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enot [183]
3 years ago
11

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

Chemistry
1 answer:
Nitella [24]3 years ago
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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Answer:

To 318.18 mL would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN

Explanation:

Dilution is the reduction of the concentration of a chemical in a solution and consists simply of adding more solvent.

In a dilution the amount of solute does not vary. But as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.

In a solution it is fulfilled:

Ci* Vi = Cf* Vf

where:

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  • Vi: initial volume
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In this case:

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Replacing:

1.40 M* 20 mL= 0.088 M* Vf

Solving:

Vf=\frac{1.40 M* 20 mL}{0.088 M}

Vf= 318.18 mL

<u><em>To 318.18 mL would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN</em></u>

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