In 200 ml. water was dissolved in 40 g. na2co3 * 10 * h2o. What is the mass fraction of na2co3 in the resulting solution?

Given:
V (H2O) = 200 ml
m (Na2CO3 * 10H2O) = 40 g

To find:
ω (Na2CO3) -?

1) n (Na2CO3 * 10H2O) = m / M = 40/286 = 0.14 mol;
2) M (Na2CO3 * 10H2O) = 286 g / mol;
3) n (Na2CO3) = n (Na2CO3 * 10H2O) = 0.14 mol;
4) M (Na2CO3) = 106 g / mol;
5) m (Na2CO3) = n * M = 0.14 * 106 = 14.84 g;
6) m (H2O) = ρ * V = 1 * 200 = 200 g;
7) m solution (Na2CO3) = m (H2O) + m (Na2CO3 * 10H2O) = 200 + 40 = 240 g;
8) ω (Na2CO3) = m * 100% / m solution = 14.84 * 100% / 240 = 6.18%.

Answer: The mass fraction of Na2CO3 is 6.18%.