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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.
| 951. |
In which mode of expression, the concentration of a solution remains independent of temperature?A. (a)MolarityB. (b)NormalityC. (c )FormalityD. (d)Molality |
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Answer» Correct Answer - D Molality of a solution remains independent of temperature. |
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| 952. |
The density `("in" g mL^(-1))` of a `3.60 M` sulphuric acid solution that is `29% H_(2)SO_(4)` (Molar mass `=98 g mol^(-1)`) by mass will be:A. 1.22B. 1.45C. 1.64D. 1.88 |
| Answer» Correct Answer - A | |
| 953. |
Magnetite, `Fe_(3)O_(4)`, can be converted into metallic iron by heating with carbon monoxide as represented by this equation: `Fe_(3)O_(4)(s)+CO(g) rarr Fe(s) +CO_(2)(g)` The kilograms of `Fe_(3)O_(4)` which must be processed in this way to obtain `5.00 kg` of iron, if the process is `85%` efficient is closest to? `[M: = Fe = 56]`A. `6.92 kg`B. `8.15 kg`C. `20.8 kg`D. `24.4 kg` |
| Answer» Correct Answer - B | |
| 954. |
Hematitic `(Fe_(2) O_(3))` is an important ore of iron. The free metal `(Fe)` is obtained by reducing hematite with carbon monoxide `(CO)` in a blast furnace: `Fe_(2)O_(3)(s) + 3CO(g) rarr 2Fe(s) + 3CO_(2) (g)` How many grams of `Fe` can be produced from `1.00 kg` `Fe_(2) O_(3)` ?A. `698 g`B. `786 g`C. `896 g`D. `968 g` |
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Answer» Correct Answer - A Mass of `Fe_(2) O_(3) rarr` Moles of `Fe_(2) O_(3) rarr` Moles of `Fe rarr` Mass of `Fe` `1.00xx10^(3)g Fe_(2) O_(3) xx (1 mol Fe_(2) O_(3))/(160g Fe_(2) O_(3)) xx (2 mol Fe)/(1 mol Fe_(2) O_(3))` `xx (55.8g Fe)/(1 mol Fe) = 698g Fe` |
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| 955. |
Oxidation states of the metal in the minerals haematite and magnetite, respectively, areA. II, III in haematite and III in magnetiteB. II, III in haematite and II in magnetiteC. II in haematite and II, III in magnetiteD. III in haematite and II, III in magnetite |
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Answer» Correct Answer - D In haematite `(Fe_(2)O_(3))` Fe is present (III) oxidation state and in magnetic `(Fe_(3)O_(4))` Fe is present in (II) and (III) oxidation state |
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| 956. |
Magnetite, `Fe_(3)O_(4)`, can be converted into metallic iron by heating with carbon monoxide as represented by this equation: `Fe_(3)O_(4)(s)+CO(g) rarr Fe(s) +CO_(2)(g)` The kilograms of `Fe_(3)O_(4)` which must be processed in this way to obtain `5.00 kg` of iron, if the process is `85%` efficient is closest to? `[M: = Fe = 56]`A. 8.12 kgB. 4.14 kgC. 6.94 kgD. 16.8 kg |
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Answer» `underset(232g)(Fe_3O_4)+underset(56xx3=168g)(4CO to 3Fe)+4CO_2` 3 moles of Fe is produced from 1 mole of `Fe_3O_4` 168g of Fe will be produced from 232 g of `Fe_3O_4` 3kg of Fe will be produced from `232/168xx3000g` =4142.8g of 4.14kg of `Fe_3O_4` |
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| 957. |
Which has maximum number of atom ?A. 24 g of C (12)B. 56 g of Fe (56)C. 27 g of Al (27)D. 108 g of Ag (108) |
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Answer» Correct Answer - A 224g of C-12 contain `=6.022xx10^(23)xx10^(23)`atoms 56g of Fe-56 contain `=6.022xx10^(23)atoms` 27g of Al-27 contain `=6.022x10^(23)`atoms 108g of Ag-108 contain `=6.022xx10^(23)` atoms Thus, 24g of C-12 contains maximum number of atoms. |
