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