Chemical Forums
Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: payata on February 07, 2019, 03:47:34 AM
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How many g K2SO4 is required to prepare a 200ml solution of 190 mg K / Kg (ρ = 1.014 g/ml).
Can someobody explain it step by step?
Thank you!
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1. Convert the volume to mass by using the density.
2. Calculate the mole of potassium and convert this to mole of the potassium sulfate.
3. Calculate back to the mass of potassium sulfate what gives the amount per kg
4. Convert this mass according to the mass of your solution calculated in 1.
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Ok.
So in:
1. I calculated Mass= denisty* volume = 202.8 g
2. 0.19 g of potassium sulfate = 0.0011 mol
and I am lost again :(
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@. ?
How many milligrams of potassium is in 202.8 g? Convert milligrams to grams.
Then try using this with molecular formula (in masses).
Of course, you can use moles of potassium, convert them to moles of potassium sulfate and then calculate the mass of potassium sulfate ( a bit longer way)
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Hi!
Thank you for reply :)
Is it possible to show that calculation, on some kind example .. I do not want be judge but chem is not my strong side, but I really try :(
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200 ml = 202,8 g
0,19 g K = 0,0048 mol, this corresponds to 0,0024 mol K2SO4 and this is equal to 0,42 g
This mass is for 1 kg . For 202,8 g you need then 0,085 g
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Use 3 significant digits in the final result ad at least 4 in the intermediate results.
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Thank you very much, now I can handle another 19th exercises ;D