Question

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

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