![]() |
| |||||||
| Student Chat You'll find a list of Schools of pharmacy here, and general student topics. |
![]() |
| | LinkBack | Thread Tools | Display Modes |
| |||
|
What weight of the following soluts is required to produce a 1 molar solution of each. CaCL2 KCL MGCL2 C12H22O11 Ca= 40, CL= 35.5, K=39, Mg=24, C=12, O=16, H=1 do you just use moles=mass/mr? |
| |||
|
A 1 molar solution is equivalent to the molecular weight of the constituent ions. I.e. For CaCl2 it would be 40 + 35.5 + 35.5 = 111g Thus a 1 molar solution of CaCl2 will have 111g of CaCl2 It's as simple as that. You could use the equation as mentioned: Moles = Mass/Mr 1 = ?/111 And ? = 111 |
| |||
| If you know it, you can account for the water of crystallisation. Apologies to Admin for advertising, but as I don't have a financial interest here goes... You need introduction to pharmaceutical calculations Amazon.co.uk: Introduction to Pharmaceutical Calculations: Books: Judith A. Rees,Ian Smith,Brian Smith There is also a work book. If you are still at Uni they should have it in the library. If you are a pre-reg you can borrow books by "mail order" from the RPSGB library. RPSGB.org: Using the library Make sure you send them back though - they check if you have any outstanding or overdue books when you try to register! Alternatively, try looking around the internet its got loads of 'science and stuff' on it <g> |
| |||
| Quote:
|
| ||||
| Quote:
It's been so long since I had to know exactly what it meant that I no longer know with any certainty what "diluted 40 fold " means Quote:
(and it's been years since I did any of this so be patient) So.... First thought An equal mixture of each would give a syrup of 1.30g/ml So we need a little bit more of batch x - the more concentrated. We need Xml of syrup x and Yml of syrup y X + Y = 400 (X * 1.32 ) + (Y*1.28) = 400 * 1.305 Then it's just simultaneous equations (see GCSE maths) The answer should have X slightly more than 200 and Y slightly less than 200 Jeff |
| ||||
| Quote:
so take 1.111g and put it in 250mls to work it back: - 1.111/250 = 0.00444g/ml or 4.44mg/ml 0.00444 x 9000 = 40 = dilution factor 2. If we assume we want equal quantities of actual syrup from each batch, split in two initially: - 400 / 2 = 200mls Then calculate how much syrup is needed to give 200mls at each strength: - (strength required/strength we have) x final volume required = volume needed of batch (1.305/1.32) x 200 = 197.73mls (less then 200 as is stronger then solution required) (1.305/1.28) x 200 = 203.90mls (more then 200 as is weaker then solution required) adding these two together gives us more then 400mls (1.97.73 + 203.90) = 401.63mls so we have to reduce by a factor of (400/401.63) = 0.99594 so just multiply each volume by this factor 197.73 x 0.99594 = 196.92 203.90 x 0.99594 = 203.07 then add together 196.92 + 203.07 = 400 to check it equals the volume we want so we can use 196.92mls of solution X and 203.07mls of solution Y = 400mls in total I hope this is right - I'm tired! No 2 is a bit long but that's how my brain works I'm afraid! I tried to write the steps down. In reality you can do this a lot quicker then it looks. It's also easier if you stick to your calculator as you can use the memory (and thus the same number of decimal places) for each calculation.
__________________ Please never reveal personal details on the forum. Keep it clean because I'll be watching ! |
| ||||
| Quote:
A simpler way to do it would be: - in 200mls of a 1.305g/ml solution there is (1.305/2) x 200 = 130.5g of syrup for solution x we thus need (130.5/1.32) x 2 = 197.73mls for solution Y we thus need (130.5/1.28) x 2 = 203.90mls These add up to 401.63 so we still need to reduce by 400/401.63 to get a final volume of 400mls The same but maybe a bit shorter........
__________________ Please never reveal personal details on the forum. Keep it clean because I'll be watching ! Last edited by admin; 21st, January 2008 at 11:31 PM. |