Question

A silver nitrate solution contains 14.77 g of primary-standard AgNO, in 1.00 L. What volume of this solution will be needed to react with a 0.2527 g of NaCI? Volume 60.2527 g of NaCrO ? Volume ML C 83.43 mg of Naz AsO? Volume = mL d 445.5 mg of BaCl2H,O? Volume e 25.00 mL of 0.05605 M Na3PO,? Volume = 50,00 mL of 0.01681 M H,S? Volume

Answer 1

Ties, we shall WWII 114.99 :: Molenity of solution Nowy; Woutmelden each otion a wou i Agatan agar menning this in 1:1, the nos mo shall ho namh nu hequired equimolm Agnag 4. Nach . 0.2529 9 NA mol e .. Moles of Non rewired. D.0013 males in 1000 me one present 11 of Agasug 000 notes 00043 0.0869 1.5 14.788 m2 u volume = 49.9104 Forume: 94.91 m2 ~99.71.72 b)" Nag Corbia + 2 Ang wong --2nanong * , ananong & Ag, croq "Thus; no of moles of Aginoong required a. &x no of more of Nagerog wi oq Nancron : 02527 g ? : NO of moles of Hol wt of Magtrogo 161.97 g moll) Nazer04 a 0.00156 molch NO of moles Ag Nog hequired: 0.00312 moles. Thus, 0.0869 moles 0.00312 moles of Ag Non present in of AgNog": " 1000 mL 1000 x 0.00312 0.0864 : Volume : 35.91 me s : 35.909 mL 35.41 m2 () . Nag Asog + 3Ag Nog 3 na Nay it Ag. 3 AsO 4. : no of moles of Agnog required - 3x no of moles [ Of Nag Asun Moliwt of Naz ASOA : 207. 8 885 g mot wi of Naz Ason = 83.43 mg = 0.08343 g -

No of moles of Nag 0901: Ovono 1013 Henu ; no of moles of Mignon loquired,' 0.001201 Thus, Diogon moles of Manon friend in 1000 mL .;0.001204 moles of my woa . 120 1000 x 600|104 i'' ? 19.45 mL 'volume : T13.85 met ml - - d) Mal wil of Baelg. 2H20 e 214.26 g molt - W1 of Bach: 2 12.0 = 445.6 mg Back trag nag - Ba(NO3)2 + 2 Agel since w Moles of Bach .2 H2O 0.00 18 23 moles Now ; in 3 moles of Bull .2 H2o Baca present I moles i. 9n 0.001823 moles of 1 Bada present : 0.001823 0.000 6099 moles 3 from balanced equation. Agnon required a 2x moles of I Budz . Agnos required a no 0.001215 moles . in Thus; 0:001215 mores 0.001215 X 1000 ml of present Agnoz in 0.08.69 = 13.99 mL ~ 14.00 ML Peltos :. volume = /1400 ml e) Nazpoq + 3 Ag Nog 3 NaNO3 + Aga pog - In 25.00 mL 010 560 5 (M) Nag pou: Naz pou present 0.05605 X25.00 moles 1000 = 0.0014 01 moles . Ag Noz required : 3x01001901 moles = 0.084203 moles Thus, volume of Ag Nog required = 9:00 9203 x 1000 0.0869 = 48.37 mL R & volume = 148.37 mil

0 = I 50.00 m2, of 0.01681cmz' Has constit contains .. 50.4 0.6 1681 moles of H₂S . 0.0008u moles : C . a 2 Ag No3 + H₂S . 2A Noz & Ag 23 . . . Thus ; moles of Agnog required 2x moles of H₂5 = ax 0.00084 moles 0.001681 moles of Agnoz present in in a 0.001681 x 1000 m = 0.08169. = 19.34 ML . .! & volume. = 19.34 m2