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

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Answer is to avoid experimental error