Question

Hòa tan 12,5 gam CuS\(O_4\) 5\(H_2O\) vào 87,5 gam \(H_2O\) .

Tính C% và Cm

Answer 1

n_CuSO4.5H2O=\(\dfrac{12,5}{250}=0,05\left(mol\right)\)

n_CuSO4=n_CuSO4.5H2O=0,05(mol)

m_CuSO4=0,05.160=8(g)

C% dd CuSO₄=\(\dfrac{8}{12,5+87,5}.100\%=8\%\)

\(\sum\)m_H2O=12,5-8+87,5=92(g)

V_H2O=92.1=92(ml)

C_M dd CuSO₄=\(\dfrac{0,05}{0,092}=0,54M\)