`a)`

`A_K={39.93,258+40.0,012+41.6,73}/{100}=39,13472(a m u)`

`b)`

`n_K={420}/{39,13472}=10,73(mmol)`

\(n_{^{39}K}=10,73.93,258\%=10(mol)\)

\(\to m_{^{39}K}=10.39=39(g)\)

\(n_{^{40}K}=10,73.0,012\%=0,0013(mol)\)

\(\to m_{^{40}K}=0,0013.40=0,052(g)\)

\(n_{^{41}K}=10,73-10-0,0013=0,7287(mol)\)

\(\to m_{^{41}K}=0,7287.41=29,8767(g)\)

`c)`

`0,69<m_K<0,986(g)`

`->{0,69}/{39,13472}<n_K<{0,986}/{39,13472}`

`->0,018<n_K<0,025(mol)`

`->18<n_K<25(mmol)`

`->{18}/5<C_{M\ K}<{25}/5(mmol//L)`

`->3,6<C_{M\ K}<5(mmol//L)`