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

Explanation:

1) Data:

a) V = 500.0 ml = 0.500 liter

b) M = 0.100

c) mass = ?

2) Formulae:

a) Molarity: M = n / V

b) Number of moles: n = mass in grams / molar mass

3) Solution

a) M = n / V ⇒ n = M × V = 0.1000 M × 0.500 liter = 0.0500 mol

b) molar mass NaCl = 58.443 g/mol

c) mass = n × molar mass = 0.0500 mol × 58.443 g/mol = 2.92 g

2.92 g