InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 851. |
56V,500ml`H_2O_2` solution is kept in an open contaier due to which some `H_2O_2` is decomposed and evolves 8gm `O_2`. Simultaneously some `H_2O` also vapouries. Due to all these changes ,final volume is reduced by `20%`. Find final volume strength of `H_2O_2(aq)`. Assume 1 mole of an ideal gas occupies 22.4 L at STP)A. 56 VB. 44.8 VC. 11.2 VD. 33.6 V |
|
Answer» Correct Answer - A |
|
| 852. |
Potassium hydroxide solutions are used to absorb `CO_2`.How many litres of `CO_2` at 1.00 atm and `22^@C` would be absorbed by an aqueous solution containing 15.0 g of KOH ? (Take R=`1/12` L atm/K/mole) `KOH+CO_2 to K_2CO_3+H_2O`A. 3.29 LB. 1.65 LC. 6.58 LD. 0.329 L |
|
Answer» Correct Answer - A |
|
| 853. |
A mixture contains `40%` Cr and 60% Zn by mass. Find the volume of `(H_(2)(ml))` at 760mm Hg at `31^(@)C` which will be produced from 1gm of mixture and sufficient amount of HCl solution (Cr is in `+3 oxidation state in salt). [Cr=52, Zn=65]A. 500mlB. 416mlC. 552mlD. 620ml |
|
Answer» Correct Answer - A |
|
| 854. |
20 ml of a mixture of `CO_2 and C_2H_4` was mixed with excess of `O_2` gas and was exploded.On bringing the solution back to the original room temperature a contraction of 12 ml was observed. What is the volume percentage of `CO_2` in the original mixture ?A. `6%`B. `14%`C. `70%`D. `30%` |
|
Answer» Correct Answer - C |
|
| 855. |
100 gm mixture of Co and `CO_(2)` is mixed with 30 mL of `O_(2)` and sparked in eudiometer tube. The residual gas after treatment with aq. KOH has a volume of 10 mL which remains unchanged when treated with alkline pyrogallol . If all the volume are under the same conditions, point out the correct option(s)A. The volume of CO that reacts is 60 mLB. The volume of CO that remains unreacted is 10 mLC. The volume of `O_(2)` that remains unreacted is 10 mLD. The volume of `CO_(2)` that gets absorbed y aqueous KOH is 90 mL. |
|
Answer» Correct Answer - A::B::D |
|
| 856. |
How many moles of `P_(4)O_(6)` and `P_(4)O_(10)` will be produced by the combustion of 12.4gm of phosphorus in 12.8gm of `O_(2)`, leaving no `P_(4)` or `O_(2)`? [Atomic wt. P=31]A. 0.11 mol and 0.3 molB. 0.15mol and 0.25molC. 0.05 mol eachD. 0.1 mol each |
|
Answer» Correct Answer - C |
|
| 857. |
Human lungs can absorb 8gm `O_(2)` per hour by respiration. If all oxyggen atoms are converted to carbohydrates `(C_(6)H_(12)O_(6)` how long will it take to produce 180 gm `C_(6)H_(12)O_(6)` ?A. 8 hoursB. 12 hoursC. 10 hoursD. 6 hours |
|
Answer» Correct Answer - B |
|
| 858. |
X mL of a `60%` w/w alcohol by weight `(d = 0.6gm//mL)` must be used to prepare 200 mL of `12%` alcohol by weight `(d=0.6gm//mL)`. Then the value of X will be :A. 20B. 40C. 60D. 80 |
|
Answer» Correct Answer - C |
|
| 859. |
An unknown compound A `(Mn_(x)O_(y))` composed of manganese and oxygen , has 36.7% oxygen by weight. When 8.7 g of A is heated with HCl it liberates `Cl_(2)` gas as per the following reaction: `Mn_(x)O_(y) +HCl rarr MnCl_(2)+Cl_(2)+H_(2)O` (unbalanced)The volume of `Cl_(2)` gas at STP obtained when 8.7 gm of compound A is heated with excess of HCl , (assume molecular formula and , empirical formula to be same):A. 1.135 LB. 2.27 LC. 4.54 LD. 0.567 L |
