`A" " 0.2g` sample , which is mixture of `NaCl, NaBr` and `NaI` was dissolved in water and excess of `AgNO_(3)` was added. The precipitate containing `AgCl,AgBr` and `AgI` was filtered, dried and weighed to be `0.412g`. The solid was placed in water and treated with excess of `NaBr`, which converted all `AgCl` into `AgBr`. The precipitate was then weighed to be `0.4881 g`. It was then placed into water and treated with excess of `NaI`, which converted all `AgBr` into `AgI`. The precipitate was then weighed to be `0.5868 g`. What was the percentage of `NaCl, NaBr` and `NaI` in the original mixture :

`A" " 0.2g` sample , which is mixture of `NaCl, NaBr` and `NaI` was di

1.	`A" " 0.2g` sample , which is mixture of `NaCl, NaBr` and `NaI` was dissolved in water and excess of `AgNO_(3)` was added. The precipitate containing `AgCl,AgBr` and `AgI` was filtered, dried and weighed to be `0.412g`. The solid was placed in water and treated with excess of `NaBr`, which converted all `AgCl` into `AgBr`. The precipitate was then weighed to be `0.4881 g`. It was then placed into water and treated with excess of `NaI`, which converted all `AgBr` into `AgI`. The precipitate was then weighed to be `0.5868 g`. What was the percentage of `NaCl, NaBr` and `NaI` in the original mixture :
Answer» Correct Answer - `50,20.23,29.77` Working in backward direction In the last step moles of `(AgBr+Agl)`= moles of `AgI` `implies (0.4881-x)/(188)+(x)/(235)=(0.5868)/(235)impliesx=0.0933g` Mass `%` of `NaI` `=(0.0933)/(235)xx150xx(100)/(0.2)=29.77` Now subtracting mass of `AgI` from `1st` and `2nd` precipitate gives Mass of `(AgCl+AgBr)=0.3187 g` and mass of `AgBr=0.3948 g` Again `(y)/(143.5)+(0.3187-y)/(188)=(0.3948)/(188)impliesy=0.245g` implies Mass `%` of `NaCl` `=(0.245)/(143.5)xx58.5xx(100)/(0.2) = 50` Mass `%` of `NaBr = 20.23`

Discussion

No Comment Found

Related InterviewSolutions

A compound `H_(2)X` with molar mass of 80 g is dissolved in a solvent having density of `0.4 g mL^(-1)`. Assuming no change in volume upon dissolution, what is the molality of 3.2 M solution of the compound ?
Fill in the blanks. a. The equivalent weight of `NaHCO_(3)` is ……..and of `SO_(2)` is ……… b. 2 mol of 50% pure `Ca(HCO_(3))_(2)` on heating forms 1 mol of `CO_(2)`. The % yield of `CO_(2)` is …….. c. `5 g` of `K_(2) SO_(4)` was dissolved in `250 mL` of solution. The volume of this solution that should be used so that `1.2 g` of `BaSO_(4)` be precipitated fromk `BaCl_(2)` is ....... (molecular mass of `K_(2) SO_(4) = 174` and `BaSO_(4) = 233`) d. The residue obtained on strongly heating `2.76 g Ag_(2) CO_(3)` is ........ `[Ag_(2)CO_(3) overset(Delta)rarr Ag + CO_(2) + O_(2)]` Atomic mass of `Ag = 108`
`5.6 g` of a metal forms `12.7 g` of metal chloride. Hence equivalent weight of the metal isA. 127B. 254C. 56D. 25
Assertion: The average mass of one Mg atom is `24.305 amu`, which is not actual mass of one Mg atom. Reason: Three isotopes, `24 Mg, 25 Mg and 26 Mg`, of Mg are found in nature.A. Statement-1 : is True, Statement-2 : is True, Statement-2 : is a correct explanation for Statement-1 :B. Statement-1 : is True, Statement-2 : is True, Statement-2 : is NOT a correct explanation for Statement-1 .C. Statement-1 : is Ture, Statement-2 : is False.D. Statement-1 : is False, Statement-2 : is True.
` 1 g` of impure `Na_(2)CO_(3)` is dissolved in water and the solution is made upto `250 mL`. To `50 mL` of this solution, `50 mL` of `0.1 N HCl` is added and the mixture after shaking well required `10 mL` of `0.16 N NaOH` solution for complete neutralization. Calculation percent purity of the sample of `Na_(2) CO_(3)`.
`0.7 g` of iron reacts directly with `0.4 g` of sulphur to form ferrous sulphide. If `2.8 g` of iron is dissolved in dilute `HCl` and excess of sodium sulphide solution is added, `4.4 g` of iron sulphide is precipitated. Prove the law of constant composition.
`4.0 g` of a mixture of `Nacl` and `Na_(2) CO_(3)` was dissolved in water and volume made up to `250 mL`. `25 mL` of this solution required `50 mL` of `N//10 HCl` for complete neutralisation. Calculate the percentage composition of the original mixture.
Molarity and Molality of a solution of a liquid (molecular weight = 50) in aqueous solution is 9 and 18 respectively. What is the density of solution?A. 1 g/ccB. `0.95` g/ccC. `1.05` g/ccD. `0.66` g/cc
Objective question (single correct answer). i. `H_(3) PO_(4)` is a tribasic acid and one of its salt is `NaH_(2) PO_(4)`. What volume of `1M NaOH` solution should be added to `12 g` of `NaH_(2) PO_(4)` to convert in into `Na_(3) PO_(4)` ? a. `100 mL` b. 2 mol of `Ca (OH)_(2)` c. Both d. None iii. The normality of a mixture obtained mixing `100 mL` of `0.2 m H_(2) SO_(4)` with `100 mL` of `0.2 M NaOH` is: a. `0.05 N` b. `0.1 N` c. `0.15 N` d. `0.2 N` iv `100 mL` solution of `0.1 N HCl` was titrated with `0.2 N` `NaOH` solutions. The titration was discontinued after adding `30 mL` of `NaOH` solution. The reamining titration was completed by adding `0.25 N KOH` solution. The volume of `KOH` required from completing the titration is: a. `70 mL` b. `35 mL` c. `32 mL` d. `16 mL`
`H_(2) SO_(2)` solution `(20 mL)` reacts quantitatively with a solution of `KMnO_(4) (20 mL)` acidified is just dilute `H_(2) SO_(4)`. The same volume of the `KMnO_(4)` solution is just decolourised by `10 mL` of `MnSO_(4)` in neutral medium. simulataneously forming a dark brown precipitate of hydrated `MnO_(2)`. The brown precipitate is dissolved in `10 mL` of `0.2 M` sodium oxalate under boiling condition in the presence of dilute `H_(2) SO_(4)`. Write the balanced equations involved in the reactions and calculate the molarity of `H_(2) O_(2)` solution.

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment