Question

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?

Answer 1

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