How many grams of sodium chloride are required to prepare 500.0 ml of a 0.100 m solution?

a. 1.46 g strike reset

b. 2.93 g strike reset

c. 29.3 g strike reset

d. 58.5 g strike reset?

Molarity is the number of moles of solute in 1 L solution.

molarity of solution to be prepared is 0.100 M

volume of solution to be prepared is 500.0 mL

the number of moles in 1 L - 0.100 mol

therefore number of moles in 500.0 mL - 0.100 mol/L x 500.0 x 10⁻³ L = 0.0500 mol

mass of NaCl required - 0.0500 mol x 58.5 g/mol = 2.925 g

mass of NaCl needed is therefore 2.93 g

answer is B. 2.93 g