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A 0.200 M K 2SO 4 solution is produced by ________. dilution of 250.0 mL of 1.00 M K2SO4 to 1.00 L dissolving 43.6 g of K2SO4 in water and diluting to a total volume of 250.0 mL diluting 20.0 mL of 5.00 M K2SO4 solution to 500.0 mL dissolving 20.2 g of K2SO4 in water and diluting to 250.0 mL, then diluting 25.0 mL of this solution to a total volume of 500.0 mL dilution of 1.00 mL of 250 M K2SO3 to 1.00 L

Respuesta :

bhuvna789456

A 0.200 M of K2SO4 solution is produced by diluting 20.0 mL of 5.00 M K2SO4 solution to 500.0 mL.  

Explanation:

  •  When dealing with dilution we will use the following equation:

                              M1 V1 = M2 V2

where,

                      M1 = initial concentration

                      V1 = initial volume

                      M2 = final concentration

                      V2 = final volume

  • By diluting 20.0 mL of 5.00 M K2SO4 solution to 500.0 mL, we get

                            M1 V1 = M2 V2

     20.0 mL [tex]\times[/tex]    5.00 M = M2 [tex]\times[/tex] 500.0 mL

                               M2 = (20.0 mL [tex]\times[/tex]    5.00 M) / 500.0 mL

                              M2 =  0.200 M.

Hence A 0.200 M of K2SO4 solution is produced by diluting 20.0 mL of 5.00 M K2SO4 solution to 500.0 mL.  

okpalawalter8

The complete statement is as follows A 0.200 M of K2SO4 solution is produced by the dilution of 20.0 mL of 5.00 M K2SO4 solution to 500.0 mL.

M2 =  0.200 M.

What is a 0.200 M K 2SO 4 solution is produced by?

Question Parameter(s):

A 0.200 M K 2SO 4 solution

dilution of 250.0 mL of 1.00 M K2SO4 to 1.00 L dissolving 43.6 g of K2SO4 in wate

Generally, the equation for the Mass volume relationship   is mathematically given as

M1 V1 = M2 V2

Therefore

M1 V1 = M2 V2

20.0 mL*5.00 M = M2 500.0 mL                                

M2 = (20.0 mL5.00 M) / 500.0 mL                              

M2 =  0.200 M.

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