Question

NGUR 8. Calculate the new molarity that results when each of the following solutions is diluted to a final total volume of 2.50 L. a. 425 mL of 0.104 M HCI b. 10.5 mL of 12.1 M HCI c. 25.2 mL of 14.9 M HNO3 d. 6.25 mL of 18.0 M H2SO4

Answer 1

Final volume (12) for each case is 2.5L ie 2500ml I MV = M₂ V, (a) 425 ml of 0.104 M HCl Mi = 0-104 M v = 425ml as MV = M2 V 2 M₂ = MIVI = 0.10 4MX 425 me = 44.2 M 2500 me 2500 | M₂ = 0.017 68M / Hence New Holanty M₂ = 0.0 17 68 M (6) 10.5me of 12-1 MHO & M = 12. 12 MIVI = M₂ Y₂ (4 = 10 sme M2 = MIVI - 12-18 10 sme – 127.05 M 2500 25oo me [M2 = 0.05 M | New Molocity (C) 25.2 ml of 14.9M HHOZ M, = 14.9 M VI25-2 me MIV, = M₂ V2 M₂ = Mvi = 14.9 MX 25.2 ml = 375.48 M = 0 • 150 M 2 asoome 2500 New rolarity [M2 = 0.150 M ]

(d) 6.25 ml of 18.0 MH₂ Soy Mi = 18.0M V, = 6.25 MIVI = M₂ V2 M₂= HIVI = 18M X 6-25ml = 112- 5M =0.04SM 25ooml 2500 | M₂ = 0.045 Hence New Molaeity = 0.045M