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| 958. |
A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. `H_(2)SO_(4)`. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will beA. (a)`1.4`B. (b)`3.0`C. (c )`2.8`D. (d)`4.4` |
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Answer» Correct Answer - a `underset(0.05)(HCOOH)overset(H_(2)SO_(4))(rarr)underset(0.05) (H_(2)O)+underset(0.05)(CO)` `underset(0.05)(H_(2)C_(2)O_(4))overset(H_(2)SO_(4))(rarr)underset(0.05)(H_(2)O)+underset(0.05)(CO)+underset(0.05)(CO_(2))` Total moles of `CO=0.05+0.05=0.10` KOH absorbs `CO_(2)(g)` `:.` Mass of `CO=0.1xx28=2.8 g`. |
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| 959. |
Find the Faharenhit temperature when the abosuoulte temperature is `400 K` Strategy: First convert Kelvin to degress Celsius using to Eq. (1.2). Then carry ou t the next converstion from degree Celsius to Fahrenheit. |
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Answer» `? .^(@)C = (400 K - 273 K) (1.0^(@)C)/(1.0 K) = 127^(@) C` As a shortcut, we just subtract 273 from 400 but scientifically it is required to follow the concept of units. We as indians just hate to use units as we alaways in a hury ! `? .^(@)F = (127^(@) C xx (1.8^(@) F)/(1.0^(@) C)) + 32^(@) F = 261^(@) F` |
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| 960. |
In a certain car battery, the density of s8lphuric acid is `1.41 g ML^(-1)`. Caluculate of mass of `242 mL` of the acid. Strategy: Reaarange the densiy equation (1.1). |
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Answer» Mass (m) = Density(d) `xx` Volume (V) `= (1.41 (g)/(mL)) (242 mL)` `= 341 g` |
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| 961. |
Denstity conversion : The density of liquid metal, mercucy,n is `13.6 g cm^(-3)`. What is the density in `kg m^(-3)` ? Strategy : We need two unit factors-one to convert gram to kilogram and the other to convert cubic meter to cubic centimeter. Use the relationships `1 kg = 1000g` and `1 cm = 1xx10^(-2) m`. |
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Answer» `? Kg^(-3) = ((13.6g)/(cm^(3))) xx ((1kg)/(1000 g)) xx ((1cm)/(1xx10^(-2) m))^(3)` `= 13.6xx10^(3) "kg m"^(-3)` `= 1.36xx10^(4) "kg m"^(-3)` Observation: Units kg `m^(-3)` provide inconvenly large value for densiy. |
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| 962. |
Molarlity: An acqueous solution contains `128g` of mehanol `(CH_(2) OH)` in `108g` of water. Calculate the molarity of the solution. Strategy: Convert the grams of `CH_(3) OH` to moles of `CH_(3) OH` express the mass of `H_(2) O` in kilogram and apply the definition of molality. |
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Answer» Step 1: Calculate the moles of `CH_(3) OH` Number of moles of `CH_(3) OH = ("Mass of" CH_(3) OH)/("Molar mass of" CH_(3) OH)` `= (128g)/(32.0 g mol^(-1))` `= 4.00 mol CH_(3) OH` Step 2: Convert the mass of `H_(2) O` into kilograms. Kilogram of `H_(2) O = (108g)/(1000g kg^(-1)) = 0.108 kg` Step 3: Apply the definition of molality. Molatiy `= ("Molar of solute")/("Mass of solvent(kg)")` `= (4.00 mol)/(0.108 kg) = 37.0m CH_(3) OH` Alternatively, in single setup, `(? mol CH_(3) OH)/(kg H_(2) O) = (128g CH_(3) OH)/(0.108 kg H_(2) O) xx (1 mol CH_(3) OH)/(32.0g CH_(3) OH)` `= (37.0 mol CH_(3) OH)/(kg H_(2) O)` `= 37.0m CH_(2) OH` |
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| 963. |
Calculate the density of silver if a silver coin has a mass of `16.6 g` and occupies a volume of `1.58 cm^(2)` Strategy: Divide the mass of object by its volume. |