|
Answer» Correct Answer - B |
|
| 860. |
In 20 ml of a gaseous mixture containing `N_2 and H_2` gases , 5 ml `O_2` gas is added and the mixture is exploded. If the final volume becomes 13ml, then the only incorrect statement is : (All the volumes are measured at the same pressure and temperature )A. The initial mixture contains 10 ml `N_2` gasB. The initial mixture contains 8 ml `H_2` gasC. The final mixture contains 1 ml `O_2` gasD. The final mixture contains 12 ml `N_2` gas. |
|
Answer» Correct Answer - A |
|
| 861. |
20 ml of pure acetic acid (density = 0.75 gm `ml^(-1)`) is mixed with 50 gm of water (density = 1 gm`ml^(-1)`) at a certain temperature. Calculate the molality of acetic acid in the final solution. |
|
Answer» Correct Answer - 5 |
|
| 862. |
Calculate molality of one litre of solution containing 200 gm `CaBr_(2)`. Given density of solution equal to `1.0` gm/ml. (Atomic mass of Ca = 40, Br = 80)A. 1 mB. `1.25` mC. `0.8` mD. `1.4` m |
|
Answer» Correct Answer - B |
|
| 863. |
A mineral consists of an equimolar mixture of the carbonates of two bivalent metals. One metal is present to the extent of 12.5% by weight. 2.8gm of the mineral on heating lost 1.32gm of `CO_(2)`. What is the % by weight of the other metal?A. 87.5B. 23.21C. 65.11D. 40.35 |
|
Answer» Correct Answer - B |
|
| 864. |
A 5 L vessel contains 2.8 g of `N_(2)` only, when heated to 1800 K 30% molecules are dissociated into atoms.A. Total no, of moles N in the container will be 0.12B. Total no. of molecules in the container will be close to `0.421 xx 10^(23)`C. Total no. of moles in the container will be 0.098D. Pressure in the container decreased |
|
Answer» Correct Answer - A::B |
|
| 865. |
If a pure compound is composed of `X_(2)Y_(3)` molecules and consists of `60%` X by weight what is the atomic weight of Y in term of atomic weight of X (Atomic mass of `X=M_(x)`)?A. `(9)/(4)M_(x)`B. `(4)/(9)M_(x)`C. `(2)/(3)M_(x)`D. `(3)/(2)M_(x)` |
|
Answer» Correct Answer - B |
|
| 866. |
An ideal gaseous mixture of ethane `(C_(2)H_(6))` and ethene `(C_(2)H_(4))` occupies `28` litre at `1atm` `0^(@)C`. The mixture reacts completely with `128 gm O_(2)` to produce `CO_(2)` and `H_(2)O`. Mole of fraction at `C_(2)H_(6)` in the mixtture is-A. 0.6B. 0.4C. 0.5D. 0.8 |
|
Answer» Correct Answer - B |
|
| 867. |
12 mol gaseous mixture of an alkane and an alkene (containing same number of carbon atoms) require exactly 285 ml of air (containing 20% v/v `O_(2)` and rest `N_(2)` ) for complete combustion at 200K . After combustion when gaseous mixture is passes through KOH solution it shows volume contraction of 36 ml. Mole fraction of `CO_(2)` in final gaseous sample:A. `(6)/(51)`B. `(6)/(44)`C. `(6)/(45)`D. `(6)/(13)` |
|
Answer» Correct Answer - B |
|
| 868. |
12 mol gaseous mixture of an alkane and an alkene (containing same number of carbon atoms) require exactly 285 ml of air (containing 20% v/v `O_(2)` and rest `N_(2)` ) for complete combustion at 200K . After combustion when gaseous mixture is passes through KOH solution it shows volume contraction of 36 ml. Calculate mol % of oxygen which is converted into `H_(2)O`A. 0.3684B. 0.7368C. 0.2061D. 0.2563 |
