1.
Đặt C\%_A=3x\%->C\%_B=x\%
Cho m_{dd\ A}=50(g);m_{dd\ B}=20(g)
->m_A=50.3x\%=1,5x(g);m_B=20.x\%=0,2x(g)
->C\%_C={1,5x+0,2x}/{50+20}.100=20
->x={140}/{17}
->C\%_A={420}/{17}\%≈24,71\% và C\%_B={140}/{17}\%≈8,24\%
2.
Có m_1+m_2=m_{dd\ MSO_4}=166,5(g);m_1-m_2=6,4
->m_1=86,45(g);m_2=80,05(g)
Ở 100^oC:
m_{MSO_4}=166,5.41,561\%≈69,2(g)
Ở 20^oC:
S_{MSO_4}={m_{MSO_4}}/{m_{H_2O}}.100=20,92
->m_{MSO_4}=0,2092m_{H_2O}
Mà m_{MSO_4}+m_{H_2O}=80,05
->m_{MSO_4}≈13,85(g);m_{H_2O}≈66,2(g)
->m_{MSO_4(kt)}=69,2-13,85=55,35(g)
Bảo toàn M: n_{MSO_4(kt)}=n_{MSO_{4}.5H_2O}
->{55,35}/{M+96}={86,45}/{M+96+18.5}
->M\approx 64
->MSO_4 là CuSO_4.