Using these charts 1) I need to find the molarity of the diluted KSCN solution.

Question

Using these charts 1) I need to find the molarity of the diluted KSCN solution. 2) Also need to find the initial molarities for Fe3+ ion and SCN- ion in test tubes 1 through 5. Chart 1 Test tube Diluted KSCN (mL) 0.25M Fe(NO3)3 (mL) 0.1M HNO3 (mL) 1 1.0 5.0 4.0 2 2.0 5.0 3.0 3 3.0 5.0 2.0 4 4.0 5.0 1.0 5 5.0 5.0 0 Chart 2 Test tube 0.0025M Fe(NO3)3 (mL) 0.0025 KSCN(mL) 0.1M HNO3 (mL) 6 1.0 1.0 5.0 7 1.0 1.5 4.5 8 1.0 2.0 4.0 9 1.0 2.5 3.5 10 1.0 3.0 3.0 11 2.0 1.0 4.0 12 2.0 1.5 3.5 13 2.0 2.0 3.0 14 2.0 2.5 2.5 15 2.0 3.0 2.0

Explanation / Answer

density of KSCN = 1.89 g/cm^3 = 1.89 g/ml and molar mass of KSCN = 97.18 g/mol => weight of KSCN = 1.89V => moles of KSCN = (1.89V)/97.181 moles If the total volume in the test tube is V1 => molarity of KSCN = (1.89*V)/(97.18*V1) So, for test tube 1: V = 1 ml and V1 = 1+5+4 = 10 ml => molarity of KSCN = (1.89*1)/(97.18*10) = 0.0019 M In test tube 2: V = 2 ml and V1 = 2+5+3 = 10 ml => molarity of KSCN = (1.89*2)/(97.18*10) = 0.0038 M for test tube 3: V = 3 ml and V1 = 3+5+2 = 10 ml => molarity of KSCN = (1.89*3)/(97.18*10) = 0.0058 M for test tube 4: V = 1 ml and V1 = 1+5+4 = 10 ml => molarity of KSCN = (1.89*4)/(97.18*10) = 0.0078 M for test tube 5: V = 5 ml and V1 = 5+5+0 = 10 ml => molarity of KSCN = (1.89*5)/(97.18*10) = 0.0097 M b) 1 mole KSCN liberates 1 mole of SCN- ion => initial molarity of SCN- ion = 0.0019 M 1 mole of Fe(NO3)3 liberates 1 mole of Fe3+ ion since,, initial moles of Fe3+ present = 0.25*5 => initial molarity of Fe3+ ion = (0.25*5)/10 = 0.125 M hope this helps you.. If any doubts, ask me in comments..

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