|
Answer» Correct Answer - A |
|
| 869. |
12 mol gaseous mixture of an alkane and an alkene (containing same number of carbon atoms) require exactly 285 ml of air (containing 20% v/v `O_(2)` and rest `N_(2)` ) for complete combustion at 200K . After combustion when gaseous mixture is passes through KOH solution it shows volume contraction of 36 ml. Formula of alkane is:A. `C_(5)H_(12)`B. `C_(3)H_(8)`C. `C_(2)H_(6)`D. `C_(4)H_(10)` |
|
Answer» Correct Answer - B |
|
| 870. |
Equal volume of liquid A (d=0.8 gm/ml) and liquid B (d=1.2 gm/ml) are mixed to form a solution. Calculate mole fraction of A in solution. [`M_(A)=16,M_(B)=32`]A. `(3)/(8)`B. `(2)/(3)`C. `(4)/(7)`D. `(3)/(4)` |
|
Answer» Correct Answer - C |
|
| 871. |
A solution of `KCl` has a density of `1.69 g mL^(-1)` and is 67% by weight. Find the denisty of the solution if it is diluted so that the percentage by weight of `KCl` in the diluted solution is 30%` |
|
Answer» Let the volume of the `KCl` solution be `100 mL`. Weight of `KCl` solution `= 100 xx 1.69 = 169 g` `100 g` of solution contains `= 67 g` of `KCl` `169 g` of solution `= (67)/(100) xx 169 = 113.23 g`. Let `x mL` of `H_(2) O` be added. New volume fo soltuion `= (100 + x) mL` New weight of solution `= (169 + x) g` (since `x mL of H_(2)O = x of H_(2) O, d_(H_(2)O) = 1)` New percentage of the solution = 30% % be weight ` = ("Weight of solute" xx 100)/("Weight of solution")` `30 = (113.23)/((169 + x)) xx 100` `X = 208.43 mL = 208.43 g` New density `= ("New weight fo solution")/("New volume of Solution")` `= ((169 + x))/((100+ x))` `((169 + 208.43))/((100 + 208.43)) = (377.43)/(308.43)` `:. d = 1.224` |
|
| 872. |
100 ml aqueous solution containing equimolar mixtrue of `Ca(OH)_(2)` and `Al(OH)_(3)` requires `0.5` litre of 4M HCl for complete neutralisation. Molarity of `Ca(OH)_(2)`, in the original solution is :A. 2 MB. 4 MC. 8 MD. 10 M |
|
Answer» Correct Answer - B |
|
| 873. |
In the above diagram, the paired open spheres represent `H_(2)` molecules and the paired solid spheres represent `N_(2)` molecules. When the molecules in the box react to form the maximum possible amount of ammonia `(NH_(3)` molecules, what is the limiting reactant and how many molecules of `NH_(3)` can be formed?A. `N_(2)` is limiting. 5 molecules of `NH_(3)` can be formed.B. `N_(2)` is limiting. 10 molecules of `NH_(3)` can be formed.C. `H_(2)` is limiting. 8 molecules of `NH_(3)` can be formed.D. `H_(2)` is limiting, 12 molecules of `NH_(3)` can be formed. |
|
Answer» Correct Answer - C |
|
| 874. |
In which of the following pairs do 1 g of each have an equal number of molecules .A. `N_2O and CO`B. `N_2 and C_3O_2`C. `N_2 and CO`D. `N_2O and CO_2` |
|
Answer» Correct Answer - C::D |
|
| 875. |
Assertion: A molecule of butane, `C_(4)H_(10)` has a mass of `58.12 amu`. Reason: One mole of butane contains `6.022xx10^(23)` molecules and has a mass of `58.12 g`.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. |
|
Answer» Correct Answer - A |
|
| 876. |
A gaseous mixture contains `SO_(3)(g)` and `C_(2)H_(6)(g)` in a 16:15 ratio by mass. The ratio of total number of atoms present in `C_(2)H_(6)(g)` and `SO_(3)(g)` is:A. `2:5`B. `1:5`C. `5:1`D. `5:2` |
