Answer:
123 g/mol
Explanation:
Let's consider the neutralization reaction.
H₂Z(aq) + 2 NaOH(aq) → Na₂Z(aq) + 2 H₂O(ℓ)
The moles of NaOH that reacted are:
[tex]40.50 \times 10^{-3} L \times \frac{0.100molNaOH}{1L} =4.05\times 10^{-3} molNaOH[/tex]
The molar ratio of H₂Z to NaOH is 1:2. The moles of H₂Z that reacted are:
[tex]4.05\times 10^{-3} molNaOH \times \frac{1molH_{2}Z}{2molNaOH} =2.03\times 10^{-3} molH_{2}Z[/tex]
The molar mass of H₂Z is:
[tex]\frac{0.250gH_{2}Z}{2.03\times 10^{-3} molH_{2}Z} =123g/mol[/tex]
