Lecture 11 - Solutons

# Lecture 11 - Solutons - Lecture 11 Solutions Lecture 11...

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Lecture 11: Solutions •S o l u t i o n s a r e homogeneous mixtures of two or more substances. •Re la t ive amoun ts o f so lu te and lven t . There are several concentration units. •ppm /ppb : s tudy these w i th you r homewo rk Solution Concentration 1-2 1-3 1-4

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M HCl moles of HCl liters of solution 3 6 2 Molarity = moles solute liters of solution = mol L The brackets [ ] represent “molarity of ” The brackets [ ] represent “molarity of ” •S h o r t h a n d : [N aOH ] = 1 . 0 0 M Molarity •H ow m a n y m o l e s o f D a r e i n s o l u t i o n A ? •W h a t i s t h e m o l a r i t yo f a 2 5 0 0m L s o l u t i o n m a d e b y dissolving 5 grams of A? m a n y g r am s o f Z s h o u l d b e d i s s o l v e d t o m a k e 500mL of 0.25M solution? h a t v o l um e o f 0 . 2 5M X c o n t a i n s 0 . 0 1 2 m o l e s X ? m a n y m Lo f a 1 . 0 M s o l u t i o n o f X s h o u l d b e diluted to make 1L of 0.197M solution? Molarity = moles solute liters of solution 2-3 2-5
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 3-4

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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) ĺ A l 3+ (aq) + 3 3 NO 3 - (aq) 1 A l (NO 3 ) 3 Ł 1 A l 3+ 1 A l (NO 3 ) 3 Ł 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 - 1 A l (NO 3 ) 3 •n = M x volume (in Liters)
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## This note was uploaded on 03/12/2012 for the course CHEMISTRY 131 taught by Professor Lacey during the Fall '11 term at SUNY Stony Brook.

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Lecture 11 - Solutons - Lecture 11 Solutions Lecture 11...

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