|
Answer» Correct Answer - C |
|
| 877. |
Ethanol is the substance commonly called alcohol. The denisty of liquid alcohol is `0.8 gm//ml` at 293 K. if 1.2 mole of ethanol are needed for a particular experiment, what volume of ethanol should be measured out?A. 55.2 mlB. 57.5 mlC. 69mlD. 47.9ml |
|
Answer» Correct Answer - C |
|
| 878. |
The density of air is `0.001293g//ml` at S.T.P. Its vapour density will be:A. 10B. 15C. 1.468D. 14.68 |
|
Answer» Correct Answer - D |
|
| 879. |
100 ml aqueous solution `("density"(5)/(3)gm//ml)` contains `40%` by weight NaOH. The number of molecules of NaOH dissolved in the above solution is : `("Use" N_(A)=6xx10^(23))`A. `2xx10^(22)`B. `3.33xx10^(22)`C. `10^(24)`D. `3.33xx10^(23)` |
|
Answer» Correct Answer - C |
|
| 880. |
Which of the following has the smallest number of molecules?A. 22.7 mL of `CO_(2)` gas at STPB. 22 g opf `CO_(2)` gas at STPC. 11.35L of `CO_(2)` gas at STPD. 0.1 mole of `CO_(2)` gas |
|
Answer» Correct Answer - A |
|
| 881. |
Calculate the density (in gm//ml) of aqueous `NaOH` solution of which molarity and (% w//w) are equal. |
|
Answer» Correct Answer - 4 |
|
| 882. |
What is the molarity of a hydorchlric acid solution if `20.00` mL of it neutralizes `18.46` mL of a 0.0420M `Ba(OH)_(2)` solution?A. 0.0194MB. 0.0388MC. 0.0455MD. 0.0775M |
|
Answer» Correct Answer - D |
|
| 883. |
What approximate volume of 0.40 M `Ba(OH)_(2)` must be added to 50.0 mL of 0.30 M NaOH to get a solution in which the molarity of the `OH^(-)` ions is 0.50 M?A. 33 mLB. 66 mLC. 133 mLD. 100 mL |
|
Answer» Correct Answer - A |
|
| 884. |
`35 % w//v 400 ml` of `NH_(4)OH` is mixed with `12 M, 600 ml` of `H_(2)SO_(4)`. Find the molarity `[NH_(4)^(+)]` in solution. |
|
Answer» Correct Answer - 4 |
|
| 885. |
What volume of 0.2M `Ba(OH)_(2)` must be added to 300 mL of 0.08 M HCl solution to get a solution in which the molarity of hydroxyl `(OH^(-))` ions is 0.8 M?A. 375 mLB. 300 mLC. 225 mLD. 150 mL |
|
Answer» Correct Answer - D |
|
| 886. |
300 gm, `30%`(w/w) NaOH solution is mixed with 500 gm `40%`(w/w) NaOH solution. What is `%` (w/v) NaOH if density of final solution is 2 gm/mL?A. `72.5`B. 65C. `62.5`D. None of these |
|
Answer» Correct Answer - A |
|
| 887. |
A solution is prepared by mixing `25.0` mL of `6.0` M HCI with `45.0` mL of `3.0` M `HNO_(3)`. What is `[H^(+)]` in the resulting solution?A. 1.9MB. 2.1MC. 4.1MD. 4.5M |
|
Answer» Correct Answer - C |
|
| 888. |
400 ml 0.1 M `BaCL_(2)` is mixed with 600 ml 0.1 M `H_(2)SO_(4)` to form products according to following reaction: `BaCl_(2) + H_(2)SO_(4) rarr BaSO_(4) + 2HCl` Select the correct option(s) after reaction is completed.A. Molarity of `Ba^(+2)` ions in final solution is 0.04 MB. Molarity of `SO_(4)^(2-)` ions in solution is 0.02 MC. Molarity of `H^(+)` ions do not change on mixingD. Final molarity of `H^(+)` ions in solution is 0.12 M |
|
Answer» Correct Answer - B::D |
|
| 889. |
500 ml of 2M `AlCl_(3)` solution is mixed with 200 ml of `58.5 % w//v` NaCl solution and 300 ml of 50% `w//w` `BaCl_(2)` solution `(d=2.08 g//ml)`. Calculate molarity of `Cl^(-)` in the final solution. |
|
Answer» Correct Answer - 8 |
|
| 890. |
