Question

C6H5NH3Cl is a chloride salt with an acidic cation. If 46.3 g of C6H5NH3Cl is dissolved in water to make 150 mL of solution, what is the initial molarity of the cation

Answer 1

Answer: The initial molarity of cation is 2.38 M

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Given mass of $C_6H_5NH_3Cl$ = 46.3 g

Molar mass of $C_6H_5NH_3Cl$ =  129.6 g/mol

Volume of solution = 150 mL

Putting values in above equation, we get:

$\text{Molarity of }C_6H_5NH_3Cl=\frac{46.3\times 1000}{129.6\times 150}\\\\\text{Molarity of }C_6H_5NH_3Cl=2.38M$

1 mole of $C_6H_5NH_3Cl$ produces 1 mole of $C_6H_5NH_3^+$ cation and 1 mole of $Cl^-$ anion

So, molarity of $C_6H_5NH_3^+$ cation = (1 × 2.38) = 2.38 M

Hence, the initial molarity of cation is 2.38 M