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

Step-by-step explanation:

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

`\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \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=(46.3* 1000)/(129.6* 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