What is the final `[Na^(+)]` in a solution prepared by mixing `70.0`mL of `3.00` M `Na_(2)SO_(4)` with `30.0` mL of `1.00` M NaCl?A. `2.00` MB. `2.40` MC. `4.00` MD. `4.50` M |
|
Answer» Correct Answer - D |
|
| 891. |
`100mL` of `H_(2)SO_(4)` solution having molarity `1M` and density `1.5g//mL` is mixed with `400mL` of water. Calculate final m plarity of `H_(2)SO_(4)` solution, if final density is `1.25g//mL`?A. `4.4M`B. `0.145M`C. `0.52M`D. `0.227M` |
|
Answer» Correct Answer - 4 Total moles of `H_(2)SO_(4)=0.1` mole Total volume `=(150+400)/(1.25)=-(550)/(1.25)=440` `:." "M=(0.1)/(440)xx1000=(1)/(4.4)=0.227M` |
|
| 892. |
50 mL solution of `BaCl_(2)` (20.8% w//v) and 100 mL solution of `H_(2)SO_(4)` (9.8% w//v) are mixed (Ba = 137, Cl = 35.5, S=32) `BaCl_(2) + H_(2)SO_(4) rightarrow BaSO_(4) darr 2HCl`Weight of `BaSO_(4)` formed is:A. 23.3gB. 46.6gC. 29.8gD. 11.65g |
|
Answer» Correct Answer - D |
|
| 893. |
If 200 ml of `0.1` M `Na_(2)SO_(4)` is mixed with 100 ml of `0.2` M `NA_(3)PO_(4)` solution, molarity of `Na^(+)` in the final solution, if final solution has density `1.2gm//ml`, will be :A. `0.196` MB. `0.33` MC. `0.5` MD. none of these |
|
Answer» Correct Answer - B |
|
| 894. |
300 ml, 2 M `H_(2)SO_(4)` solution is mixed with 200 ml, 2 M Ba`(OH)_2)` solution then find final molarity of sulphate ion in diluted to 4 times. (`BaSO_(4)` formed is precipitated)A. `0.2` MB. `0.8` MC. `0.3` MD. `0.1` M |
|
Answer» Correct Answer - D |
|
| 895. |
100 ml of `0.2"M H"_(2)SO_(4)` solution is mixed with 400ml of `0.05"M Ba"_(3)(PO_(4))_(2)`. Calculate the concentration of `[Ba^(+2)]` ion in resulting solution.A. `0.08` MB. `0.04` MC. `0.4` MD. `0.8` M |
|
Answer» Correct Answer - A |
|
| 896. |
A solution is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9. What is the molality `(m)` of the solution. b. Water is added to the above solution such that the mole fraction of water in the solution becomes 0.9. What is the molality `(m)` of the solution? |
|
Answer» a. Molar mass of `H_(2) O = 18 mol^(-1)` Molar mass of ethanol `(C_(2) H_(5) OH)` `= 2 xx 12 + 5 + 16 + 1` `= 46 g mol^(-1)` Mole fraction of ethanol is 0.9 and is greater than the mole fraction of water (i.e., 1 - 0.9 = 0.1), so ethanol is solvent and water is solute. Thus `X_(1) = 0.9 ` and `X_(2) = 0.1 (Mw_(1) = 46` and `Mw_(2) = 18)`. `:. m = (X_(2) xx 1000)/(X_(1) xx Mw_(1)) = (("Mole fraction of solute" xx 1000)/("Mole fraction of solvent" xx "Molar mass of solvent"))` `(0.1 xx 1000)/(0.9 xx 46) = 2.415 m` b. Now mole fraction of water is `0.9` and is greater than the mole fraction of ethanol (i.e, 1 - 0.9 = 0.1), so water is solvent and ethanol is solute. Thus, `X_(1) = 0.9, X_(2) = 0.1 (Mw_(1) = 18` and `Mw_(2) = 46)`. `:. m = (X_(2) xx 100)/(X_(1) xx Mw_(1)) = (0.1 xx 1000)/(0.9 xx 18) = 6.17 m` |
|
| 897. |
An organic base is tetraacidic. If from every 10gm of the chloroplatinate salt of the base 3.9 gm of the residue of platinum is obtained, then what will be the molecualr mass of the base? [Pt = 195]A. 180B. 360C. 90D. 270 |
|
Answer» Correct Answer - A |
|
| 898. |