| Answer» Density `(d) = ("Mass(m)")/("Volume (V)") = (16.6 g)/(1.58 cm^(3)) = 10.5g cm^(-3)` | |
| 964. |
0.5 gm of fuming `H_(2)SO_(4)` (Oleum) is diluted with water. This solution is completely neutralised by 26.7 ml of 0.4 M `NaOH` solution. Calculate the percentage of free `SO_(3)` in the given sample. Give your answer excluding the decimal places.A. Mass of `SO_(3)` is 0.104 gB. % of free `SO_(3) = 20 : 7`C. Normality of `H_(2) SO_(4)` for neutralization is 0.2 ND. Weight of `H_(2) SO_(4)` is 0.104 g |
| Answer» Correct Answer - A::B | |
| 965. |
Calculate the % of free `SO_(3)` in an oleum, that is labelled 118%. |
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Answer» Oleum `=H_(2)SO_(4)+SO_(3)=H_(2)S_(2)O_(7)` If initial weight of labelled `H_(2)S_(2)O_(7)=100 g` Weight of `H_(2)SO_(4)`, after dilution = 118 g Wt. of `H_(2)O=18 g` Moles of `H_(2)O` = moles of `SO_(3)=(18)/(18)=1` Wt. of `SO_(3)=80` `therefore %` of free `SO_(3)=("Wt. of "SO_(3)xx100)/(100)=80%` |
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| 966. |
In the titration of a certain `H_(2)SO_(4)` solution, `60 mL` of `5.0 M NaOH` solution was used to completely neutralise `75 mL` of the acid. The molarity of the acid solution may be expressed asA. `(5M xx 60 mL)/(2 xx 75 mL)`B. `(5M xx 75 mL xx 2)/(60 mL)`C. `(75 mL xx 2)/(5.0 xx 60 mL)`D. `(60 mL xx 75 mL xx 2)/(5.0 M)` |
| Answer» Correct Answer - A | |
| 967. |
How much of NaOH is reuired to neutralise 1500 `cm^(3)` of 0.1 N HCl (Na=23)?A. (a)40 gB. (b)4 gC. (c )6 gD. (d)60 g |
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Answer» Correct Answer - C `N=(W(g)xx1000)/(VxxEq. Wt.)` `1500 ml of 0.1 N HCl=150 ml(N)` `1=(W(g)xx1000)/(150xx40), W(g)=(150xx40)/1000=6 g` |
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| 968. |
Calculate the normality of HCl solution whose 500 ml is utillised to neutralise the 1500 ml of `(N)/(10)` NaOH solution. |
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Answer» According to law of equivalence Equivalent of HCl - Equivalent of NaOH `N_(1)V_(1)=N_(2)V_(2)` `rArr N_(1)xx(500)/(1000)=(1)/(10)xx(1500)/(1000)` `rArr N_(1)=(150)/(500)=(3)/(10)` . |
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| 969. |
How much of NaOH is reuired to neutralise 1500 `cm^(3)` of 0.1 N HCl (Na=23)?A. 40 gB. 4 gC. 6 gD. 60 g |
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Answer» Correct Answer - C `N=(W(gm)xx1000)/(Vxx"Eq. wt.")` 1500 ml of `0.1 N HCl =150 ml` (N) `1=(W(gmxx1000))/(150xx40), W(gm)=(150xx40)/1000=6 gm`. |
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| 970. |
1.5 mol of `O_(2)` combines with Mg to form oxide MgO. The mass of Mg (at. Mass 24) that has combined isA. 72 gB. 36 gC. 48 gD. 24 g |
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Answer» Correct Answer - A `underset("1 mole")(Mg)+underset("0.5 mole")(1/2 O_(2)) rarr MgO` 0.5 mole of oxygen react with 1 mole of Mg 1.5 mole of oxygen react with `1.5/0.5=3` mole `24xx3=72 gm`. |
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| 971. |
An ion is reduced to the element when it absords `6xx10^(20)` electrons. The number of equivalents of the ion is:A. (a)`0.1`B. (b)`0.01`C. (c )`0.001`D. (d)`0.0001` |
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Answer» Correct Answer - C `6xx10^(23) "electrons"-=1` equivalent `:. 6xx10^(20)` electrons`-=(6xx10^(20))/(6xx10^(23))=0.001` equivalent |
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| 972. |
Insulin contains `3.4%` sulphur. Calculate minimum mol.wt. of insulin.A. `94*117`B. `1884`C. `941*176`D. `976*` |
