Question

1. Starting with a 16M H2SO4, you remove 15ml and you are wanting a 6M solution. How many liters of water do you need? 2. If you have a 6M hydrochloric acid solution and you want at 1M solution in 1L. How much concentrated HCl do you need? 3. If you have a 5M solution of nitric acid and if you dilute 20ml in IL solution, what is the molarity of the solution? 4. If you have 10M solution of acetic acid and you want a 2M solution in 3L. How much of the 10M solution do you need to add? 5. Starting with a 4M solution of boric acid and you remove 15ml and add it to a 2L solution, what is the new molarity? 6. If you have a 3M solution of sodium hydroxide and you want a 0.5M solution in 2L, how much of the original solution do you need? 7. If you begin with a 3M solution of ammonium hydroxide and you decant 37ml into a IL of water, what is the new molarity? 8. If you begin with a 4M solution of calcium hydroxide and you decant 22mL and you final solution is at 0.25M. How much water did you add the 22mL to? 9. If you have a solution at 1M sodium hydroxide and then final molarity is 0.5M in 2 L how much of the original solution did you add to the 2L?

Answer 1

Solution, 1. 16 (M) Hq Soy sou M Hasoy solution. che ebon (the) roones doo 1000 me solution Conterin 16 moles of Hasoy 15 " " " = 1000 x 15 x 15 enstorogorot . rodila nevip = 0.24 moles of H₂ soy est Remain H₂SO4 = (16-0.24) = 15.76 moles of Hasoy. we one Prepared 6(M) Hq soy solution. Therefore, 6 moles of Hesoy Present 1800 me aqueous solen 15.76 - 1080 X 15.76 .. = 2626.67 me a Hence, we one prepared 6(M) izsoy solution, the as water need 2.63 liters. 2. 10

2. we com use simple - 0.0- Vis, = V₂ S2 formula biss sind risind P Vo(Inition volume) = 2 ore S, (Initial Concentration) = 6(M) V2 (final volume) = 1 L = 1000 m S2 (finon Concentration] =1(M). via V2S2 = 1x 1000 - 166.67 m. i S 6 61m) Hel 166.67 mi need goiz m o su 3. 1000 mm Nitric acid solution Contain 5 kindle of Hiroz 20 - - - 20/= 0.1 molen Therefore, 0.1 mole of Hroz Present in 12 (ar 1000 m) Solution. Hence, the molarity of the solution o. 1(M) .. 4. we can use simple was Vis, = V2S2. formula. vi (Inition volume) = ? sit Inition Concentration = 10 cm) V2 (final rolume) = 32 à 3000 me. S2 (final Concentration) = 2 m in to drlar f 10 (m) o na S V2S2 2004 2006 X 2 10 Hence, 600 ml of the 10(M) Solution we are add. 1000 me bonic acid 15 - 20 solution - Contains 4 moles of B(OH)₂ - = = 0.06 moles .

Remaing Bonic acid = (4-0.06 ) = 3.94 moles BLOH)₂. gain, 3:gy moles of Bonic acid added at of the solution Therefor, al solution Contain 3.gy boric acid. 1 3 .94-1.97 - - The new molanity = 1.97(M). No we Cem use Simple Vis = 232 font v = ? 5,= 3 V2 = 2L 2000ml S₂ = 0.5 (M) (m) Vi - VS2, 2000, X0.5 = 333-333.m · Home, 333:333 me of the oniginal solution we one need. 7. 1000 me of solution Contain 3 moles of ammonium hydronice 37 - - - * 27 = 0.111. . Remering ammonium lydronide = (3-0) 2 2889 moles Of MH OH : 2.889 moles of Mayot Present in 1L solution. Hence, The new molarity of the solution 2.889 laporation 8. 1000 m solution Contain y moles of calcium hydronice. 22 4 8 22 1000 - 0.088

0.088 - 0.25 Again, 0.25 mole calcium ly dronide Present 1000 me solution. - 0.x0,088 - = 352 me u. Hence, the water added (352-22) = 330 me off to the 22on 4 (m) Ca(OH)2 solution to Prepared 0.25 (M) Ca (014) Solution tonmula. 9. we can ge V = ? s = 1(M) simple vs. = Vasa a 2 =26= 2000mL S = 0. 5 (m). V2 S2 2000 x 0.5 - s. - - 1000 me. Hence, 1000 m (14) aoriginal Solution we one added to the al Solution to Prepared 0.5 (m) Solution. - o-