Answer:
1. Molecular weight = 333.3 g/mol
2. Strenght = 2222.2 mg/L
Explanation:
First, we need to find molarity (M) from normality (N) as follows:
[tex] N = nM [/tex]
Where n is the number of OH⁻ = 2
[tex] M = \frac{N}{2} = \frac{0.03}{2} = 0.015 mol/L [/tex]
Now, when the H₃X solution is neutralized with Ca(OH)₂ we have:
[tex]\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[/tex]
Hence, the molecular weight (MW) of H₃X is:
[tex] MW = \frac{m_{H_{3}X}}{\eta_{H_{3}X}} = \frac{0.2 g}{6 \cdot 10^{-4} moles} = 333.3 g/mol [/tex]
Finally, we can calculate the strength (S) of this solution in ppm as follows:
[tex] S = \frac{m}{V} = \frac{200 mg}{0.050 L + 0.040 L} = 2222.2 mg/L [/tex]
I hope it helps you!
