A) 49
B) 98
C) 196
D) 9.8
Correct Answer: B
Solution :
Equation of orthophosphoric acid with sodium hydroxide is given as : \[\underset{\begin{smallmatrix} \text{1}\,\text{mole} \\ \text{= 3}\,\,\text{+}\,\text{31}\,\text{+}\,\text{64} \\ \text{=}\,\,\text{98}\,\text{g} \end{smallmatrix}}{\mathop{{{H}_{3}}P{{O}_{4}}}}\,+\underset{\begin{align} & \text{3}\,\text{mole} \\ & \text{3}\,\,\times \,\,(23\,+\,16\,+\,1) \\ & =3\,\,\times \,\,40 \\ & =120\,g \\ \end{align}}{\mathop{3NaOH}}\,\xrightarrow{{}}N{{a}_{3}}P{{O}_{4}}+3{{H}_{2}}O\] It is clear from the above equation that 120 g of NaOH is neutralised by 98 g of \[{{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}\text{.}\]
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