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Published byStephany Marsh Modified over 5 years ago

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**Percent by mass, mole fraction, molarity, and molality**

Concentration Units Percent by mass, mole fraction, molarity, and molality

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**Percent by mASS PERCENT BY MASS = MASS OF SOLUTE X 100%**

MASS OF SOLUTION A sample of 0.892g of potassium chloride is dissolved in 54.6g of water. What is percent by mass of KCl in the solution? 0.892g Percent by mass of KCl = X 100% 0.892g g = 1.61%

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**Percent by mass, mole fraction, molarity, and molality**

Concentration Units Percent by mass, mole fraction, molarity, and molality

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**Mole fraction (x) MOLE FRACTION OF COMPONENT A = XA = Moles of A**

Σ all moles in solution Your solution contains 5 moles of KCl and 15 moles of water. What is the mole fraction of KCl? 5 moles KCl XKCl = = 0.25 5 moles KCl + 15 moles H2O What is the mass percent of KCl? 74.55 g 5 mol KCl x Mass percent of KCl = 1 mol KCl = 58% 74.55 g 18.02 g 5 mol KCl x 15 mol H2O x + 1 mol KCl 1 mol H2O

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**Percent by mass, mole fraction, molarity, and molality**

Concentration Units Percent by mass, mole fraction, molarity, and molality

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**Molarity and molality MOLARITY = M = MOLALITY = m = Moles of solute**

Liters of solution Kg of solvent Calculate the molality of a sulfuric acid solution containing 24.4g of sulfuric acid in 198g of water. The molar mass of sulfuric acid is g/mol. 1 mol H2SO4 24.4 g H2SO4 x mH2SO4 = 98.09 g = 1.26m = 1.26 “molal” 0.198 kg H2O What is the molarity of the solution? (d = 1.84 g/mL) 1 mL Volume of solution = (24.4g + 198g) = mL 1.84g mol H2SO4 Molarity = M = = 2.06M = 2.06 “molar” L solution

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