The equivalent weight of H_3PO_4 in the follo | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given reaction is $H_3PO_4 + Ca(OH)_2 	o CaHPO_4 + 2H_2O$. In this reaction,$H_3PO_4$ loses $2$ hydrogen ions to form $HPO_4^{2-}$. Thus,the

Vedclass · Accepted Answer

$49$

Vedclass · Answer

(B) The given reaction is $H_3PO_4 + Ca(OH)_2 	o CaHPO_4 + 2H_2O$. In this reaction,$H_3PO_4$ loses $2$ hydrogen ions to form $HPO_4^{2-}$. Thus,the valency factor ($n$-factor) of $H_3PO_4$ is $2$. The molecular weight of $H_3PO_4 = (3 	imes 1) + 31 + (4 	imes 16) = 3 + 31 + 64 = 98$. The equivalent weight is calculated as $\frac{	ext{Molecular Weight}}{n	ext{-factor}} = \frac{98}{2} = 49$.

The equivalent weight of $H_3PO_4$ in the following reaction is: $H_3PO_4 + Ca(OH)_2 \to CaHPO_4 + 2H_2O$

Similar Questions

Calculate the molecular mass of a glucose $C_{6}H_{12}O_{6}$ molecule.

Which of the following statements is correct about the equivalent weight of $KMnO_4$?

The approximate atomic weight of an element is $26.89$. If its equivalent weight is $8.9$,the exact atomic weight of the element would be:

The weight of $H_2SO_4$ in $150 \ mL$ of $N = 1/7$ solution is ........ $g$.

The atomic mass of noble gases is determined by which of the following relationships?

Vedclass Test Series

Exam Paper Generator

Online Exam Module