Question

What is the molarity of a solution composed of $5.85g$ of potassium iodide, $KI$ , dissolved in enough water to make $0.125L$ of solution?

Answer 1

Hint : To solve the given question, first we will find the molar mass of the given compound and then we will conclude the number of moles of given compound. Now, we can find the molarity with the help of the molar mass and the number of moles.

Complete Step By Step Answer:
Molarity tells you how many moles of solute, which in your case is potassium iodide, $KI$ , are dissolved in exactly one litre of solution.
The first thing to do here is use the molar mass of potassium iodide to figure out how many moles you have in that sample
$5.85g\dfrac{{1mole\,KI}}{{166.0g}} = 0.03524moles\,KI$
Now, you know that this many moles of potassium iodide are dissolved in $0.125L$ of solution. Since you're interested in figuring out how many moles would be present in $1L$ of solution, all you have to do is scale up your current volume by a factor of
$\dfrac{{1\,L}}{{0.125}} = 8$
The volume goes up by a factor of 8, which can only mean that the number of moles of solute must go up by the same factor
$1\,L\,solution\dfrac{{0.03524moles\,KI}}{{0.125\,L\,solution}} = 0.282\,moles\,KI\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$
Since you have $0.282moles$ of potassium iodide in $1L$ of solution, it follows that the molarity will be
$\therefore molarity = 0.282\,mol\,{L^{ - 1}}\, = 0.282M$

Note :
Molarity can be used to calculate the volume of solvent or the amount of solute. The relationship between two solutions with the same amount of moles of solute can be represented by the formula ${c_1}{V_1}\; = {\text{ }}{c_2}{V_2}$ , where $c$ is concentration and $V$ is volume.