Hi, I'm wondering if anyone could help me with a question I'm really stuck on please.
What weight of potassium chloride is needed to prepare 500ml of a solution containing 6mmol of potassium ions per 5ml spoonful?
(Atomic weight of potassium = 39g
Atomic weight of chloride = 35.5g)
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Pharmacy Calculations help watch
- Thread Starter
- 06-03-2018 14:49
- 06-03-2018 16:29
Let me guide you through this calculation step by step (I will try and get you to think and understand,so you can do similar Qs yourself in future).
1. We need 6 mmol of of potassium ions (K+ - why is it "one plus"?.......................... Yes well done!: because a potassium atom has one electron in outer shell (it is a Group 1 metal in periodic table just like sodium) so it loses this electron (negative) to reach the stable state of eight electrons in outer shell - if you lose one electron i.e. one minus, what charge are you left with?.......................Yes cool! One plus) in 5 ml of solution.
2. Ok so far so good? But we need 500 ml, which do you agree is greater than 5ml (sorry for treating you like baby haha!) Ok how much greater? Yes, you did it once again - 100 times greater!
3. So how many mmoles of potassium ions do we need in 500 ml of sol-n? ..........................I can't believe it! You got it right once again! 6 X 100 = 600 mmoles = how many moles? 0.6 moles (because there are 1000 millimoles in one mole [milli = one-thousandth part])
To have 0.6 moles of K+ (potassium ions) you need 0.6 moles of potassium chloride (KCl) - as there is one K in KCl (other example there are two potassium-s in one molecule of potassium sulphate (K2SO4) - the K2 part - are u with me? Good!)
So you need to work out the mass of 0.6 moles of potassium chloride to get your answer.
4. Add 39g to 35.5g to give mass of one mole of KCl = 74.5g
5. 0.6 moles is a bit more than half (0.5) moles so it should weigh less than the weight of one mole (bit more than half of 75 roughly so 40 something - yes?)
= 0.6 X 74.5g
= 44.7g (ANSWER).
Easy peasy, lemon squeezy, agreed?