You wish to make 250. mL of 0.20 M KCl from a stock solution of 1.5 M KCl and DI water.

1 answer:

8 0

The question requires us to use the dilution formula $M_iV_i=M_fV_f$ , where $M_i$ and $V_i$ are the stock concentration and volume respectively, then $M_f$ and $V_f$ are the dilute concentration and volume respectively.

a. $C_s_t_o_c_k$ = 1.5 M KCI, $C_d_i_l_u_t_e$ =0.20M KCl, $V_d_i_l_u_t_e$ =250ml KCl

b. $M_iV_i=M_fV_f\\\implies V_i= \frac{M_fV_f}{M_i} = \frac{0.20M \times 250ml}{1.5M} = 33.3\ ml$

To prepare the solution 33.3ml of 1.5M KCl is diluted to a total final volume of 250ml.

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