Lec-31-Chap5-3

Lec-31-Chap5-3 - Solution Concentration Relative amounts of...

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Relative amounts of solute and solvent. There are several concentration units. Most important to chemists: Molarity Molarity ± solute solute – substance dissolved. ± solvent solvent – substance doing the dissolving. Solution Concentration Molarity = moles solute liters of solution = V of solution not not solvent. mol L The brackets [ ] represent “molarity of ” The brackets [ ] represent “molarity of ” • Shorthand: [NaOH] =1.00 M Molarity
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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 , (b) the molar concentration of A l 3+ and NO 3 - ions in solution. (b) Molarity of A l 3+ , NO 3 - ? A l (NO 3 ) 3 (aq) A l 3+ (aq) + 3 3 NO 3 - (aq) 1 A l (NO 3 ) 3 1 A l 3+ 1 A l (NO 3 ) 3 3 NO 3 - [A l 3+ ] = 0.120 M A l (NO 3 ) 3 = 0.120 M A l 3+ 1 A l 3+ 1 A l (NO 3 ) 3 [NO 3 - ] = 0.120 M A l (NO 3 ) 3 = 0.360 M NO 3 - 3 NO 3 -
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This note was uploaded on 11/14/2010 for the course CHE 131 taught by Professor Kerber during the Spring '08 term at SUNY Stony Brook.

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Lec-31-Chap5-3 - Solution Concentration Relative amounts of...

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