Potassium selenate is isomorphous with potassium sulphate and contains `50.0%` of `Se`. The atomic weight of `Se` is a. `142`, b. `71`, c. `47.33`, d. `284`A. Oxidation state of Se in the given compound is `+6`B. Atomic weight of Se is `118.6`C. Equivalent weight of potassium selenate is `130.3`D. All the above

Potassium selenate is isomorphous with potassium sulphate and contains

1.	Potassium selenate is isomorphous with potassium sulphate and contains `50.0%` of `Se`. The atomic weight of `Se` is a. `142`, b. `71`, c. `47.33`, d. `284`A. Oxidation state of Se in the given compound is `+6`B. Atomic weight of Se is `118.6`C. Equivalent weight of potassium selenate is `130.3`D. All the above
Answer» Correct Answer - D Potassium selenate is isomorphous to `K_(2)SO_(4)` and thus its molecular formula is `K_(2)SeO_(4)`. Now molecular weight of `K_(2)SeO_(4)` `= (39 xx 2 +a +4 xx 16) = (142 +a)` where a is atomic weight of Se `(142 +a)g K_(2)SeO_(4)` has `Se = ag` `100g K_(2)SeO_(4)` has `Se = (a xx 100)/(142 +a)` `:. %` of `Se = 45.52` `= 45.52 a = 118.6` Also equivalents of `K_(2)SeO_(4) =` `("Mol.wt")/(2) = (2 xx 39 +118.6 +64)/(2) = 130.1`

Discussion

No Comment Found

Related InterviewSolutions

Silver metal in ore is dissolved by potassium cyanide solution in the presence of air by the reaction `4 Ag + 8KCN +O_(2)+2H_(2)O rarr 4K[Ag(CN)_(2)]+4KOH`A. The amount of KCN required to dissolve 100g of pure Ag is 120gB. The amount of oxygen used in this process is `0.742g`C. The mount of oxygen used in this process is `7.40g`D. The volume of oxygen used at STP is `5.20` lit.
1 g of a sample of NaOH was dissolved in 50 " mL of " 0.33 M alkaline solution of `KMnO_(4)` and refluxed till all the cyanide was converted into `OCN^(ɵ)`. The reaction mixture was cooled and its 5 mL portion was acidified by adding `H_(2)SO_(4)` in excess and then titrated to end point against 19.0 " mL of " 0.1 M `FeSO_(4)` solution. The percentage purity of NaCN sample isA. `55.95%`B. `65.95%`C. `75.95%`D. `85.95%`
A sample containing `0.496 gm` of `(NH_(4))_(2) C_(2)O_(4) (MW = 124)` and inert material was dissolved in water and made strongly alkaline with KOH which converts `NH_(4)^(+)` into `NH_(3)`. The liberated `NH_(3)` was distilled into exactly 50ml of `0.05M H_(2)SO_(4)`. The excess `H_(2)SO_(4)` was back titrated with 10ml of `0.1MNaOH`. The percentage of `(NH_(4))_(2) C_(2)O_(4)` with sample isA. `40%`B. `50%`C. `60%`D. `75%`
1.575 g of oxalic acid `(CO OH)_(2).xH_(2)O` are dissolved in water and the volume made up to 250 mL. On titration 16.68 mL of this soltuion requires 25 mL of `(N)/(15)NaOH` solution for complete neutralisation. Calculate x.
`A^(n+1)` is maximum oxidised by acidified `KMnO_(4)` solution into `AO_(3)^(-)`. If `2.68` m moles of `A^(+(n+1))` requires `32.16 mL` of a `0.05M` acidified `KMnO_(4)` solution for complete oxidation,value of n is
X litres of carbon monoxide is present at S.T.P. It is completely oxidised to `CO_(2)`. The volume of `CO_(2)` formed is 11.207 L at S.T.P. What is the value of X in litres?A. `22.414`B. `11.207`C. `5.6035`D. `44.828`
Certain " mol of "HCN is oxidised completely by 25 " mL of " `KMnO_(4)`. The products are `CO_(2)` and `NO_(3)^(ɵ)` ion. When all `CO_(2)` is passed through lime water , 1 g of `CaCO_(3)` is obtained the molarity of the `KMnO_(4)` used isA. `1.44M`B. `0.72M`C. `0.36M`D. `0.8M`
Direct titration of `I_(2)` with a reducing agent is called iodimetry. If `I_(2)` is leberated by the oxidation of `I_(ɵ)` ion by a strong oxidising agent in neutral or acidic medium, the liberated `I_(2)` is then titrated with reducing agent. Iodometry is used to estimate the strngth of the oxidising agent. For example, in the estimation of `Cu^(2+)` with `S_(2)O_(3)^(2-)` `Cu^(2+)+I^(ɵ)toCuI_(2)+I_(2)` (iodometry) `I_(2)+S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)+I^(ɵ)` (iodimetry) Strach is used as an indicator at the end point, which forms bluecoloured complex with `I_(3)^(ɵ)` Disappearance of blue colourindicates the end point whe free `I_(2)` in not present. Q. In the reaction `2CuSO_(4)+4KItoCu_(2)I_(2)+2K_(2)SO_(4)+I_(2)` The equivalent weight of `CuSO_(4)` is `(Mw=159.5g mol^(-1))`A. `10%`B. `20%`C. `5%`D. `30%`
`KMnO_(4)` is a strong oxidising agent in acidic medium. To provide acidic medium `H_(2)SO_(4)` is used instead of HCl. This is becauseA. `H_(2)SO_(4)` is a stronger acid than HClB. `HCl` is oxidised by `KMnO_(4)` to `Cl_(2)`C. `H_(2)SO_(4)` is a dibasic acidD. rate is faster in the presence of `H_(2)SO_(4)`
The vapour density of a tribasic acid is x. the equivalent mass of that acid isA. `(x)/(3)`B. `x-3`C. `(2x)/(3)`D. `2x -3`

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment