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Calculate the final concentration of the solution when water is added to prepare each of the...

Calculate the final concentration of the solution when water is added to prepare each of the following solutions. 22.0 mL of a 10.0 %(m/v) K2SO4 solution is diluted to 60.0 mL & 80.0 mL of a 7.60 M NaOH solution is diluted to 245 mL

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Answer #1

(a). Solution.1 :- 22.0 mL of a 10.0 %(m/v) K2SO4 solution is diluted to 60.0 mL :-

10.0%(m/v) of K2SO4 means, 10.0 g of K2SO4 are present in 100.0 mL of solution

Therefore,

22.0 mL of solution contains = 10.0 g x 22.0 mL / 100.0 mL = 2.2 g of K2SO4

Number of moles of K2SO4 = Mass of K2SO4 / Gram molar mass of K2SO4

= 2.2 g / 174.259 g/mol

= 0.01262 mol of K2SO4

Total volume becomes = 22.0 mL + 60.0 mL = 82.0 mL = 0.082 L

Now,

Final concentration of K2SO4 = Number of moles of K2SO4 / Volume in L

= 0.01262 mol / 0.082 L

= 0.154 M

Therefore, final concentration of K2SO4 = 0.154 M

(b). Solution.2 :- 80.0 mL of a 7.60 M NaOH solution is diluted to 245 mL :-

Number of moles of NaOH = Molarity of NaOH x Volume in L

= 7.60 M x 0.080 L

= 0.608 mol

Total volume becomes = 80.0 mL + 245 mL = 325 mL = 0.325 L

Now,

Final concentration of NaOH= Number of moles of NaOH / Volume in L

= 0.608 mol / 0.325 L

=1.87 M

Therefore, final concentration of NaOH = 1.87 M
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