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Acid neutralized

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Question:

Jesley

Expert

Karma Points: 1,397

Date Posted: 5/12/2008 1:06:43 AM Status: Live

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Response:

given mass of oxalic acid, m= 0.900 g

molar mass of oxalic acid,H₂C₂O₄= 90.04 g/mol

moles of oxalic acid,( solute) = mass/molar mass = 0.900/90.04 =9.9956*10^-3 mol

molarityof a solution = moles of solute/volume of solution in liters

given, molarity = 0.300 M

∴volume of solution = (moles of solute)/molarity=9.9956*10^-3 mol/0.300 =0.03332 liter

∴volume of solution in (ml) = 33.318 ml

vsiddhardha

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Karma Points: 515

Date Posted: 5/12/2008 5:45:17 AM Status: Live

Asker's Rating: This answer has not been rated. If you asked this question, then please login.

Response:

Moles of oxalic acid, = mass/molar mass = 0.900/90.04 =9.9956*10^-3 mol

Molarityof a solution = moles of solute/volume of solution in liters

given, molarity = 0.300 M

∴volume of solution = (moles of solute)/molarity=9.9956*10^-3 mol/0.300 =0.03332 liter

∴volume of solution in (ml) = 33.3ml

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