P4O6 reacts with water according to equation P4O6 + 6H2O→ 4H3PO3. Calculate the volume of 0.1 M NaOH solution required to neutralise the acid formed by dissolving 1.1 g of P4O6 in H2O.

This is a short answer type question as classified in NCERT ExemplarP4O6 + 6H2O→4H3PO3 (i)Next is neutralization4 times H3PO3 + 2NaOH→Na2HPO3 + 2H2O (ii)On addition of 2 equationsP4O6+8NaOH→4Na2HPO3+2H2OP4O6 ( mol.mass )= (4×31+16×6)=220Number of moles of P4O6 = ? / V = ? The product formed is neutralized 8 moles of NaOHP4O6 = 8× ? = ? molNaOHMolarity of NaOH in 1 litre = 0.1M Molarity = M o l a r M a s s G i v e n m a s s Volume = 1.1 220 = 1.1 220 × 8.8 220 = 0.4 L

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P₄O₆ reacts with water according to equation P₄O₆ + 6H₂O→ 4H₃PO₃. Calculate the volume of 0.1 M NaOH solution required to neutralise the acid formed by dissolving 1.1 g of P₄O₆ in H₂O.

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Payal Gupta | Contributor-Level 10

3 months ago

This is a short answer type question as classified in NCERT Exemplar

P₄O₆ + 6H₂O→4H₃PO₃ (i)

Next is neutralization

4 times H₃PO₃ + 2NaOH→Na₂HPO₃ + 2H₂O (ii)

On addition of 2 equations

P₄O₆+8NaOH→4Na₂HPO₃+2H₂O

P₄O₆ ( mol.mass )= (4×31+16×6)=220

Number of moles of P₄O₆ = $? / V$ = $?$

The product formed is neutralized 8 moles of NaOH

P₄O₆ = 8× $?$ = $?$ molNaOH

Molarity of NaOH in 1 litre = 0.1M

Molarity = $\frac{M o l a r M a s s}{G i v e n m a s s}$

Volume = $\frac{1.1}{220}$

= $\frac{1.1}{220}$ × $\frac{8.8}{220}$ = 0.4 L

This is a short answer type question as classified in NCERT ExemplarP4O6 + 6H2O→4H3PO3  (i)Next is neutralization4 times H3PO3 + 2NaOH→Na2HPO3 + 2H2O  (ii)On addition of 2 equationsP4O6+8NaOH→4Na2HPO3+2H2OP4O6 ( mol.mass )= (4×31+16×6)=220Number of moles of P4O6 = <math> <mi mathvariant="bold-italic">? </mi> <mo>/</mo> <mi mathvariant="bold-italic">V</mi> </math>    = <math> <mi mathvariant="bold-italic">? </mi> </math> The product formed is neutralized 8 moles of NaOHP4O6 = 8× <math> <mi mathvariant="bold-italic">? </mi> </math>  = <math> <mi mathvariant="bold-italic">? </mi> </math>  molNaOHMolarity of NaOH in 1 litre  =  0.1M Molarity = <math> <mfrac> <mrow> <mrow> <mi mathvariant="normal">M</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">l</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">r</mi> <mi mathvariant="normal">M</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">s</mi> <mi mathvariant="normal">s</mi> </mrow> </mrow> <mrow> <mrow> <mi mathvariant="normal">G</mi> <mi mathvariant="normal">i</mi> <mi mathvariant="normal">v</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal">m</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">s</mi> <mi mathvariant="normal">s</mi> </mrow> </mrow> </mfrac> </math>  Volume = <math> <mfrac> <mrow> <mrow> <mn>1.1</mn> </mrow> </mrow> <mrow> <mrow> <mn>220</mn> </mrow> </mrow> </mfrac> </math>  = <math> <mfrac> <mrow> <mrow> <mn>1.1</mn> </mrow> </mrow> <mrow> <mrow> <mn>220</mn> </mrow> </mrow> </mfrac> </math> × <math> <mfrac> <mrow> <mrow> <mn>8.8</mn> </mrow> </mrow> <mrow> <mrow> <mn>220</mn> </mrow> </mrow> </mfrac> </math> = 0.4 L

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P₄O₆ reacts with water according to equation P₄O₆ + 6H₂O→ 4H₃PO₃. Calculate the volume of 0.1 M NaOH solution required to neutralise the acid formed by dissolving 1.1 g of P₄O₆ in H₂O.

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