A sample of $$\text{P}_2\text{O}_5$$ contains some impurity. 0.405 gm sample is reacted with water and the resulting solution requires 42.5 ml of 0.25 M NaOH solution. The salt resulted is monobasic acid salt. Calculate the percent impurity.

$$\text{P}_2\text{O}_5 + 3\text{H}_2\text{O} \rightarrow 2\text{H}_3\text{PO}_4$$, and an acid salt with one replaceable hydrogen left is $$\text{Na}_2\text{HPO}_4$$, so each $$\text{H}_3\text{PO}_4$$ uses 2 NaOH. NaOH used is $$0.0425 \times 0.25 = 0.010625$$ mol, so $$\text{H}_3\text{PO}_4$$ is 0.0053125 mol and $$\text{P}_2\text{O}_5$$ is 0.00265625 mol, i.e. $$0.00265625 \times 142 \approx 0.3772\text{ g}$$. The impurity is $$\frac{0.405 - 0.3772}{0.405} \times 100 \approx 6.87\%$$.