Molarity of SO4 is 0.10M, Volume at intersection is 10.35 mL Use the volume of s

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Molarity of SO4 is 0.10M, Volume at intersection is 10.35 mL

Use the volume of sulfuric acid from the Interpolation Text Box to compute the quantities: Show all of your calculations on a separate sheet of paper , 35 m a. mole of sulfuric acid use: n MxV b, molarity (M) of the Ba(OH)2(aq) in the unknown solution: n Ba(OH)2(aq) = n H2SO4 c. number of moles of BaO in 20.00 mL of unknown solution: n Ba(OH)2(aq) Bao d. number of grams of BaO in 20.00 mL of unknown solution: m n BaOX MMBao the % saturation of the unknown solution: % saturation = (g/Lnsat / g/Lsa) x 100% (Assume that a saturated solution would contain: 34.8 g (BaO)y/L.) Submit the Data Table, Graph and Calculations with your Lab Report. e, f. 6) No Post-Lab Questions:

Explanation / Answer

a) Moles of sulfuric acid used = n = M*V

M = molarity = 0.10 M

V = Volume = 10.35 ml = 0.01035 L

number of moles = 0.10*0.01035 = 0.001035 mol = 1.035 mmol

b) Moles of Ba(OH)2 in the unknown solution = nBa(OH)2 = nH2SO4 = 1.035 mmol

c) moles of BaO in 20.00 ml of unknown solution = nBa(OH)2 = nBaO = 1.035 mmol

d) number of grams of BaO in 20.00 ml of unknown solution = m = nBaO*MMBaO

nBaO = moles of BaO = 1.035 mmol = 1.035*10-3 mol

MMBaO = molar mass of BaO = 153 g/mol

Grams of BaO = m = 1.035*10-3 mol*153 g/mol = 158*10-3 g = 0.158 g

e) the % saturation = (g/Lunsat / g/Lsat )*100

g/Lunsat for BaO = grams of BaO/volume

grams of BaO = 0.158 g

volume = 20.00 ml = 0.020 L

g/Lunsat for BaO = 0.158/0.020 = 7.9 g/L

g/Lsat for BaO =34.8 g/L

So, % saturation = (7.9/34.8)*100 = 22.70 %

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