Molar Calculation Problem (1 Viewer)

[Derkk]

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<meta http-equiv="Content-Type" content="text/html; charset=utf-8"><meta name="ProgId" content="Word.Document"><meta name="Generator" content="Microsoft Word 10"><meta name="Originator" content="Microsoft Word 10"><link rel="File-List" href="file:///C:%5CDOCUME%7E1%5CAdmins%5CLOCALS%7E1%5CTemp%5Cmsohtml1%5C01%5Cclip_filelist.xml"><!--[if gte mso 9]><xml> <w:WordDocument> <w:View>Normal</w:View> <w:Zoom>0</w:Zoom> <w:Compatibility> <w:BreakWrappedTables/> <w:SnapToGridInCell/> <w:WrapTextWithPunct/> <w:UseAsianBreakRules/> <w:UseFELayout/> </w:Compatibility> <w:BrowserLevel>MicrosoftInternetExplorer4</w:BrowserLevel> </w:WordDocument> </xml><![endif]--><style> <!-- /* Font Definitions */ @font-face {font-family:"MS Mincho"; panose-1:2 2 6 9 4 2 5 8 3 4; mso-font-alt:"ＭＳ 明朝"; mso-font-charset:128; mso-generic-font-family:modern; mso-font-pitch:fixed; mso-font-signature:-1610612033 1757936891 16 0 131231 0;} @font-face {font-family:"\@MS Mincho"; panose-1:2 2 6 9 4 2 5 8 3 4; mso-font-charset:128; mso-generic-font-family:modern; mso-font-pitch:fixed; mso-font-signature:-1610612033 1757936891 16 0 131231 0;} /* Style Definitions */ p.MsoNormal, li.MsoNormal, div.MsoNormal {mso-style-parent:""; margin:0in; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:"Times New Roman"; mso-fareast-font-family:"MS Mincho";} @page Section1 {size:8.5in 11.0in; margin:1.0in 1.25in 1.0in 1.25in; mso-header-margin:.5in; mso-footer-margin:.5in; mso-paper-source:0;} div.Section1 {page:Section1;} --> </style><!--[if gte mso 10]> <style> /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:10.0pt; font-family:"Times New Roman"; mso-fareast-font-family:"Times New Roman";} </style> <![endif]--> 2) A sample of KOH_(s) was found to contain 8.75% water by mass. How many grams of this material are required to make up 500mL of 6.00 MolL^-1 KOH?


Just post working out and answer, ta.
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[brenton1987]

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n = C * V
n = 3.0 mol^.L^-1 * 0.5 L
n = 1.5 mol

m = n * F
m = 1.5 mol * 668.9964772 g^.mol^-1
m = 1003.4947 g

<link rel="File-List" href="file:///C:%5CDOCUME%7E1%5CAdmins%5CLOCALS%7E1%5CTemp%5Cmsohtml1%5C01%5Cclip_filelist.xml"><!--[if gte mso 9]><xml> <w:WordDocument> <w:View>Normal</w:View> <w:Zoom>0</w:Zoom> <w:Compatibility> <w:BreakWrappedTables/> <w:SnapToGridInCell/> <w:WrapTextWithPunct/> <w:UseAsianBreakRules/> <w:UseFELayout/> </w:Compatibility> <w:BrowserLevel>MicrosoftInternetExplorer4</w:BrowserLevel> </w:WordDocument> </xml><![endif]--><style> <!-- /* Font Definitions */ @font-face {font-family:"MS Mincho"; panose-1:2 2 6 9 4 2 5 8 3 4; mso-font-alt:"ＭＳ 明朝"; mso-font-charset:128; mso-generic-font-family:modern; mso-font-pitch:fixed; mso-font-signature:-1610612033 1757936891 16 0 131231 0;} @font-face {font-family:"\@MS Mincho"; panose-1:2 2 6 9 4 2 5 8 3 4; mso-font-charset:128; mso-generic-font-family:modern; mso-font-pitch:fixed; mso-font-signature:-1610612033 1757936891 16 0 131231 0;} /* Style Definitions */ p.MsoNormal, li.MsoNormal, div.MsoNormal {mso-style-parent:""; margin:0in; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:"Times New Roman"; mso-fareast-font-family:"MS Mincho";} @page Section1 {size:8.5in 11.0in; margin:1.0in 1.25in 1.0in 1.25in; mso-header-margin:.5in; mso-footer-margin:.5in; mso-paper-source:0;} div.Section1 {page:Section1;} --> </style><!--[if gte mso 10]> <style> /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:10.0pt; font-family:"Times New Roman"; mso-fareast-font-family:"Times New Roman";} </style> <![endif]-->2) A sample of KOH_(s) was found to contain 8.75% water by mass. How many grams of this material are required to make up 500mL of 6.00 MolL^-1 KOH?[<o></o>/quote]
n = C * V
n = 6 mol^.L^-1 * 0.5 L
n = 3 mol

m = n * F
m = 3 mol * 56.1771 g^.mol^-1
m = 168.5313 g

m = 168.5313 * 1.0875
m = 183.2778 g

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The formulae in blue you should know. What were you doing wrong?

 

[Derkk]

Ehh thanks. I eventually figured it out using C1V1 lol.
I didn't want to use n=cv and realised could've used C1V1.

 

[Lukybear]

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m = n * F
m = 1.5 mol * 668.9964772 g^.mol^-1
m = 1003.4947 g

Are you sure thats right? Isnt it 666.458 grams.

m = n * F
m = 3 mol * 56.1771 g^.mol^-1
m = 168.5313 g

m = 168.5313 * 1.0875
m = 183.2778 g

where n = no.moles
NOt sure thats right either. I.e. n x (100-8.75)/100 = 3
n=3 x 100/91.75
m=56.108 x n
= 184.46

I know that it is terribly a small no. but the working out wasnt correct, at least i believe. Feel free to correct me.

 

[brenton1987]

Are you sure thats right? Isnt it 666.458 grams.

I calculated again using some different tables and got
666.4094 g^.mol^-1
666.4274 g^.mol^-1
666.4277 g^.mol^-1
666.4280 g^.mol^-1

So,

m = n * F
m = 1.5 mol * 666.4277 g^.mol^-1
m = 999.6416 g

The maths was correct but the table was wrong. So clearly the online periodic table is a bag of crap and the tables from the textbooks are much better.

where n = no.moles
Not sure thats right either. I.e. n x (100-8.75)/100 = 3
n=3 x 100/91.75
m=56.108 x n
= 184.46

I dont see what youve done there.

Anyway the mass calculated when not using the online table is
m = 183.0445 g

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As long you use the values given in the data sheet provided in the exam you will be given full marks even if the data values do not agree with reality.