Question

How many grams of KI are required to make 550.0 mL of a 1.50 M KI solution?

Answer 1

Here, first we have to calculate the moles of KI using the formula of molarity, that is,
Molarity=$\dfrac{{{\text{Moles}}\,{\text{of}}\,{\text{solute}}}} {{{\text{Volume}}\,{\text{of}}\,{\text{solution}}\left( {\text{L}} \right)}}$. Then, we have to use the formula of the number of moles to calculate the mass of KI.

Complete step by step answer:
The molarity of KI solution is given as 1.50 M and the volume of KI solution is 550.0 mL. By rearranging the molarity of molarity, we get,
Moles of solute = Molarity$\times$Volume of solution
Now, we have to convert the volume into litres.
550 mL=$\dfrac{{550\,}}{{1000}} = \dfrac{{11}}{{20}}\,{\text{L}}$
Now, we have to calculate the moles of KI.
Moles of KI=$1.50\,{\text{mol}}\,\,{{\text{L}}^{ - 1}} \times \dfrac{{11}}{{20}}{\text{L}} = 0.825\,{\text{mol}}$
Now, we have to calculate the mass of KI. We know that,
Number of moles=$\dfrac{{{\text{Mass}}}} {{{\text{Molar}}\,{\text{Mass}}}}$
We calculated the moles of KI, that is 0.825 mol
The molar mass of KI is 166 g/mol.
So,
Mass of KI=Moles of KI$\times$ Molar mass of KI
$\Rightarrow$Mass of KI=$0.825 \times 166 = 136.95\,{\text{g}} \approx {\text{137}}\,{\text{g}}$

Hence, 137 grams of KI are required to make 550.0 mL of a 1.50 M KI solution.

Additional Information:
The mole or mol (abbreviation form) is the unit which measures the amount of substance present. A mole is the term that defines the number of carbon (C) atoms in 12 g of pure carbon. After so many years of experiment, it has been proved that a mole of anything contains $6.022 \times {10^{23}}$ representative particles.

Note: It is to be noted that molarity and molality both are different. Molarity is the number of moles of solute present in one litre solution and molality is the moles of solute present in one kilogram of solvent.