can rearrange this equation to solve for the number of moles of solute:
n = MV
which, in our case, is equal to (2.43 M NaNO₃)(0.3750 L) = 0.91125 moles NaNO₃.
We are asked how many grams of NaNO₃ are needed to make this solution. So, to convert from moles to grams, we multiply the number of moles of NaNO₃ we have just obtained by the molar mass of NaNO₃, or 84.9947 g/mol.
m = (0.91125 moles NaNO₃)(84.9947 g NaNO₃/moles NaNO₃)
m = 77.451 g NaNO₃
To three significant figures, we need 77.5 grams of NaNO₃ to make 375.0 mL of a 2.43 M solution of NaNO₃.