Holanty of Stcy K2CrOu: 125x 10IOrHal Buret read ing: 35.5mlFinal buret reaaingVOlume of Stock Soluton used: mLMolanty of Pinal solutonAbsorbance at 440 nm 0.051Color of Solunon: yowsh-graunish42.5mLCalcuiate the thorencal molarties of all the Pinal SOIUMONS Using the cXOct masses,volumesand molantiesof the solutons Snowall workand wafch sig fig. Volumetnc ppertcsand voiumetnic Plasks are accurate to 2 decimal places l00.00mL 25.00ml, 10.00mL,etcUSing Beers law and the exhinchon coefficents that are provided, caiculate the expenmentalalue molartics Prom the absorpance of each solution.The cel path length = 1.CmCalcviatc tne fperror betveen your theorehcal and cxperimentamolantiesCalcula hions for Pequred Aounts"Part AImmot CUCCH3COO)2 H20Cu=(03.55 (03.55C12.01x4 = 48.04H 008x8 = 8.064O lox5 80.0Pequired Amount: Copperl) Acetatem moles/LitersO.125moleslo=0.0125x 199.o5 glmol2.4910 rea amt.199. lo5 glmoleAIM.v-M2v2(0.1250)(v)E(0.0s000)25 000)NEIOML reg amt. of UtCH COD, H20Part Ci miv-m2V2Part BI: = m, v, m2vzlo.ozxxx)(V)-0.00Cco) (00)V 5mLOf StoCK SOlution(0.25 x10(V)- (8.16 x 10410.aVI-7ML StocK 2CrO4Part BIL:= m,VMV2(10V)000)(25.00)V10ML KMNO4

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Asked Oct 14, 2019
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How do I find the theoretical molarity?

Holanty of Stcy K2CrOu: 125x 10
IOrHal Buret read ing: 35.5ml
Final buret reaaing
VOlume of Stock Soluton used: mL
Molanty of Pinal soluton
Absorbance at 440 nm 0.051
Color of Solunon: yowsh-graunish
42.5mL
Calcuiate the thorencal molarties of all the Pinal SOIUMONS Using the cXOct masses,volumes
and molantiesof the solutons Snowall workand wafch sig fig. Volumetnc ppertcs
and voiumetnic Plasks are accurate to 2 decimal places l00.00mL 25.00ml, 10.00mL,etc
USing Beers law and the exhinchon coefficents that are provided, caiculate the expenmental
alue molartics Prom the absorpance of each solution.
The cel path length = 1.Cm
Calcviatc tne fperror betveen your theorehcal and cxperimentamolanties
Calcula hions for Pequred Aounts"
Part AI
mmot CUCCH3COO)2 H20
Cu=(03.55 (03.55
C12.01x4 = 48.04
H 008x8 = 8.064
O lox5 80.0
Pequired Amount: Copperl) Acetate
m moles/Liters
O.125moleslo
=0.0125x 199.o5 glmol
2.4910 rea amt.
199. lo5 glmole
AIM.v-M2v2
(0.1250)(v)E(0.0s000)25 000)
NEIOML reg amt. of UtCH COD, H20
Part Ci miv-m2V2
Part BI: = m, v, m2vz
lo.ozxxx)(V)-0.00Cco) (00)
V 5mLOf StoCK SOlution
(0.25 x10(V)- (8.16 x 10410.a
VI-7ML StocK 2CrO4
Part BIL:= m,VMV2
(10V)000)(25.00)
V10ML KMNO4
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Holanty of Stcy K2CrOu: 125x 10 IOrHal Buret read ing: 35.5ml Final buret reaaing VOlume of Stock Soluton used: mL Molanty of Pinal soluton Absorbance at 440 nm 0.051 Color of Solunon: yowsh-graunish 42.5mL Calcuiate the thorencal molarties of all the Pinal SOIUMONS Using the cXOct masses,volumes and molantiesof the solutons Snowall workand wafch sig fig. Volumetnc ppertcs and voiumetnic Plasks are accurate to 2 decimal places l00.00mL 25.00ml, 10.00mL,etc USing Beers law and the exhinchon coefficents that are provided, caiculate the expenmental alue molartics Prom the absorpance of each solution. The cel path length = 1.Cm Calcviatc tne fperror betveen your theorehcal and cxperimentamolanties Calcula hions for Pequred Aounts" Part AI mmot CUCCH3COO)2 H20 Cu=(03.55 (03.55 C12.01x4 = 48.04 H 008x8 = 8.064 O lox5 80.0 Pequired Amount: Copperl) Acetate m moles/Liters O.125moleslo =0.0125x 199.o5 glmol 2.4910 rea amt. 199. lo5 glmole AIM.v-M2v2 (0.1250)(v)E(0.0s000)25 000) NEIOML reg amt. of UtCH COD, H20 Part Ci miv-m2V2 Part BI: = m, v, m2vz lo.ozxxx)(V)-0.00Cco) (00) V 5mLOf StoCK SOlution (0.25 x10(V)- (8.16 x 10410.a VI-7ML StocK 2CrO4 Part BIL:= m,VMV2 (10V)000)(25.00) V10ML KMNO4

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Step 1

The values are given as follows:

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molar mass of Cu(CH,COo), =199.65 g/mol volume of Cu(CH,coo) , = 0.1L mass of Cu (CH,COo0), = 2.496 g

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Step 2

The number of moles of copper (II) acetate can be calculated as follows:

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mass of Cu(CH,Coo), molar mass of Cu (CH,COO), moles of Cu (CH,coo) 2.496 g 199.65 g/mol 0.0125 mol Cu(CH,COO)

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Step 3

The molarity of the copper (II) acetat...

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moles of Cu(CH,COO) volume of Cu(CH,COO) molarity of Cu(CH,coo) 0.0125 mol 0.1L =0.125 mol/L =0.125 M

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