Which volume of distilled water do 27.1 g of barium chloride need to be dissolved in to achieve a concentration of 0.100 mol/L of barium chloride?
M(BaCl2) = 208.23 g/mol
n(BaCl2) "= \\frac{27.1}{208.23} = 0.130 \\;mol"
Proportion:
0.100 mol – 1000 mL
0.130 mol – x mL
"x = \\frac{0.130 \\times 1000}{0.100} = 1300 \\;mL"
Answer: 1300 mL
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