A) 59
B) 49
C) 25
D) 98
Correct Answer: D
Solution :
\[\underset{\begin{smallmatrix} 23+16+1 \\ =40 \\ 1g.eq \end{smallmatrix}}{\mathop{NaOH}}\,+\underset{\begin{smallmatrix} 1\times 3+31+16\times 4 \\ =98 \\ 1g-eq \end{smallmatrix}}{\mathop{{{H}_{3}}P{{O}_{4}}}}\,\xrightarrow[{}]{{}}Na{{H}_{2}}P{{O}_{4}}+{{H}_{2}}O\] \[\therefore \]equivalent weight of\[{{H}_{3}}P{{O}_{4}}\]is 98 g In the above reaction no change in oxidation state of phosphorus takes place. So, equivalent weight will by equal to its molecular weight.You need to login to perform this action.
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