Question

Cho 5,72g Na2CO3.10H2O(Sô đa tinh thể) vào 200g dd Na2CO3 10%. Tính C% dd Na2CO3 thu được.

Answer 1

\(m_{Na_2CO_3}=\dfrac{5.72}{286}\cdot106=2.12\left(g\right)\)

\(m_{Na_2CO_3\left(10\%\right)}=200\cdot10\%=20\left(g\right)\)

\(m_{dd}=5.72+200=205.72\left(g\right)\)

\(C\%_{Na_2CO_3}=\dfrac{2.12+20}{205.72}\cdot100\%=10.75\%\)

Answer 2

\(n_{Na_2CO_3}=n_{Na_2CO_3.10H_2O}=\dfrac{5,72}{286}=0,02\left(mol\right)\\ m_{Na_2CO_3}=0,02.106=2,12\left(g\right)\\ m_{Na_2CO_3\text{ trong dd 10%}}=\dfrac{200.10}{100}=20\left(g\right)\\ m_{dd\text{ mới}}=5,72+200=205,72\left(g\right)\\ C\%_{dd\text{ mới}}=\dfrac{20+2,12}{205,72}.100\%=10,75\%\)