Asked by Colin
The mole fraction of an aqueous solution of sodium dichromate is 0.367. Calculate the molarity (in mol/L) of the sodium dichromate solution, if the density of the solution is 1.42 g mL-1.
Answers
Answered by
DrBob222
XNa2CrO4 = 0.367.
XH2O = 1-0.367 = 0.633
Suppose we have 1 mol of solution. That means we have 0.367 mol Na2CrO4 and 0.633 mol H2O. 0.633 mole H2O = 0.633*18 g/mol = 11.4 g H2O. g Na2CrO4 = mols x molar mass = 0.367*about 162 = about 59.5. Total mass of solution = 59.5+11.4 = about 71 grams. mass = volume x density or volume = mass/density = 71/1.42 = about 50 mL and 0.367moles/0.050 L = ??M Check my work. It's late. I've estimated so you need to clean up the numbers.
XH2O = 1-0.367 = 0.633
Suppose we have 1 mol of solution. That means we have 0.367 mol Na2CrO4 and 0.633 mol H2O. 0.633 mole H2O = 0.633*18 g/mol = 11.4 g H2O. g Na2CrO4 = mols x molar mass = 0.367*about 162 = about 59.5. Total mass of solution = 59.5+11.4 = about 71 grams. mass = volume x density or volume = mass/density = 71/1.42 = about 50 mL and 0.367moles/0.050 L = ??M Check my work. It's late. I've estimated so you need to clean up the numbers.
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