Question

5.6 g $KOH$ completely neutralises 0.1 mole of $H_{3}P{O}_x$. Thus $x$ is:

• A. 2
• B. 3
• C. 4
• D. 1

Answer 1

Answer: (C)

4

Acc. to Neutralization $r\alpha x$

$\underset{–}{H_{3}P{O}_4}+\underset{–}{3KOH}\to K_{3}P{O}_4+3H_{2}O$

any

$\downarrow$ $\downarrow$

$h_f=x$ $h_f=3$

$mole{s}_1\times h_{fi}=mole{s}_2\times h{f}_2$

$\frac{5.6}{56}\times 3=0.1\times P$

$P=3$ means protons will be release.

which is possible in $H_{3}P{O}_3$

hence $x=4$