How many grams are required to prepare a 0.500 L solution of aqueous NaCI, 0.95 M? {Ans: 27.8 g NaCI] Write the set up that justifies the answer.
To calculate the number of grams required to prepare a 0.500 L solution of aqueous NaCl with a concentration of 0.95 M, we can use the formula:
Molarity (M) = moles of solute / liters of solution
First, we need to find the moles of NaCl required:
0.500 L x 0.95 mol/L = 0.475 moles
Next, we need to convert moles to grams by multiplying by the molar mass of NaCl:
0.475 moles x 58.44 g/mol = 27.8 g
Therefore, 27.8 grams of NaCl are required to prepare a 0.500 L solution of aqueous NaCl with a concentration of 0.95 M.