lecture11 - Lecture 11: Solutions Lecture 11: Solutions...

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Lecture 11: Solutions • Solutions are homogeneous mixtures of two or more substances. • Relative amounts of solute and solvent. There are several concentration units. • ppm/ppb : study these with your homework Most important to chemists: Molarity Molarity solute solute – substance dissolved. solvent solvent – substance doing the dissolving. Solution Concentration
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M HCl moles of HCl liters of solution 3 6 2 Molarity = moles solute liters of solution = V of solution not solvent. mol L The brackets [ ] represent “molarity of ” The brackets [ ] represent “molarity of ” • Shorthand: [NaOH] =1.00 M Molarity Types of problems you need to be able to solve • How many moles of D are in solution A? • What is the molarity of a 2500mL solution made by dissolving 5 grams of A? • How many grams of Z should be dissolved to make 500mL of 0.25M solution? • What volume of 0.25M X contains 0.012 moles X? • How many mL of a 1.0 M solution of X should be diluted to make 1L of 0.197M solution? Molarity = moles solute liters of solution
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Molarity Calculate the molarity of sodium sulfate in a solution that contains 36.0 g of Na 2 SO 4 in 750.0 mL of solution. n Na2SO4 = 36.0 g 142.0 g/mol = 0.2534 mol [Na 2 SO 4 ] = 0.2534 mol 0.7500 L [Na 2 SO 4 ] = 0.338 mol/L = 0.338 M Unit change! (mL to L) Molarity (a) A l (NO 3 ) 3 FM = 26.98 + 3(14.00) + 9(16.00) [ A l (NO 3 ) 3 ] = 2.991 x 10 -2 mol / 0.250 L = 0.120 M 6.37 g of A l (NO 3 ) 3 are dissolved to make a 250. mL aqueous solution. Calculate (a) [A l (NO 3 ) 3 ] (b) [A l 3+ ] and [NO 3 - ]. n Al(NO3)3 = = 2.991 x 10 -2 mol 6.37 g 213.0 g/mol g mol = 213.0
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Molarity 6.37 g of A l (NO 3 ) 3 in a 250. mL aqueous solution. Calculate (a) the molarity of the A l (NO 3 ) 3
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This note was uploaded on 11/18/2011 for the course CHE 131 taught by Professor Kerber during the Spring '08 term at SUNY Stony Brook.

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lecture11 - Lecture 11: Solutions Lecture 11: Solutions...

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