The question requires us to use the dilution formula [tex]M_iV_i=M_fV_f[/tex], where [tex]M_i[/tex] and [tex]V_i[/tex] are the stock concentration and volume respectively, then [tex]M_f[/tex] and [tex]V_f[/tex] are the dilute concentration and volume respectively.
a. [tex]C_s_t_o_c_k[/tex]= 1.5 M KCI, [tex]C_d_i_l_u_t_e[/tex]=0.20M KCl, [tex]V_d_i_l_u_t_e[/tex]=250ml KCl
b.[tex]M_iV_i=M_fV_f\\\implies V_i= \frac{M_fV_f}{M_i} = \frac{0.20M \times 250ml}{1.5M} = 33.3\ ml[/tex]
To prepare the solution 33.3ml of 1.5M KCl is diluted to a total final volume of 250ml.
