126 mL of 0.505 M KCl B. 126 g of 0.505 m KCl 126 g of 5.1% KCl solution by mass
ID: 728192 • Letter: 1
Question
126 mL of 0.505 M KClB. 126 g of 0.505 m KCl
126 g of 5.1% KCl solution by mass
How much solvent would you take to prepare the solution in part B(four sig figs)?
How much solvent would you take to prepare the solution in part C(two sig figs)?
All answers are in grams. Please show work!
Explanation / Answer
a) Molarity 126 ml of 0.505 M KCL in 1000 ml we have 0.505 moles of KCL MW of KCL = 74.6 g/mol 0.505 moles => 37.67 g in 1000 ml in 126 ml -> 4.746 g of KCL => 4.75 g of dry solute b)Molality - : concentration measured by the number of moles of solute per kilogram of solvent. MW of KCl= 74.6 g/mol Let x mole of KCl be used. (Implies wt = 74.6x) and y be the wt of water(in gm) used. By definition of molality, 0.505 = x/(y/1000) implies 0.505y/1000 = x Also total weight of soln should by 126g. x + y = 126g Solving the equations , x=0.06g y=125.94g. c) mass percentage 5.1 % in 1000 g of solution 51 g of KCL in 126 g we have 6.426 g of KCL mass of dry solute 6.43 g mass of solvent taken = 126 - 6.426 =119.574=119.57 g Hope you got it :)
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