Fall 07 HW11 solution

# Fall 07 HW11 solution - Homework Assignment#11 Name...

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Homework Assignment #11 Name____________________________ Laboratory Section (circle one) MW1.30 MW6.30 TTh7 1. Chapter 13, Problem 57, 59, 61, 63, 65, 67, 69, 71, 75, 93, 95, 101, 105 13.57 a) Molarity = 2 5 2 5 3 2 5 0.82 g C H OH 1 mol C H OH 1 mL 10.5 mL 46.07 g C H OH 10 L -   = 1.69514 = 1.7 M C 2 H 5 OH b) Molarity = 3 3 3 3 1.22 g NH 1 mol NH 1 mL 33.5 mL 17.03 g NH 10 L -  = 2.138456 = 2.14 M NH 3 13.59 Dilution calculations can be done using M conc V conc = M dil V dil a) M conc = 6.15 M HCl V conc = 25.0 mL M dil = ? V dil = 0.500 L M dil = M conc V conc / V dil = ( )( ) ( ) 3 6.15 M 25.0 mL 10 L 0.500 L 1 mL - = 0.3075 = 0.308 M b) M conc = 2.00 x 10 -2 M KI V conc = 8.55 mL M dil = ? V dil = 10.0 mL M dil = M conc V conc / V dil = ( ) ( ) ( ) 2 2.00 x 10 M 8.55 mL 10.0 mL - = 0.0171 M 13.61 a) Find the number of moles NaCl needed to make 3.5 L of this solution. Convert moles to mass using the molar mass of NaCl (Molar mass = 58.44 g/mol) Mass NaCl = ( ) 0.55 mol NaCl 58.44 g NaCl 3.5 L L 1 mol NaCl = 112.497 = 1.1 x 10 2 g NaCl Add 1.1 x 10 2 g NaCl to enough water to make 3.5 L of aqueous solution. b) Use the relationship M conc V conc = M dil V dil to find the volume of 2.2 M urea needed. M conc = 2.2 M urea V conc = ? M dil = 0.3 M urea V dil = 17.5 L V conc = M dil V dil / M conc = (0.3 M ) (17.5 L) / (2.2 M ) = 2.38636 = 2 L Add 2 L of 2.2 M urea to enough water to make 17.5 L of solution. Note because of the uncertainty in the concentration of the dilute urea (0.3 M ), only one significant figure is justified in the answer. 13.63 a) To find the mass of Cr(NO 3 ) 3 needed, find the moles of Cr(NO 3 ) 3 in 67.5 mL of a 1.33 x 10 -3 M solution and convert to grams using molar mass of Cr(NO 3 ) 3 . Mass Cr(NO 3 ) 3 = ( ) 3 3 3 3 3 3 3 3 1.33 x 10 mol Cr(NO ) 238.03 g Cr(NO ) 10 L 67.5 mL 1 mL L 1 mol Cr(NO ) - -   = 0.0213691 = 0.0214 g Cr(NO

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Fall 07 HW11 solution - Homework Assignment#11 Name...

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