Calculate the percentage composition of: Alumina `(Al_(2) O_(3)`, potassium oxide `(K_(2) O)`, and silcia `(SiO_(2))` in the sample of clay `(Al_(2) O_(3). K_(2) O. 6SiO_(2))`. b. Potassium sulphate `(K_(2) SO_(4))`. Aluminium sulphate, and water of crystallisation in the simple of potash alum, `(K_(2) SO_(4). Al_(2) (SO_(4))_(3). 24H_(2) O)`. |
|
Answer» Molar mass of `Al_(2) O_(3). K_(2) O. 6 SiO_(2)` `= (2 xx 27 + 3 xx 16) + (2 xx 39 + 16) + 6 (28 + 2 xx 16)` `= 102 + 94 + 360 = 556 g` `Mass % of Al_(2)O_(3) = (102)/(556) xx 100 = 18.35%` Mass % of `K_(2) O = (94)/(556) xx 100 = 16.90 %` Mass % of `SiO_(2) = (360)/(556) xx 100 = 64.75%` b. Molar mass of `K_(2) SO_(4) Al_(2) (SO_(4))_(3) 23 H_(2) O` `= (2 xx 39 + 32 + 64) + [2 xx 27 + 3 (32 + 64) + 24 xx 18]` `= 174 + 342 + 432 = 948 g` Mass % of `K_(2) SO_(4) = (174)/(948) xx 100 = 18.35%` Mass % of `Al_(2) (SO_(4))_(3) = (342)/(948) xx 100 = 36.07 %` Mass % of `H_(2) O = (432)/(948) xx 100 = 45.56 %` c. Molar mass of `4 (SO_(4)^(2-)) = 4(32 + 64) = 384 g` Mass % of `SO_(4)^(2-) = (384)/(948) xx 100 40.50%` |
|
| 899. |
1 gram of pyrolusite `(MnO_(2))` was boiled with excess of concentrated `HCl` and the issuing gas was passed through a solution of potassium iodide when 1.27 g of iodide were liberated. What is the percentage of pure `MnO_(2)` in the sample ? |
|
Answer» Calculation of weight of pure `MnO_(2)` The following partial equations are involved in the problem `{:(" "MnO_(2)+4HClrarrMnCl_(2)+Cl_(2)+2H_(2)O),(" "2KI+Cl_(2)rarr2KCl+I_(2)),(bar("Add :"" "underset(55+32(=87g))(MnO_(2))+4HCl+2KIrarrMnCl_(2)+2KCl+2H_(2)underset(2xx127(=254g))(O+I_(2)))):}` 254g of `I_(2)` are obtained from `MnO_(2)=87` g 1.27 g of `I_(2)` are obtained from `MnO_(2)=(87)/(254)xx1.27 g = 0.435 g` Step II. Calculation of percentage purity of the sample Weight of pure `MnO_(2)=0.435 g` Weight of the sample `= 1.0 g` Percentage purity of the sample `= ("Weight of pure MnO"_(2))/("Weight of sample")=((0.435g))/((1.0g))xx100=43.5%`. |
|
| 900. |
A mixture of `FeO` and `Fe_(3)O_(4)` when heated in air to a constant weight, gains 5% of its weight. Find the composition of the intial mixutre. |
|
Answer» Let the total mass of mixture = 100 g Let the mass of FeO in the mixture = x g `:.` Mass of `Fe_(3)O_(4)` in the mixture = (100-x)g Both the oxides on heating in air are converted to `Fe_(2)O_(3)` Step I. Calculation of mass of `Fe_(2)O_(3)` from `FeO` `{:(2FeO+,(1)/(2)O_(2)rarr,Fe_(2)O_(3)),(2(56+16),,2xx56+3xx16),(=144g,,=160g):}` 144g of FeO upon oxidation form `Fe_(2)O_(3)=160g` `:.` xg of FeO upon oxidation form `Fe_(2)O_(3)=(160)/(144)xx xg` Step II. Calculation of mass of `Fe_(2)O_(3)` from `Fe_(3)O_(4)` `{:(2Fe_(3)O_(4)+(1)/(2)O_(2),overset("Heat")(rarr),3Fe_(2)O_(3)),(2(3xx56+4xx16),,3(2xx56+3xx16)),(=464g,,=480g):}` 464g of `Fe_(3)O_(4)` upon oxidation form `Fe_(2)O_(3) = 48` g (100-x)g of `Fe_(3)O_(4)` upon oxidation form `Fe_(2)O_(3) = (480)/(464)(100-x)g` Step III. Calculation of percentage composition of the mixture Total mass of `Fe_(2)O_(3)` actually formed `= 100 +5 = 105` g `:. (160x)/(144)+(480)/(464)(100-x)=105,1.11x+1.034(100-x)=105` `1.11x+103.4-1.034x =105` `1.11x-1.034x=105-103.4 or0.076=1.6orx=(1.6)/(0.076)=21g` % of FeO in the sample `= ((21 g))/((100 g))xx100 = 21 %` % of `Fe_(3)O_(4)` in the sample `= 100 - 21 = 79 %`. |
|