One mole of $P X_{4} O X_{10}$ and one mole of $P X_{4} O X_{6}$ are separately treated with excess water, and each product is completely neutralised with NaOH (all replaceable hydrogens). The ratio of moles of NaOH used for $P X_{4} O X_{10}$ : $P X_{4} O X_{6}$ is

Answer: C

P X_{4} O X_{10} + 6 H X_{2} O 4 H X_{3} PO X_{4}

(tribasic): 4 x 3 = 12 mol NaOH.

P X_{4} O X_{6} + 6 H X_{2} O 4 H X_{3} PO X_{3}

(dibasic): 4 x 2 = 8 mol NaOH. Ratio = 12 : 8 = 3 : 2.