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CH 301 FUND Ch4 notes part 1 answers

# CH 301 FUND Ch4 notes part 1 answers - M 1 =0.1M V 1 =5ml V...

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CH301(4) Notes part 1 answers 1. Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution. Note here we must covert moles to grams: 4 2 4 2 4 2 4 2 4 2 4 2 SO H 0728 . 0 L SO H mol 0728 . 0 SO H g 98.1 SO H mol 1 n sol' L 75 . 1 SO H g 12.5 n sol' L SO H mol ? M = = × = 2. Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M Ca(NO 3 ) 2 . 2 3 2 3 2 3 2 3 2 3 ) Ca(NO g 459 ) Ca(NO mol 1 ) Ca(NO g 164 L ) Ca(NO mol 0.800 L 50 . 3 ) Ca(NO g ? = × × = Note: you would NOT add this to 3.5L of water, but use the method shown previously: add sufficient distilled water to the dissolved solid to make the volume 3.5L Dilution example 1: We take 5ml of a 0.1M solution of HCl and place it in a 100ml volumetric flask, and dilute to the mark with distilled water. What is the concentration of the new HCl solution?
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Unformatted text preview: M 1 =0.1M, V 1 =5ml, V 2 =100ml, find M2. M 2 = 0.1M x 5ml = 0.005 M HCl 100ml Notice as LONG as you make sure the units cancel its OK to use ml not L. Dilution example 2: If 10.0 mL of 12.0 M HCl is added to enough water to give 100. mL of solution, what is the concentration of the solution? ilution example 3: sulfuric acid is required to make 2.50 L of a 2.40 M sulfuric acid solution? M M M M M M M 20 . 1 mL 100.0 mL . 10 . 12 mL 100.0 mL . 10 . 12 V V 2 2 2 2 1 1 = × = × = × = D What volume of 18.0 M V V = M M mL 333 or L 0.333 18.0 2.40 L 2.50 V V V 1 1 2 2 1 2 2 1 1 = × = × = M M M M...
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