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Estimation of phosphorous: Second meth...

Estimation of phosphorous:
Second method: A known mass of compound is heated with fuming `HNO_(3)` or sodium peroxide `(Na_(2)O_(2)` in Carius tube which converts phosphorous to `H_(3)PO_(4)`. Magnesia mixture `(MgCl_(2)+NH_(4)Cl)` is then added, which gives the percipate of magnesium ammonium phosphate `(MgNH_(4).PO_(4))` which on heating gives magnesium pyrophosphate `(Mg_(2)P_(2)O_(7))` , which is weighted.
Percentage of P`=("Atomic mass of P")/("Molecular mass of " Mg_(2)P_(2)O_(7)) xx ("Molecular mass of "Mg_(2)P_(2)O_(7) xx 100)/("Mass of compound")=(62)/(222) xx("Mass of " Mg_(2)P_(2)O_(7)xx 100)/("Mass of compound")`
An organic compound has 6.2% of phosphorus .In the reaction sequence , all phosphorous is converted to `Mg_(2)P_(2)O_(7)` . Find wt. of `Mg_(2)P_(2)O_(7)` formed

A

2.22

B

10.2

C

15

D

20

Text Solution

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The correct Answer is:
A
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In a determination of P an aqueous solution of NaH_(2)PO_(4) is treated with a mixture of ammonium and magnesium ammonium phosphate Mg(NH_(4))PO_(4).6H_(2)O . This is heated and decomposed to magnesium pyrophosphate, Mg_(2)P_(2)O_(7) which is weighed . A solution of NaH_(2)PO_(4) yeilded 1.11 g of Mg_(2)P_(2)O_(7) . What weight of NaH_(2)P_(2)O_(7) was present originally ? (P = 31)

How many moles of magnesium phosphate, Mg_(3)(PO_(4)_(2) will contain 0.25 mole of oxygen atoms?

Knowledge Check

  • Estimation of phosphorous: Second method: A known mass of compound is heated with fuming HNO_(3) or sodium peroxide (Na_(2)O_(2) in Carius tube which converts phosphorous to H_(3)PO_(4) . Magnesia mixture (MgCl_(2)+NH_(4)Cl) is then added, which gives the percipate of magnesium ammonium phosphate (MgNH_(4).PO_(4)) which on heating gives magnesium pyrophosphate (Mg_(2)P_(2)O_(7)) , which is weighted. Percentage of P =("Atomic mass of P")/("Molecular mass of " Mg_(2)P_(2)O_(7)) xx ("Molecular mass of "Mg_(2)P_(2)O_(7) xx 100)/("Mass of compound")=(62)/(222) xx("Mass of " Mg_(2)P_(2)O_(7)xx 100)/("Mass of compound") 0.12 gm of and organic compound containing phosphorus gave 0.22 gm of Mg_(2)P_(2)O_(7) by the usual analysis. Calculate the percentage of phosphorus in the compound.

    A
    25
    B
    9.25
    C
    801
    D
    `51.20`
  • (d) Estimation of phosphorous: A known mass of compound is heated with fuming HNO_(3) or sodium peroxide (Na_(2)O_(2)) in Carius tube which converts phosphorrous to H_(3)PO_(4) . Magnesia mixture (MgCl_(2) + NH_(4)Cl) is then added, which gives the precipitate of magnesium ammonium phosphate (MgNH_(4).PO_(4)) which on heating gives magnesium pyrophosphate (Mg_(2)P_(2)O_(7)) , which is weighed. 0.124 gm of an organic compound containing phosphorus gave 0.222 gm of Mg_(2)P_(2)O_(7) by the usual analysis . Calculate the percentage of phosphorous in the compound (Mg=24, P=31)

    A
    25
    B
    75
    C
    62
    D
    50
  • (d) Estimation of phosphorous: A known mass of compound is heated with fuming HNO_(3) or sodium peroxide (Na_(2)O_(2)) in Carius tube which converts phosphorrous to H_(3)PO_(4) . Magnesia mixture (MgCL_(2) + NH_(4)Cl) is then added, which gives the precipitate of magnesium ammonium phosphate (MgNH_(4).PO_(4)) which on heating gives magnesium pyrophosphate (Mg_(2)P_(2)O_(7)) , which is weighed. An organic compound has 6.2 % of phosphorus. On sequence of reaction, the phosphorous present in 10 gm of organic compound is converted of Mg_(2)P_(2)O_(7) . Find the weight of Mg_(2)P_(2)O_(7) formed.

    A
    2.22 gm
    B
    10.0 gm
    C
    4.44 gm
    D
    1.11 gm
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    How many mole of magnesium phosphate Mg_(3)(PO_(4))_(2) will contain 0.25 "mole" of oxygen atoms?

    How many moles of magnesium phosphate, Mg_(3)(PO_(4))_(2) , will contains 0.25 mole ofoxygen atoms?

    How many moles of magnesium phosphate, Mg_(3)(PO_(4))_(2) will contain 0.25 mole of oxygen atoms ?

    How many moles of magnesium phosphate, Mg_(3)(PO_(4))_(2) will contain 0.25 mole of oxygen atoms?

    0.5 g of an organic substance containing phosphorus was heated with conc HNO_(3) in the carius tube, the phosphoric acid thus formed was precipitated with magnesia mixture (MgNH_(4)PO_(4)) which on ignition gave a residue to 1.0 g of megnesium pyrophosphate (Mg_(2)P_(2)O_(7)) The percentage of phosphorous in the organic compound is