0.01 , text{mole} of H_3PO_x is completely ne | vedclass

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(B) The molar mass of $KOH$ is $56 \, 	ext{g/mol}$.Moles of $KOH = \frac{0.56 \, 	ext{g}}{56 \, 	ext{g/mol}} = 0.01 \, 	ext{mol}$.Given that $0.01

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$x = 2$ and the given acid is monobasic

Vedclass · Answer

(B) The molar mass of $KOH$ is $56 \, 	ext{g/mol}$.Moles of $KOH = \frac{0.56 \, 	ext{g}}{56 \, 	ext{g/mol}} = 0.01 \, 	ext{mol}$.Given that $0.01 \, 	ext{mol}$ of $H_3PO_x$ is neutralized by $0.01 \, 	ext{mol}$ of $KOH$,the stoichiometry of the reaction is $1:1$.This indicates that the acid $H_3PO_x$ is monobasic (it releases one $H^+$ ion).Among the phosphorus oxoacids,$H_3PO_2$ (hypophosphorous acid) is monobasic,where $x = 2$.Therefore,$x = 2$ and the acid is monobasic.

$0.01 \, \text{mole}$ of $H_3PO_x$ is completely neutralized by $0.56 \, \text{g}$ of $KOH$. Hence:

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