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 July 26th, 2017, 04:10 AM #1 Member   Joined: May 2015 From: Australia Posts: 78 Thanks: 7 What volume of water should be added? What volume of water should be added to 40.0mL acetic acid B.P. (33% w/w, density of 1.041g/mL) such that the final concentration of acetic acid in the solution is 10% w/v? (The molecular weight of acetic acid is 60.0). The answer is 97.4mL I've worked out two ways to get the answer, but I'm not sure whether the methods are correct. Solution 1: 33% w/w = 33g acetic acid per 100g solution 100g / (1.041g/mL) = 96.06mL solution Thus, 33g acetic acid per 96.06mL solution  X g acetic acid per 40.0mL solution  Using cross multiplication, I worked out x using  and  40x33=96.06x X=13.74g Thus, 13.74g acetic acid in 40.0mL solution (13.74g)/(40mL+xmL) x 100 = 10%w/v Working that out, I got x=97.4mL Solution 2: 33%w/w x 1.041g/mL = 34%w/v C1 = 34%w/v, V1= 40mL, C2= 10% w/v, V2=? Using C1V1=C2V2, I got V2=97.4mL Any input would be appreciated. =) Last edited by skipjack; July 26th, 2017 at 07:25 AM. July 26th, 2017, 08:10 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,926 Thanks: 2205 What does "w/v" in the original description mean? How exactly did you get 97.4mL in your solution 2? July 26th, 2017, 01:54 PM #3 Member   Joined: May 2015 From: Australia Posts: 78 Thanks: 7 Sorry, I should have named solution 1 and 2 as method 1 and method 2 instead. I just worked out 2 ways to find the answer, but I don't know whether it's the right approach. Last edited by skipjack; July 26th, 2017 at 02:36 PM. July 26th, 2017, 02:38 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,926 Thanks: 2205 How did you get 97.4mL using your second method? Can you give the details of your arithmetic? July 26th, 2017, 05:50 PM #5 Member   Joined: May 2015 From: Australia Posts: 78 Thanks: 7 33%w/w x 1.041g/mL =33g/100g x 1.041g/mL = 34%w/v C1 = 34%w/v, V1= 40mL, C2= 10% w/v, V2=? Using C1V1=C2V2, I got V2=97.4mL Just looking at my working out above, it doesn't look like the calculation for lines 1 and 2 are possible. Maybe method 2 isn't the correct way to solve this problem? July 27th, 2017, 07:45 AM   #6
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Quote:
 Originally Posted by pianist Using C1V1=C2V2, I got V2=97.4mL
What are the full details of the arithmetic you did for this step? Typing "I got" doesn't tell me what exactly you did. Also, you have at all stages omitted to say when you have rounded your calculated values and what the unrounded values were.

Have you checked whether "w/v" was a typo? July 27th, 2017, 04:08 PM #7 Member   Joined: May 2015 From: Australia Posts: 78 Thanks: 7 The w/v in the question isn't a typo. 33%w/w x 1.041g/mL =33g/100g x 1.041g/mL = 34.353%w/v C1 = 34.353%w/v, V1= 40mL, C2= 10% w/v, V2=? Using C1V1=C2V2, (34.353)(40)=(10)(V2) V2=137.412mL 137.412mL - 40mL = 97.412mL of water needs to be added Tags added, volume, water Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post pianist Chemistry 3 May 29th, 2017 01:32 AM Ganesh Ujwal Physics 2 January 3rd, 2015 06:47 AM bree05 Algebra 1 May 9th, 2012 04:47 AM kittycub4 Calculus 2 October 21st, 2011 03:55 PM kevin Algebra 14 August 8th, 2009 10:27 AM

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