Question

3 attempts left Check my work Enter your answer in the provided box. points Water is added to 75.0 mL of a 0.992 M KNO, solution until the volume of the solution is exactly 250 mL. What is the concentration of the final solution? eBook Print References

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Answer 1

MI? a 0.992 M V of KNO₂ 75.0mL + water U2= 250mL M₂=2 a) As per mot balance Miv = M₂ V₂ 0.992 x 75 = 0.2976 250 Bus: [ 0.298 m M = 0.468 M Gineu! Y =54.2 mL Ca (NO3)2 M₂ = 1.296M V₂ = 75.5 m2 I Ca(NO3)2 hew Volume v= 54.2 mL +75.5mL = 129,7 ml mivi + M₂ V₂ = MV -) So, Scanned with CamScanner

(0.468 x 5412) + (1.296x75,5) = Mx 129.7. 3 29.9916 + 97.848 - Mi 12917 M = 0.9856 Aus: 10. 986M 14 @ Nat following in each Sodium of the sulfate 0 3,65 M =) 3.65 M Na, so 3 [Na, 50, 7 = [Nq+] . [504") - 3,651 7.30 m 7 [Nat] =2x3,65 = Ô 1.18 M. sodium carbonate 3 [Na, 602] [Na t] - [603"): 1:18 Cs camped with Scanned with CamScanner Aud: 2:36M 12.30 M Dus :

6 0.585 M sodium bicarbonate 3 [Na Mcoy ] - [ Nat] -1°) - (notos] = 0.585 Aue! 10.585M ® concentration of lithium carbonate in it in 0.295 M 0. 295 m - » Licon] - [11. 6205 2 Ou: 95.0 mL of moles of No 0.942 M (KNO) = 95.0 x0,942 89.49 mmols + 125.5 mL of 1.570M (Ca(NO3)2) moles of Noz = 2x125.5 81.570 = 394.07 umer Ics Scanned with CamScanner

и с Now Tosal volume - - = 3, owL + 12 . gs.oml + 12 z 220.5 mL Total notes of Noj = 89, 49 mmol +394.07 cm - 493.56 nomor Resulting No is concento a him in moly Volume Bo 483, SG z 2.193 220.5 Aue : 2. 193 M=/2.19 M Scanneenith CamScanner

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