How many grams of HNO3 are required to prepare 500Ml of a 0.601M HNO3 Solution?
First, we need to calculate the number of moles of HNO3 in 500mL of 0.601M solution:
0.601 moles/L x 0.5 L = 0.3005 moles
Next, we can use the molar mass of HNO3 (63.01 g/mol) to convert moles to grams:
0.3005 moles x 63.01 g/mol = 18.95 g
Therefore, 18.95 grams of HNO3 are required to prepare 500mL of a 0.601M solution.
To determine the number of grams of HNO3 required to prepare a 0.601M solution, you need to use the formula:
Molarity (M) = moles of solute / volume of solution (L)
First, convert the volume of the solution from milliliters (mL) to liters (L):
500 mL = 500/1000 = 0.5 L
Now, rearrange the formula to solve for moles of solute:
moles of solute = Molarity (M) × volume of solution (L)
moles of solute = 0.601 mol/L × 0.5 L
moles of solute = 0.3005 mol
Finally, use the molar mass of HNO3 to convert moles to grams:
molar mass of HNO3 = 1(atomic mass of H) + 1(atomic mass of N) + 3(atomic mass of O)
= 1(1.01 g/mol) + 1(14.01 g/mol) + 3(16.00 g/mol)
= 1.01 g/mol + 14.01 g/mol + 48 g/mol
= 63.03 g/mol
grams of HNO3 = moles of solute × molar mass of HNO3
grams of HNO3 = 0.3005 mol × 63.03 g/mol
grams of HNO3 = 18.93 grams
Therefore, you would require 18.93 grams of HNO3 to prepare 500 mL of a 0.601M HNO3 solution.