4.648 gm of solute is needed to make 37.5 mL of 0.750 M KI solution.
Solution:
We will start with the Molarity
![\text { Molarity }=\text { Mole of solute } \div \text { liter of solution }](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Molarity%20%7D%3D%5Ctext%20%7B%20Mole%20of%20solute%20%7D%20%5Cdiv%20%5Ctext%20%7B%20liter%20of%20solution%20%7D)
Also we know 1000 ml = 1 L
Therefore 37.5 ml by 1000ml we obtained 0.0375L
Equation for solving mole of solute
![\text { Mole of solute }=\text { Molarity } \times \text { Liters of solution }](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Mole%20of%20solute%20%7D%3D%5Ctext%20%7B%20Molarity%20%7D%20%5Ctimes%20%5Ctext%20%7B%20Liters%20of%20solution%20%7D)
Now, multiply 0.750M by 0.0375
Substitute the known values in the above equation we get
![0.750 \times 0.0375=0.0281](https://tex.z-dn.net/?f=0.750%20%5Ctimes%200.0375%3D0.0281)
Also we know that Molar mass of KI is 166 g/mol
So divide the molar mass value to get the no of grams.
![0.028 \times 166=4.648](https://tex.z-dn.net/?f=0.028%20%5Ctimes%20166%3D4.648)
So 4.648 gm of Solute is required for make 37.5 mL of 0.750 M KI solution.