lab report 2- individual - Rylee Saunders Chem 403...

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Rylee Saunders Chem 403 9/23/18 Quantitative Aspects of Working With Solutions- Individual 15.) 150g sucrose × 1 mol sucrose 342.196 g = 0.438 moles sucrose 0.438 moles 250 mL × 1000 mL 1 L = 1.752 moles per liter 16.) mass= density × volume 0.792 g/mL × 125 mL= 99.0 g 99.0 g × 1 molmethanol 32.0326 g = 3.09 moles molarity= moles liters 3.09 moles 15 L = 0.206 moles per liter 17.) 250 mL= 0.250 L 0.250 L × 0.4 mNaOH 1 L = 0.1 mol NaOH × 40 g 1 mol NaOH = 4.0 g NaOH 18.) M 1 V 1 =M 2 V 2 (0.650 M)(v1) = (0.423 M)(250.0 mL) 0.650(v1) = 105.75 v1= 105.75 M x mL 0.650 M V 1 = 162.69 mL 19.) a) 100 mL of 0.30 M AlCl
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