Question

Calculate the molecular weight of 0.2g H3X solution if 50ml of this acid neutralized with 40ml of 0.03N Ca(oH)2 then calculate strength of this solution in ppm

Answer 1

1. Molecular weight = 333.3 g/mol

2. Strenght = 2222.2 mg/L

Step-by-step explanation:

First, we need to find molarity (M) from normality (N) as follows:

`N = nM`

Where n is the number of OH⁻ = 2

`M = (N)/(2) = (0.03)/(2) = 0.015 mol/L`

Now, when the H₃X solution is neutralized with Ca(OH)₂ we have:

`\eta_{H_(3)X} = \eta_{Ca(OH)_(2)} = M_{Ca(OH)_(2)}*V_{Ca(OH)_(2)} = 0.015*0.04 = 6 \cdot 10^(-4) moles`

Hence, the molecular weight (MW) of H₃X is:

`MW = \frac{m_{H_(3)X}}{\eta_{H_(3)X}} = (0.2 g)/(6 \cdot 10^(-4) moles) = 333.3 g/mol`

Finally, we can calculate the strength (S) of this solution in ppm as follows:

`S = (m)/(V) = (200 mg)/(0.050 L + 0.040 L) = 2222.2 mg/L`

I hope it helps you!