<u>Answer:</u>
<u>For A:</u> The molarity of solution is 0.218 M.
<u>For B:</u> The molarity of solution is 0.532 M.
<u>For C:</u> The molarity of solution is ![8.552\times 10^{-4}M](https://tex.z-dn.net/?f=8.552%5Ctimes%2010%5E%7B-4%7DM)
<u>Explanation:</u>
Molarity is defined as the number of moles present in one liter of solution.
Mathematically,
![\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20the%20solution%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20solute%7D%7D%7B%5Ctext%7BVolume%20of%20solution%20%28in%20L%29%7D%7D)
Or,
![\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20the%20solution%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20solute%7D%5Ctimes%201000%7D%7B%5Ctext%7BMolar%20mass%20of%20solute%7D%5Ctimes%20%5Ctext%7BVolume%20of%20solution%20%28in%20mL%29%7D%7D)
- <u>For A:</u> 0.12 mol of
in 5.5 L of solution
We are given:
Moles of
= 0.12 moles
Volume of the solution = 5.5 L
Putting values in above equation, we get:
![\text{Molarity of }LiNO_3=\frac{0.12}{5.5L}\\\\\text{Molarity of }LiNO_3}=0.0218M](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20%7DLiNO_3%3D%5Cfrac%7B0.12%7D%7B5.5L%7D%5C%5C%5C%5C%5Ctext%7BMolarity%20of%20%7DLiNO_3%7D%3D0.0218M)
Hence, the molarity of solution is 0.0218 M.
- <u>For B:</u> 60.7 g
in 2.48 L of solution
We are given:
Given mass of
= 60.7 g
Molar mass of
= 46 g/mol
Volume of the solution = 2.48 L
Putting values in above equation, we get:
![\text{Molarity of }C_2H_6O=\frac{60.7g}{46g/mol\times 5.5L}\\\\\text{Molarity of }C_2H_6O}=0.532M](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20%7DC_2H_6O%3D%5Cfrac%7B60.7g%7D%7B46g%2Fmol%5Ctimes%205.5L%7D%5C%5C%5C%5C%5Ctext%7BMolarity%20of%20%7DC_2H_6O%7D%3D0.532M)
Hence, the molarity of solution is 0.532 M.
- <u>For C:</u> 14.2 mg KI in 100 mL of solution
We are given:
Given mass of KI = 14.2 mg =
(Conversion factor: ![1mg=10^{-3}g](https://tex.z-dn.net/?f=1mg%3D10%5E%7B-3%7Dg)
Molar mass of KI = 166 g/mol
Volume of the solution = 100 L
Putting values in above equation, we get:
![\text{Molarity of KI}=\frac{14.2\times 10^{-3}g\times 1000}{166g/mol\times 100mL}\\\\\text{Molarity of KI}=8.552\times 10^{-4}M](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20KI%7D%3D%5Cfrac%7B14.2%5Ctimes%2010%5E%7B-3%7Dg%5Ctimes%201000%7D%7B166g%2Fmol%5Ctimes%20100mL%7D%5C%5C%5C%5C%5Ctext%7BMolarity%20of%20KI%7D%3D8.552%5Ctimes%2010%5E%7B-4%7DM)
Hence, the molarity of solution is ![8.552\times 10^{-4}M](https://tex.z-dn.net/?f=8.552%5Ctimes%2010%5E%7B-4%7DM)