Question

5. (1pt) How much solid NaOH will you need to prepare 300 ml of an approximately 0.1 M solution? 6. (2pt) If you have 5.00 ml of a 5.5% acetic acid solution, how many grams and moles of acetic acid are in that solution? (Assume that the solution has a density of 1.00 g/ml).

Answer 1

cue Cluse 5 Answert given doda : Vm= 300ml molevitysm) .ings i we lerous that in .. Maria SX1000:-D M molarity , " . W molar mass of solute in glmol m → mass of solute in qram ?. . . Yml -> o volume of solution in ml we know that molat mass of NaOH = termic mass of No, od .: Na + 0 +1 putting o4-valuerin eyno -23 +1641 equation => " (W = 40g/mol Sum of Oil. = m o x adet 1844 390 0.1 = m 4*3 m= 12x00 g : 2 1-2 gram of solid maoy will you required forproept prepare soome to of dolay solution. 4.6 Answers Giyen data: volume of solution - 5 ml. of acetic acid solution : 5.5% (w/w ). density of solution - 1.00 g/ml ; solut e Acetic acid. - To find no. of grams of acetic acid in Solution we know that I w/w% = mass of solute in gram . mass of solution in gram - 100 /ml Solution : solute't solvent - X100

I have denidos ene .. mass hove, density of solution = ma alto mass of solution .. - volume of solution > Ig/ml... mass of solution : mass of solution = loglist x 5 mx = 5g of solution .: egne = - mass of solute in from xizoo 5.5 * - . $9 mass of solutie in gram = 5.5 = 0.275 ejram -To find no. of moles of acetic acido 9 mass of substance in gran we know that no. of moles = imalar mass of substance in g/mol we know molar mass of scuzeoon cacetic acid) = 60 g/mol ... no. of moles = 0.275. g. Go gimo how of moles = 0.0046 mole we have 8. 275 grams and 0.0046 mole of acetic acid inl given solution I