a)
m CuSO4 = 0,3.160 = 48(gam)
C% CuSO4 = 48/250 .100% = 19,2%
b)
n FeCl2 = 63,5.127 = 0,5(mol)
CM FeCl2 = 0,5/0,4 = 1,25M
a, Ta có : nCuSO4 = 0,3 mol
=> mCuSO4 = n.M = 48g
\(\Rightarrow C\%=\dfrac{m}{m_{dd}}.100\%=19,2\%\)
b, Ta có : mFeCl2 = 63,5g
=> nFeCl2 = m/M = 0,5mol
=> \(C_M=\dfrac{n}{V}=1,25M\)