Answer 3
Solubility equilibrium of oxalic acid dihydrate (H2C2O4.2H2O)
H2C2O4.2H2O (s) + Heat <------> C2O42- (aq) + 2H+ (aq) + 2H2O (l)
Answer 4
Density of water = 1.00 g/mL
Mass of 33.30 mL of water = 33.30g
Mass of Na2SO4 in 33.30 g of water = 6.51 g
Mass of Na2SO4 in 100 g of water
= (6.51 g*100 g/33.30 g)
= 19.5 g
Solubility = 19.5 g/100 g of water.
Molar mass of solute, Na2SO4 = 142.04 g/mol
Moles of solute = 19.5 g/142.04 g/mol = 0.137 mol
Moles of water = 100 g/18 g/mol = 5.56 mol
Total moles = 0.137 + 5.56 = 5.70
Mole fraction of solute = 0.137/5.70 = 0.024
earchSelect 559 your on the 8.) Dipole and its geome. determine that t. Laboratory Experiments 113 QUESTIONS 1. Do...
please heeelp me compute the concentration mass solute/100 g H2O and mole fraction please. i don't know how to do it. thank you A. Determination of Solubility Above Ambient Temperature Temperature 24°C Mass of apparatus 117. 33 g Mass of apparatus with oxalic acid dehydrate 123.33 Mass of oxalic acid dihydrate used 6 g Mole fraction Crystal formation Temperature (°C) Concentration mass solute/100 H2O Total Volume (mL) 10 ml S m mu 480 45°C 417°C 10 ml 24° C B....
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