Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Firoj on October 30, 2020, 05:29:17 PM
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The question ask to find the number of particles in 1L of a 1% solution of KCl.
OK, so my understanding is that I start up with 1L=1000mL and 1% KCl = 1g KCl/100mL
Therefore,
(1g KCl/100mL) x (1000mL) x (1mol KCl/75g KCl) = 0.1333 mol KCl. And since KCl dissociates (let's assume completely) into K+(aq) and Cl-(aq), so the amount of particle solutes is 0.267 mol. Is my calculation correct?
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Depends on what you mean by "particles".
What you did for KCl is more or less right (density of the solution is not 1g/mL, but the difference is quite small), but do water molecules count?
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That's probably what they meant. If you were given a solubility product (Ksp) as well, be careful assuming it dissociates completely. If that's the case, you need to make an ICE table and go through the calculations.
If you want the actual number of particles, then you need to multiply your final answer by Avogadro's constant.