gwiyomi17 Posted June 16, 2013 Share Posted June 16, 2013 Calculate the mass of impure KOH needed to make up 1.20 L of 0.60 mol/L KOH(aq)? Assume the impure KOH is 84% KOH by mass and 16% water. mimpure KOH=?...........84% VKOH=1.20L CKOH=0.60 mol/L H2O=16% I tried to do this but i don't think this is right, I don't know what should I do first... nimpure KOH=m/M= 84/56.1056 = 1.497 mol nH2O= m/M = 16/18 = 0.89 mol How to solve this?? Link to comment Share on other sites More sharing options...
studiot Posted June 16, 2013 Share Posted June 16, 2013 (edited) I should start by deciding the mass of KOH in 1.2L of 0.6M solution. Then see if you can write an equation connecting this to the mass of KOH in the impure source. Does the inclusion of water make any difference? Why did they tell you it was 16%? Edited June 16, 2013 by studiot Link to comment Share on other sites More sharing options...
cresol Posted June 17, 2013 Share Posted June 17, 2013 we dnt do home work here...but if i will help, try to rearrange the values and write the equation..it works with the KOH mass Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted June 17, 2013 Share Posted June 17, 2013 Don't worry so much about the water component of the KOH. It's really only the % by mass of KOH that you need to worry about. If it helps, another way of expressing the amount of KOH you have is by saying that in every gram of your impure sample, you have 0.84 grams of KOH. If you know the mass required to make 1.2 L of 0.60 M KOH, then knowing how much of the impure stuff you need is just a simple conversion. For instance, if say I need 0.84 g of KOH to make up my solution, then I would need 1 g of the impure sample. If I need 1.68 g of KOH, I would use 2 g of the impure sample (1.68 / 0.84 = 2). Your first step is, as studiot said, to work out the mass of KOH required to make your solution. Link to comment Share on other sites More sharing options...
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