If a student have 0.15 equivalents of $H_{3} P O_{4}$ (full reaction or full neutralization occurs), then how many moles of $H_{3} P O_{4}$ do you have?

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Solution

Gram equivalents $= \frac{no. of moles}{n_{f}}$
As $H_{3} P O_{4}$ is a tribasic acid so, its n-factor is $3$ .
No. of moles of $H_{3} P O_{4}$ $= n = 0.15$ $\times$ $3 = 0.45$ moles

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