100 , mL of Na_{3}PO_{4} solution contains 3. | vedclass

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(B) The dissociation of $Na_{3}PO_{4}$ is given by: $Na_{3}PO_{4} \rightarrow 3Na^{+} + PO_{4}^{3-}$.Number of moles of $Na$ atoms = $\frac{3.45 \, g}

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$50$

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(B) The dissociation of $Na_{3}PO_{4}$ is given by: $Na_{3}PO_{4} ightarrow 3Na^{+} + PO_{4}^{3-}$.Number of moles of $Na$ atoms = $\frac{3.45 \, g}{23.0 \, g \, mol^{-1}} = 0.15 \, mol$.Since $1 \, mol$ of $Na_{3}PO_{4}$ contains $3 \, mol$ of $Na$,the moles of $Na_{3}PO_{4}$ = $\frac{1}{3} 	imes 0.15 \, mol = 0.05 \, mol$.Volume of solution = $100 \, mL = 0.1 \, L$.Molarity $(M)$ = $\frac{	ext{moles of solute}}{	ext{Volume of solution in } L} = \frac{0.05 \, mol}{0.1 \, L} = 0.5 \, mol \, L^{-1}$.Expressing in terms of $10^{-2}$: $0.5 = 50 	imes 10^{-2} \, mol \, L^{-1}$.Thus,the nearest integer is $50$.

$100 \, mL$ of $Na_{3}PO_{4}$ solution contains $3.45 \, g$ of sodium. The molarity of the solution is $..... \times 10^{-2} \, mol \, L^{-1}$ $(Nearest \, integer)$.
[$Atomic \, Masses - Na: 23.0 \, u, O: 16.0 \, u, P: 31.0 \, u$]

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$100 \, mL$ of $Na_{3}PO_{4}$ solution contains $3.45 \, g$ of sodium. The molarity of the solution is $..... \times 10^{-2} \, mol \, L^{-1}$ $(Nearest \, integer)$.[$Atomic \, Masses - Na: 23.0 \, u, O: 16.0 \, u, P: 31.0 \, u$]