How many milliliters of 16% bacl2 solution do you need to take to prepare 0.25L of 0.01N solution?

Given:
ω (BaCl2) = 16%
ρ solution (BaCl2) = 1.156 g / cm3 = 1.156 g / ml
V solution (BaCl2) = 0.25 l
SN (BaCl2) = 0.01 n

To find:
V solution (BaCl2) -?

Decision:
1) n (BaCl2) = CH (BaCl2) * V solution (BaCl2) / z = 0.01 * 0.25 / 2 = 0.00125 mol;
2) m (BaCl2) = n (BaCl2) * M (BaCl2) = 0.00125 * 208 = 0.26 g;
3) m solution (BaCl2) = m (BaCl2) * 100% / ω (BaCl2) = 0.26 * 100% / 16% = 1.625 g;
4) V solution (BaCl2) = m solution (BaCl2) / ρ solution (BaCl2) = 1.625 / 1.156 = 1.406 ml.

Answer: The volume of the BaCl2 solution is 1.406 ml.