Question

18°C ; 250(g) H2O là tan tối đa 53(g) Na2CO3. Tính S;C%

Answer 1

\(S=\dfrac{53}{250}\cdot100=21,2\left(g\right)\\ C_{\%}=\dfrac{53}{53+250}\cdot100\%=17\%\)

Answer 2

\(S=\dfrac{53}{250}\cdot100=0,212\left(g\right)\\m_{dd}=53+250=303\left(g\right)\\ C_{\%}=\dfrac{53}{303}\cdot100\%=17\%\)

Answer 3

`S_{Na_2CO_3}` = `53/250` . 100 = 21,2g

`C%` = `53/{53 + 250}` . 100`%` = 17,492`%`