1) what is the moler consentration of a salutioncontaining 10g/L of (a) NaOH, (b
ID: 684838 • Letter: 1
Question
1) what is the moler consentration of a salutioncontaining 10g/L of (a) NaOH, (b)Na2SO4, (c)K2Cr2O7 ,(d) KCL? 2) Awater sample contains 44 mg/L of calciumion and 19 mg/Lof magnesium ion what is the hardness expressed asmg/Lof CaCO3?note that the hardness is sum of themultivalent cations. 3)A non-aqueous-phase liquid (NAPL) mixture contains 100 kg oftetrachloroethene [C2Cl4: also calledperchloroethene (PCE), 100 kg of benzene(C6H6) kg of toluene(C7H8),80kg of ethylbenzene(C8H10), and 60 kg of xylene(C8H10). calculate the mole fraction ofeach compound in the NAPL. 2) Awater sample contains 44 mg/L of calciumion and 19 mg/Lof magnesium ion what is the hardness expressed asmg/Lof CaCO3?note that the hardness is sum of themultivalent cations. 3)A non-aqueous-phase liquid (NAPL) mixture contains 100 kg oftetrachloroethene [C2Cl4: also calledperchloroethene (PCE), 100 kg of benzene(C6H6) kg of toluene(C7H8),80kg of ethylbenzene(C8H10), and 60 kg of xylene(C8H10). calculate the mole fraction ofeach compound in the NAPL.Explanation / Answer
1) what is the moler consentration of asalution containing 10g/L of (a) NaOH, (b)Na2SO4, (c)K2Cr2O7 ,(d) KCL? (a) Mol. wt . of NaOH = 23 + 16 +1= 40 g No. of moles = weight / Mol. wt = 10g / 40g = 0.25 moles Molar concentration = No. of moles / Volume inL = 0.25 moles/ 1L = 0.25 mol/L (b) Mol. wt . of Na2SO4= 2(23) + 32 +4(16) = 142 g No. of moles = weight / Mol. wt = 10g / 142 g = 0.0704 moles Molar concentration = No. of moles / Volume inL = 0.0704 moles/ 1L = 0.0704 mol/L (c) Mol. wt .of K2Cr2O7 = 2(39)+ 2*52 +7(16) = 294 g No. of moles = weight / Mol. wt = 10g / 294 g = 0.03401 moles Molar concentration = No. of moles / Volume inL = 0.03401 moles/ 1L =0.03401 mol/L (d) Mol. wt . of KCl = 39 +35.5 = 74.5 g No. of moles = weight / Mol. wt = 10g / 74.5 g = 0.1342 moles Molar concentration = No. of moles / Volume inL = 0.1342 moles/ 1L =0.1342 mol/L 2) Awater sample contains 44 mg/L of calciumion and 19 mg/Lof magnesium ion what is the hardness expressed asmg/Lof CaCO3?note that the hardness is sum of themultivalent cations 44 mg/L of calcium ion = 44 ppm of calciumion 19 mg/Lof magnesium ion = 19 ppm of magnesiumion 1 MgSO4 identical to 1 CaCO3 120 ppm identical to 100 ppm 19 ppm identical to ( 100 * 19 )/ 120 = 15.833 ppm 1 CaSO4 identical to 1 CaCO3 136 ppm identical to 100 ppm 44 ppm identical to ( 100* 44 ) / 136 = 32.353 ppm Total hardness in water = Hardness due to CaSO4 + hardness dueto MgSO4 = 32.353ppm + 15.833ppm = 48.186 ppm =48.186 mg / L 3)A non-aqueous-phase liquid (NAPL) mixture contains 100 kg oftetrachloroethene [C2Cl4: also calledperchloroethene (PCE), 100 kg of benzene(C6H6) kg of toluene(C7H8),80kg of ethylbenzene(C8H10), and 60 kg of xylene(C8H10). calculate the mole fraction ofeach compound in the NAPL Here the weigth of Toluene is not given If it was given then we proceed like this manner No. of moles of each component = weight of the component /Mol. wt. of the component From this we can get the no. of moles of each component Combain all the no. of moles resulting to the total no . ofmoles Mole fraction of each component = No. of moles of thatcomponent / Total no. of moles (b) Mol. wt . of Na2SO4= 2(23) + 32 +4(16) = 142 g No. of moles = weight / Mol. wt = 10g / 142 g = 0.0704 moles Molar concentration = No. of moles / Volume inL = 0.0704 moles/ 1L = 0.0704 mol/L (c) Mol. wt .of K2Cr2O7 = 2(39)+ 2*52 +7(16) = 294 g No. of moles = weight / Mol. wt = 10g / 294 g = 0.03401 moles Molar concentration = No. of moles / Volume inL = 0.03401 moles/ 1L =0.03401 mol/L (d) Mol. wt . of KCl = 39 +35.5 = 74.5 g No. of moles = weight / Mol. wt = 10g / 74.5 g = 0.1342 moles Molar concentration = No. of moles / Volume inL = 0.1342 moles/ 1L =0.1342 mol/L 2) Awater sample contains 44 mg/L of calciumion and 19 mg/Lof magnesium ion what is the hardness expressed asmg/Lof CaCO3?note that the hardness is sum of themultivalent cations 44 