Answer to Question #70902 in General Chemistry for Angel

Question

2. How many grams of NaCl are required to prepare 200 mL of a solution of M NacL? (molar mass of NaCl=58.44g/mol)

Answer 1

C M (NaCl) = n(NaCl) / V(solution)
n(NaCl) = V(solution)  C M (NaCl) = 0.2 L  0.3 mol/L = 0.06 mol
n(NaCl) = m(NaCl) / M(NaCl)
m(NaCl) = n(NaCl)  M(NaCl) = 0.06 mol  58.44 g/mol = 3.5