52-Mixing-Prob-Ans

# 52-Mixing-Prob-Ans - MOLARITY MIXING GCHEM II LAB CONCEPT...

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MIX MIX 20.0 mLs H 2 O 60.0 mls H 2 O 40.0 mLs H 2 O 0.525 M NaCl 0.724 CaCl 2 BEAKER “A” BEAKER “C” BEAKER “B” moles NaCl = ( M) (Volume) M NaCl = moles NaCl/Volume moles CaCl 2 = (M) (Volume) = (0.525) (0.020) = 0.011/0.060 = (0.724) (0.040) = 0.011 M = 0.183 M = 0.029 M NaCl(aq)  Na +1 (aq) + Cl -`1 (aq) CaCl 2 (aq) Ca +2 (aq) + 2Cl -1 (aq) 0.011 0.011 0.011 M CaCl 2 = moles CaC 2 l/Volume 0.029 0.029 0.058 = 0.029/0.060 = 0.483 M M Cl -1 ion =(mole Cl -1 from NaCl + moles Cl -1 from KCl) / Volume = 0.011 + 0.058 / 0.060 = 1.115 M 1) A reaction mixture was made by mixing 655.0 mls of 0.540 M KNO 3 with 375.0 mls of 0.251 M NaBr. What is the initial molarity of the KNO 3 and NaBr in the mixture ? moles KNO 3 = M*Volume M KNO 3 = moles KNO 3 /Volume Total = (.540)(.6550) = .3537/(.6550 + .3750) = .3537 = .343 M M KNO 3 = .343 M moles NaBr = (.251)(.3750) M NaBr = .09413/(.6550 + .3750) M NaBr = .0914 M = .09413 = .0914 M

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52-Mixing-Prob-Ans - MOLARITY MIXING GCHEM II LAB CONCEPT...

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