what is the number of grams needed to prepare a 1.23 M solution of nitric acid if 250.0 mL are needed?
To find the number of grams needed to prepare the given concentration of nitric acid solution, you need to use the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
First, let's convert the volume of the solution from mL to L:
250.0 mL = 250.0 mL / 1000 mL/L = 0.250 L
Now, we rearrange the formula to solve for moles of solute:
moles of solute = Molarity × volume of solution
Substituting the given values:
moles of solute = 1.23 M × 0.250 L = 0.3075 moles
Next, we need to convert moles of solute to grams using the molar mass of nitric acid (HNO3) which is 63.01 g/mol. The conversion formula is:
grams = moles × molar mass
grams = 0.3075 moles × 63.01 g/mol = 19.38 grams
Therefore, you would need approximately 19.38 grams of nitric acid to prepare a 1.23 M solution with a volume of 250.0 mL.