Question

Calculate the required volume needed to make a 1.65 M solution consisting of 33.8 moles.

Calculate the required volume needed to make a 3.14 M solution consisting of 2.53 g of H2SO4

Calculate the number of moles of barium sulfide required to make 200 mL of a 0.53 M solution.

Calculate the mass of sodium chloride in a 0.0854 M solution, given 625 mL.

511 mL of a 9.22 M solution was diluted to 18.5 L. What is the new concentration?

A chemist prepared 112.8 L of a 0.0373 M solution using a 12.4 M stock. What volume of stock solution was needed?

A chemist combines 140. mL of a 1.25 M solution of sodium carbonate with 100. mL of a 0.750 M solution of silver nitrate. What is the mass of precipitate generated? What is the concentration of sodium nitrate generated?

Answer 1

Solution @ molarity = (mole)solute volume of solution 1165 M = (33.8) O solution Vsolution = 33.8 =( 20.48 litre) 165 ( molarity = (moles) solute I vsolution solution = (moles)solute = 2.539m моt avity L 98 Х3. 0.0082 litre = molarity = (moles)solute 1 Vsolution 0.53 M = moles) Baricem xloro - sulfide 200 (moles barium = 0.53 = 0.106 moles į sulfide S on D molarity = 0.0854 M Vsolution = 625 me. molarity = Wtsolute B My X solvh ) CSscanned with CamScanner

0.0854 = (w+) plass X loco 58.5 X 625 (ct)noni = 0.0854 * 58.5x625 ovo = 3.122 gm As the solution is getting Diluted, so apply Dilution law formula, mu, a M2 V2 o final volume Initial : Initial final concentation volume concentration → (9.22 M) (511 meo) = m2 (1805) x loro 2 eit ] mg = 9,22 X 511 M 1815x loro = 00255 m molarity - 12*4 m I molcerity -0.0373 m Prepared Solution of stock solution volume of = Stock Solution ? Volume of 11218 lit I prepared' = Solution m, vi= mave (12.4) x V = 0.0373 m) X 112.8 ei't _ 0.0373 X 112.8 Solution = volume of stock 9 124 CSScanned with CamScanner = Bob 0.34 litre)

3 Nacou (.) +2 ag 103 (anyo Ag, costvarez [240 me, 1.25m) [come, 0.75m] [PP+] = 1:25 X 140 low 0.75 X 100 loco moles = 0.175 moles 0.075. moles Here Agrioz is the limiting Reagent 80 2 mole of Agnoz forms. I moled of A92603 0.075 mole of Agraz form 1x0:075 mole - AgaCO3 moles of AgaCo3 (Precipitate) formed = 10:079) cats male ... mass of Precipitate = (0.075 x 276) gm | = 1035 m) ::) a moles of Agro form & 2 mole of Nado, 80, 0.075 moles of Agro, form 0.075 mole of varios i moles of NQ Noz formed = 0.075 moles volume of solution = 140 mei + 100 mli - 240 mlo CS Scanned with CamScanner

concentration of 0.075 moles y love 240 me. Nanos = (0-3125 M) CS Scanned with CamScanner