Let's solve this problem step by step:
1. Identify the Given Values:
- Mass of dibasic acid = 0.315 grams
- Volume of sodium hydroxide (NaOH) solution = 50 milliliters (ml)
- Concentration of NaOH solution = 0.1 Normal (decinormal solution)
2. Convert Volume of NaOH Solution to Liters:
[tex]\[
\text{Volume of NaOH in liters} = \frac{\text{Volume in ml}}{1000}
\][/tex]
[tex]\[
\text{Volume of NaOH in liters} = \frac{50}{1000} = 0.05 \text{ liters}
\][/tex]
3. Calculate the Moles of NaOH:
The formula to find the moles of NaOH is:
[tex]\[
\text{Moles of NaOH} = \text{Volume of NaOH in liters} \times \text{Concentration (Normality)}
\][/tex]
[tex]\[
\text{Moles of NaOH} = 0.05 \text{ liters} \times 0.1 \text{ N} = 0.005 \text{ moles}
\][/tex]
4. Determine the Moles of Dibasic Acid:
According to the stoichiometry of the reaction, 2 moles of NaOH react with 1 mole of dibasic acid (H[tex]\(_2\)[/tex]A):
[tex]\[
\text{Moles of dibasic acid} = \frac{\text{Moles of NaOH}}{2}
\][/tex]
[tex]\[
\text{Moles of dibasic acid} = \frac{0.005}{2} = 0.0025 \text{ moles}
\][/tex]
5. Calculate the Molecular Mass of the Dibasic Acid:
The formula for molecular mass is:
[tex]\[
\text{Molecular mass} = \frac{\text{Mass of acid}}{\text{Moles of acid}}
\][/tex]
[tex]\[
\text{Molecular mass} = \frac{0.315 \text{ grams}}{0.0025 \text{ moles}} = 126 \text{ grams/mole}
\][/tex]
Therefore, the molecular mass of the acid is approximately [tex]\( 126 \)[/tex] grams/mole.
