ࡱ> ?A>y ~bjbj 0${{~ ,,,,H$,z lllll$`"% @ lls4 +++ ll+++rTlz~ )R J 0z z%{j%%P+  Fz % : Molarity Notes H. Chemistry Name ____________________________________ Reference the text pp. 418-421 of our text (omit Molality). Sample problems A, B and C are appropriate for our class as well as the practice problems at the bottom of p. 421. Molarity means moles of solute dissolved per liter of solution = mol/L, (symbol = M) Define solvent: Define solute: When you are calculating the concentration of a solution, you must do so by calculating the number of moles of solute dissolved into the total volume of solution. Example 1: Molar mass of salt (NaCl) is 58.5 g/mol, so if you dissolve 58.5g of salt in enough water to make 1.00 liter of solution, you have one mole of solute dissolved per liter of solution. The concentration of this solution is one molar (1.00 mol/L or 1.00M) Example 2: What is the concentration of a solution made by dissolving 175.5g of NaCl in enough water to make 2.00 liters of solution? Answer: 175.5g NaCl 1 mol = 3.00 mol of salt dissolved in 2.00 liters so 3.00mol = 1.50 M NaCl 58.5g 2.00L Practice Problems: SHOW ALL WORK AND USE PROPER SIG FIGS AND UNITS!!! Calculate the molarity of a solution made by dissolving 29.25g of NaCl in enough water to make 2.00 L of solution. Calculate the molarity of a solution made by dissolving 502g of ammonium nitrate in enough water to make 5.00 L of solution. Calculate the molarity of a solution made by dissolving 25.5g of sodium hydroxide in enough water to make 10.5L of solution. Extension problem: How many grams of NaCl must have been dissolved in 1.50L of water to yield a solution whose concentration is 4.50M? (use unit cancellation to solve this one) Dilutions- these problems involve the dilution of a solutiongenerally resulting in a less concentrated solution. Calculate the following problem (use dimensional analysis and show all work, including units) You are given beaker A which contains 1.00L of a 5.00M solution. How many moles of solute do you have in beaker A? You pour .250L of the original 5.00M solution into beaker Bhow many moles are in this beaker? You now add 2.50L of pure water to beaker B. Determine the new concentration in beaker B.  EQUATION TO BE USED FOR ALL DILUTIONS: Minitial molarityVinitial volume = Mfinal molarityVfinal volume M1V1 = M2V2 Just rearrange the above formula to solve for the unknown! Redo the question from above using this new formulashow all work and include units! You are given 125 mL of a 1.50M solution of sodium hydroxide. 100. mL of this solution is removed and added to 75.0mL of pure water for a final volume of 175mL. Determine the new concentration. The remaining 25.0 mL of original solution is placed on a hot plate and 13.0 mL of pure water is vaporized out of it. Determine the new concentration of the resulting solution. A solution is made by dissolving 55.8-g of copper (II) nitrate in enough water to make 450. mL of solution. Calculate the molarity of this solution. Determine the new molarity if 300. mL of this solution is poured into a beaker and 1.50L of pure water is added to this beaker.  +JLcj}   3 ;   ! # $ % . νίƯ{{{e{{*jhl2hCJOJQJUaJmHnHuhl2hhl2h>*CJOJQJaJ"hl2hhl2h6>*CJOJQJaJ"hl2hhl2h5>*CJOJQJaJhRhR56OJQJhR6OJQJhROJQJhl2hhl2hOJQJh CJOJQJaJhl2hhl2hCJOJQJaJhl2hCJOJQJaJ&LR S c d s t   # $  D E F        & Fgdl2h^gdl2hgdl2h       8 > Z 5>    mnop॓ĥĥĥm-jh 5CJOJQJUaJmHnHuhl2hh CJOJQJaJ"hl2hhl2h5>*CJOJQJaJh CJOJQJaJ%hl2hhl2h56>*CJOJQJaJhl2hCJOJQJaJhl2hhl2h>*CJOJQJaJhl2hhl2hCJOJQJaJhl2hhl2h6CJOJQJaJ'       5    m & Fgdl2h & Fgdl2hgdl2hmnopqN%& & Fgdl2h & Fgdl2h & Fgdl2hh`hgdl2hgdl2h & Fgdl2hh^hgdl2h#12>?BCDEFJKLMNP}~²Δvhl2hh)irCJOJQJaJhl2hh CJOJQJaJhl2hCJOJQJaJ"hl2hh 5CJH*OJQJaJhl2hh 5CJOJQJaJh CJOJQJaJhl2hhl2hCJOJQJaJ"hl2hhl2h5CJH*OJQJaJhl2hhl2h5CJOJQJaJabcde~ & F & Fgdl2h & Fgdl2hgdl2h.:pl2h/ =!`"`#`$`% ^ 666666666vvvvvvvvv666666>6666666666666666666666666666666666666666666666666hH6666666666666666666666666666666666666666666666666666666666666666662 0@P`p2( 0@P`p 0@P`p 0@P`p 0@P`p 0@P`p 0@P`p8XV~_HmH nH sH tH 8`8 l2hNormal_HmH sH tH DA D Default Paragraph FontRiR  Table Normal4 l4a (k (No List H@H q Balloon TextCJOJQJ^JaJPK![Content_Types].xmlN0EH-J@%ǎǢ|ș$زULTB l,3;rØJB+$G]7O٭V$ !)O^rC$y@/yH*񄴽)޵߻UDb`}"qۋJחX^)I`nEp)liV[]1M<OP6r=zgbIguSebORD۫qu gZo~ٺlAplxpT0+[}`jzAV2Fi@qv֬5\|ʜ̭NleXdsjcs7f W+Ն7`g ȘJj|h(KD- dXiJ؇(x$( :;˹! I_TS 1?E??ZBΪmU/?~xY'y5g&΋/ɋ>GMGeD3Vq%'#q$8K)fw9:ĵ x}rxwr:\TZaG*y8IjbRc|XŻǿI u3KGnD1NIBs RuK>V.EL+M2#'fi ~V vl{u8zH *:(W☕ ~JTe\O*tHGHY}KNP*ݾ˦TѼ9/#A7qZ$*c?qUnwN%Oi4 =3N)cbJ uV4(Tn 7_?m-ٛ{UBwznʜ"Z xJZp; {/<P;,)''KQk5qpN8KGbe Sd̛\17 pa>SR! 3K4'+rzQ TTIIvt]Kc⫲K#v5+|D~O@%\w_nN[L9KqgVhn R!y+Un;*&/HrT >>\ t=.Tġ S; Z~!P9giCڧ!# B,;X=ۻ,I2UWV9$lk=Aj;{AP79|s*Y;̠[MCۿhf]o{oY=1kyVV5E8Vk+֜\80X4D)!!?*|fv u"xA@T_q64)kڬuV7 t '%;i9s9x,ڎ-45xd8?ǘd/Y|t &LILJ`& -Gt/PK! ѐ'theme/theme/_rels/themeManager.xml.relsM 0wooӺ&݈Э5 6?$Q ,.aic21h:qm@RN;d`o7gK(M&$R(.1r'JЊT8V"AȻHu}|$b{P8g/]QAsم(#L[PK-![Content_Types].xmlPK-!֧6 0_rels/.relsPK-!kytheme/theme/themeManager.xmlPK-!0C)theme/theme/theme1.xmlPK-! ѐ' theme/theme/_rels/themeManager.xml.relsPK] ~ $ ~  m~ 8@*(  D2    "?D2    "?B S  ?$~ 8"|'txe%t ~!+ - ! # 333333cj}#F& e cj} b@`x:|(LQJ s  $~Zr2 h^`OJQJo(hHh^`OJQJ^Jo(hHohpp^p`OJQJo(hHh@ @ ^@ `OJQJo(hHh^`OJQJ^Jo(hHoh^`OJQJo(hHh^`OJQJo(hHh^`OJQJ^Jo(hHohPP^P`OJQJo(hHh^`OJQJo(hHh^`OJQJ^Jo(hHohpp^p`OJQJo(hHh@ @ ^@ `OJQJo(hHh^`OJQJ^Jo(hHoh^`OJQJo(hHh^`OJQJo(hHh^`OJQJ^Jo(hHohPP^P`OJQJo(hHhh^h`o(.^`o(. ^`hH. pLp^p`LhH. @ @ ^@ `hH. ^`hH. L^`LhH. ^`hH. ^`hH. PLP^P`LhH. sb|( $~                             YC&"4kZdcl2h)irgx;N ^q*R~ @RRRR~ @UnknownG*Ax Times New Roman5Symbol3. *Cx Arial;. Tw Cen MT5. .[`)Tahoma?= *Cx Courier New;WingdingsA$BCambria Math"qh++&  q$``x24x x 3QHP ?kZ2!xx Molarity Notes  H Scott Wolfrey Scott Wolfrey    Oh+'0`    ( 4@HPXMolarity Notes HScott WolfreyNormalScott Wolfrey2Microsoft Office Word@@{78@Et @Et  ՜.+,0 hp  CCSDx  Molarity Notes H Title  !"#$%&'()*+,-/012345789:;<=@Root Entry F`/ BData 1Table%WordDocument0$SummaryInformation(.DocumentSummaryInformation86CompObjr  F Microsoft Word 97-2003 Document MSWordDocWord.Document.89q