mg/L of calcium ion = 44 ppm of calciumion 19 mg/Lof magnesium ion = 19 ppm of magnesiumion 1 MgSO4 identical to 1 CaCO3 120 ppm identical to 100 ppm 19 ppm identical to ( 100 * 19 )/ 120 = 15.833 ppm 1 CaSO4 identical to 1 CaCO3 136 ppm identical to 100 ppm 44 ppm identical to ( 100* 44 ) / 136 = 32.353 ppm Total hardness in water = Hardness due to CaSO4 + hardness dueto MgSO4 = 32.353ppm + 15.833ppm = 48.186 ppm =48.186 mg / L 3)A non-aqueous-phase liquid (NAPL) mixture contains 100 kg oftetrachloroethene [C2Cl4: also calledperchloroethene (PCE), 100 kg of benzene(C6H6) kg of toluene(C7H8),80kg of ethylbenzene(C8H10), and 60 kg of xylene(C8H10). calculate the mole fraction ofeach compound in the NAPL Here the weigth of Toluene is not given If it was given then we proceed like this manner No. of moles of each component = weight of the component /Mol. wt. of the component From this we can get the no. of moles of each component Combain all the no. of moles resulting to the total no . ofmoles Mole fraction of each component = No. of moles of thatcomponent / Total no. of moles (c) Mol. wt .of K2Cr2O7 = 2(39)+ 2*52 +7(16) = 294 g No. of moles = weight / Mol. wt = 10g / 294 g = 0.03401 moles Molar concentration = No. of moles / Volume inL = 0.03401 moles/ 1L =0.03401 mol/L (d) Mol. wt . of KCl = 39 +35.5 = 74.5 g No. of moles = weight / Mol. wt = 10g / 74.5 g = 0.1342 moles Molar concentration = No. of moles / Volume inL = 0.1342 moles/ 1L =0.1342 mol/L 2) Awater sample contains 44 mg/L of calciumion and 19 mg/Lof magnesium ion what is the hardness expressed asmg/Lof CaCO3?note that the hardness is sum of themultivalent cations (d) Mol. wt . of KCl = 39 +35.5 = 74.5 g No. of moles = weight / Mol. wt = 10g / 74.5 g = 0.1342 moles Molar concentration = No. of moles / Volume inL = 0.1342 moles/ 1L =0.1342 mol/L 2) Awater sample contains 44 mg/L of calciumion and 19 mg/Lof magnesium ion what is the hardness expressed asmg/Lof CaCO3?note that the hardness is sum of themultivalent cations 44 mg/L of calcium ion = 44 ppm of calciumion 19 mg/Lof magnesium ion = 19 ppm of magnesiumion 1 MgSO4 identical to 1 CaCO3 120 ppm identical to 100 ppm 19 ppm identical to ( 100 * 19 )/ 120 = 15.833 ppm 1 CaSO4 identical to 1 CaCO3 136 ppm identical to 100 ppm 44 ppm identical to ( 100* 44 ) / 136 = 32.353 ppm Total hardness in water = Hardness due to CaSO4 + hardness dueto MgSO4 = 32.353ppm + 15.833ppm = 48.186 ppm =48.186 mg / L 3)A non-aqueous-phase liquid (NAPL) mixture contains 100 kg oftetrachloroethene [C2Cl4: also calledperchloroethene (PCE), 100 kg of benzene(C6H6) kg of toluene(C7H8),80kg of ethylbenzene(C8H10), and 60 kg of xylene(C8H10). calculate the mole fraction ofeach compound in the NAPL Here the weigth of Toluene is not given If it was given then we proceed like this manner No. of moles of each component = weight of the component /Mol. wt. of the component From this we can get the no. of moles of each component Combain all the no. of moles resulting to the total no . ofmoles Mole fraction of each component = No. of moles of thatcomponent / Total no. of moles 1 CaSO4 identical to 1 CaCO3 136 ppm identical to 100 ppm 44 ppm identical to ( 100* 44 ) / 136 = 32.353 ppm Total hardness in water = Hardness due to CaSO4 + hardness dueto MgSO4 = 32.353ppm + 15.833ppm = 48.186 ppm =48.186 mg / L 3)A non-aqueous-phase liquid (NAPL) mixture contains 100 kg oftetrachloroethene [C2Cl4: also calledperchloroethene (PCE), 100 kg of benzene(C6H6) kg of toluene(C7H8),80kg of ethylbenzene(C8H10), and 60 kg of xylene(C8H10). calculate the mole fraction ofeach compound in the NAPL Here the weigth of Toluene is not given If it was given then we proceed like this manner No. of moles of each component = weight of the component /Mol. wt. of the component From this we can get the no. of moles of each component Combain all the no. of moles resulting to the total no . ofmoles Mole fraction of each component = No. of moles of thatcomponent / Total no. of molesRelated Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.