Microsoft Word - 10-18a,b Molarity wkst-Key PDF

Preview

Summary

Work on Practice problems with Molarity ...

Description



KEY 



 quantitative •  M •     moles of solute  liters of solution          !""  #          $" %&  4.00 M =

moles of solute 12.0 L

moles of solute = 48.0 mol



  $ '$(") "("& 1L = 0.25 L  1000 mL moles of solute moles of solute = 0.038 mol 0.150 M = 0.25 L ? L = 250 mL ×









 * +',-*$./',-* ")&  ? mol HNO3 = 12.6 g HNO3 ×







M=

0.200 mol HNO3 1.0 L

1 mol HNO3 63.0 g HNO3

= 0.200 mol HNO3 

= 0.200 M 

 ! '  /     %   "$(" )   "0""&  0.700 M =







moles of solute 0.250 L

moles of solute = 0.175 mol 

? g KNO3 = 0.175 mol KNO3 ×

101.1 g KNO3 1 mol KNO3

= 17.7 g KNO3

CHEMISTRY: A Study of Matter © 2004, GPB

10.18a

 ( $(**(+ & ? g KNO3 = 0.175 mol KNO3 ×







M=

101.1 g KNO3 1 mol KNO3

= 17.7 g KNO3 

3.5 mol = 28 M  0.125 L



. +&12(""/33-* (""" )     142  ."   '$-!  !" )   '-++-567

 ? mol CaCO3 = 50.0 g CaCO3 × “A”:

? L = 500.0 mL ×

“B”:

M=

1 mol CaCO3 100.0 g CaCO3

1L = 0.500 L 1000 mL

6.0 mol = 1.5 M  4.0 L

M=

= 0.500 mol CaCO3

0.500 mol = 1.00 M 0.500 L

“B” is more concentrated:

1.5 M

  0 '$($" &  2.0 M =

2.5 moles liters of solution

liters of solution = 1.25 L = 1.3 L



  8 +""/,3 """)&  ? mol = 1.00 g NaCl ×

1 mol NaCl = 0.0171 mol NaCl 58.5 g NaCl

1L ? L = 100.0 mL × = 0.1000 L 1000 mL

M=

0.0171 mol = 0.171 M 0.1000 L

  

CHEMISTRY: A Study of Matter © 2004, GPB

10.18b

...