1, Isotonic solution is a 0.15 M aqueous solution of NaCl the simulates the tota

Question

1, Isotonic solution is a 0.15 M aqueous solution of NaCl the simulates the total concentration of ions found in many cellular fluids. Therefore, common saline IV fluids given in the hospitlal are 0.15 M NaCl solutions. How would you prepare 800 ml of 0.15 M NaCl from a 6.0 M stock solution?  (give correct units and sig figs)

2. Using the balanced chemical equation, What volume of 0.250 M Na2S203(aq) is needed to completely react with 12.44 ml of 0.125 M KI(aq)?  (give correct units and sig figs)

Na2S2O3(aq) + 2KI(aq) ------> K2S2O3(aq) + NaI(aq)

Explanation / Answer

answer 1
solution =

.80 L solutio * .15 mole nacl = .12 mol nacl

.12 mole nacl / 6 mol = .020 L solution

answer 2

find moles KI3
0.01244 L of 0.125 mol/Litre KI3 =0.001555 moles KI3

find moles Na2S2O3
0.001555 moles KI3 requires twice as much Na2S2O3
0.00311 moles

find volume of Na2S2O3
0.00311 moles @ 0.250 mol/Litre Na2S2O3 =
0.01244 litres

answer could be 12.44 ml of Na2S2O3

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