1.00ml of 3.90*10^-4M of oleochemicals acid is diluted with 9.00mL of petroleum
ID: 901734 • Letter: 1
Question
1.00ml of 3.90*10^-4M of oleochemicals acid is diluted with 9.00mL of petroleum ether, forming solution A. Then 2.00mL of solution A is diluted with 8.00mL of petroleum ether, forming solution B. How many grams of oleic acid are 5.00mL of solution B? (Molar mass for oleic acid=282g/mol) 1.00ml of 3.90*10^-4M of oleochemicals acid is diluted with 9.00mL of petroleum ether, forming solution A. Then 2.00mL of solution A is diluted with 8.00mL of petroleum ether, forming solution B. How many grams of oleic acid are 5.00mL of solution B? (Molar mass for oleic acid=282g/mol)Explanation / Answer
no. of moles = molarity x volume in liters = 3.9E-4M x 0.001L = 3.9E-7moles
Molarity M1 = molarity/ volume in liters = 3.9E-7 moles / 0.009L = 4.33E-5M
no. of moles = molarity x volume in liters = 4.33E-5M x 0.002L = 8.6E-8 moles
Molarity M2 = 8.6E-8 moles / 0.008L = 1.08E-5M
no. of moles = molarity x volume in liters = 1.08E-5M x 0.005L = 5.4E-8 moles
mass = moles x molar mass = 5.4E-8 moles x 282g/mol = 1.525E-5g = 0.0152mg
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