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 August 21st, 2017, 11:25 PM #1 Member   Joined: May 2015 From: Australia Posts: 77 Thanks: 7 How to find how much needs to be added to get final concentration 150 mL of a solution contains 250 mg of Magnesium Chloride B.P. (MgCl2.6H2O, M.Wt. 203.3). What mass of sodium chloride (NaCl, M.Wt. 58.5) must be added such that when 100 mL of the solution is mixed with water and made up to 250 mL, the final concentration of chloride is 12 mmol/L? Select one: a. 4.7 g b. 39.8 g c. 79.6 mg d. 159.2 mg The answer is 79.6mg. My attempted answer: 0.250g / 203.3 g/mol = 0.00123mol Since the ratio is 1:2 for the magnesium chloride to chloride ions, 0.00123mol x 2 = 0.00246mol Thus, there are 0.00246 mol chloride ions in 150mL. This means there are 0.00164mol chloride ions in 100mL (0.00164 + x)/ 1L = 0.012mol/L x=0.01036mol of NaCl needed 0.01036mol x 58.5g/mol = 0.60606g NaCl I got the incorrect answer; I'm not sure how to get the correct one. Last edited by skipjack; August 22nd, 2017 at 08:32 AM.
 August 22nd, 2017, 09:53 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,110 Thanks: 1909 Retaining more decimal places, 2(0.250/203.3)(100/150) = 0.0016396. If x g of NaCl is added, 0.0016396 + x/58.5 = 0.012*0.25 (as you have only 0.25 L), so x = 58.5(0.012*0.25 - 0.0016396) = 0.079583..., and 0.079583... g = 79.6 mg to 3 significant figures.
 August 23rd, 2017, 02:41 AM #3 Newbie   Joined: Aug 2017 From: Australia Posts: 4 Thanks: 0 Thank you so much. Could you solve this? Potassium Citrate Mixture has the following formula: Potassium citrate 2 g Citric acid monohydrate 400 mg Lemon Syrup 1 mL Methyl hydroxybenzoate Solution 0.1 mL Purified Water to a final volume of 10 mL What volume of the mixture should be given to a patient if the amount of potassium to be administered is 7.5 mmole? The molecular weight of Potassium Citrate B.P. (K3C6H5O7.H2O) is 324.4. Last edited by skipjack; August 23rd, 2017 at 03:10 AM.
 August 23rd, 2017, 03:38 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,110 Thanks: 1909 As 2000 mg / (324.4 mg/mmol) = 6.1652 mmol (of potassium citrate) is in 10mL of the mixture, 12.165 mL of the mixture would contain 7.5 mmol of potassium citrate.
 August 23rd, 2017, 04:20 AM #5 Newbie   Joined: Aug 2017 From: Australia Posts: 4 Thanks: 0 the answer here is 4.06 ml. I don't know how is that.
 August 23rd, 2017, 08:32 AM #6 Senior Member   Joined: Apr 2014 From: UK Posts: 895 Thanks: 328 I note that the chemical formula starts K3, https://www.google.co.uk/search?q=Po...hrome&ie=UTF-8 It's a little 3, meaning there are 3 K's in it, thus divide 12.162 by 3 to get 4.06 give or take the rounding errors.
 August 23rd, 2017, 08:55 AM #7 Newbie   Joined: Aug 2017 From: Australia Posts: 4 Thanks: 0 Thank you so much. Last edited by skipjack; August 24th, 2017 at 12:04 AM.
 August 23rd, 2017, 10:48 PM #8 Member   Joined: May 2015 From: Australia Posts: 77 Thanks: 7 Hi amas4ever, I also have that exact question that you posted in my pharmaceutical calculations homework. Do we go to the same university?!?
 August 24th, 2017, 12:11 AM #9 Senior Member   Joined: Apr 2014 From: UK Posts: 895 Thanks: 328 No problem, though I should point out that I stopped learning any chemistry at 16

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