Chemistry Practice Problems Dilution Practice Problems Solution: A 0.200 M K2SO4 solution is produced by ____.A) di...

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# Solution: A 0.200 M K2SO4 solution is produced by ____.A) dilution of 250.0 mL of 1.00 M K2SO4  to 1.00 LB) dissolving 43.6 g of K2SO4 in water and diluting to a total volume of 250.0 mLC) diluting 20.0 mL of 5.00 M K2SO4 solution to 500.0 mLD) 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 mLE) dilution of 1.00 mL of 250 M K2SO4 to 1.00 L

###### Problem

A 0.200 M K2SO4 solution is produced by ____.

A) dilution of 250.0 mL of 1.00 M K2SO4  to 1.00 L

B) dissolving 43.6 g of K2SO4 in water and diluting to a total volume of 250.0 mL

C) diluting 20.0 mL of 5.00 M K2SOsolution to 500.0 mL

D) 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

E) dilution of 1.00 mL of 250 M K2SO4 to 1.00 L

###### Solution

We are asked which of the following options would yield 0.200 M K2SO4 solution

When we are adding water (or solvent) to a solution to decrease its concentration, we are diluting the solution.

When dealing with dilution we will use the following equation:

$\overline{){{\mathbf{M}}}_{{\mathbf{1}}}{{\mathbf{V}}}_{{\mathbf{1}}}{\mathbf{=}}{{\mathbf{M}}}_{{\mathbf{2}}}{{\mathbf{V}}}_{{\mathbf{2}}}}$

M1 = initial concentration
V1 = initial volume
M2 = final concentration
V2 = final volume View Complete Written Solution

Dilution

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