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Answer» Correct Answer - C For minimum molecular mass, insulin must have at least one sulphur atom in its one molecule. If it has 3.4 g S, the mol. Mass = 100 If it has 32 g S, the mol mass `= (100xx32)/(3.4)` = 941.176 |
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| 973. |
Assertion: In the titration of `Na_(2)CO_(3)` with `HCl` using methyl orange indicator, the volume of acid required is twice that of the acid required using phenolphthalein as indicaton. Reason: Two moles of `HCl` are required for the complete neutralisation of one mole of `Na_(2)CO_(3)`.A. Statement I is true, Statement II is true, Statement II is the correct explanation of Statement I.B. Statement I true, Statement II is true, Statement II is not the correct explanation of Statement I.C. Statement I is true, Statement II is falseD. Statement I is false, Statement II is true. |
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Answer» Correct Answer - B Both assertion and reason are factually true but the reason does not exactly explain the assertion. The correct explanation is, methyl orange and phenolphthalein changes their colour at different `pH`. |
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| 974. |
Statement `1 "mole" O_(3)=N "molecule"O_(3)=3N` atoms of `O=48 g` Explanation A mole is the amount of matter that contains as many as objects as the amount of atoms exactly in `12 g C^(12)`.A. (a)If both assertion and reason are true and the reason is the correct explanation of the assertion.B. (b)If both assertion and reason are true and the reason is not the correct explanation of the assertion.C. (c )If assertion is true but reason is false.D. (d)If assertion is false but reason is true. |
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Answer» Correct Answer - A Explanation is correct reason for the statement. |
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| 975. |
`Mg+2HCl toMgCl_(2)+H_(2)`. The ration of Mg used to `H_(2)` produced by weight isA. `1:12`B. `12:1`C. `24:1`D. `1:6` |
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Answer» Correct Answer - B `underset(24g)(Mg)+2HCl to MgCl_(2)+underset(2g)(H_(2))` Mass of Mg used: Mass of hydrogen produced `= 24:2=12:1` |
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| 976. |
Assertion (A): pure water obtained from different states of india always contains hydrogen and oxygen in the ration of 1:8 by mass. Reason (R ): Total mass of reactants and products during chemical change is always the same.A. (a)If both assertion and reason are true and the reason is the correct explanation of the assertion.B. (b)If both assertion and reason are true and the reason is not the correct explanation of the assertion.C. (c )If assertion is true but reason is false.D. (d)If assertion is false but reason is true. |
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Answer» Correct Answer - B Both assertion and reason correct statement but reason is not the correct explanation of assertion. The correct explanation is based on the law of constant composition. |
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| 977. |
Statement `H_(3)BO_(3)` is monobasic Lewis acid but its salt `Na_(3)BO_(3)` exist. Explanation `H_(3)BO_(3)` reacts with `NaOH` to give `Na_(3)BO_(3)`.A. (a)If both assertion and reason are true and the reason is the correct explanation of the assertion.B. (b)If both assertion and reason are true and the reason is not the correct explanation of the assertion.C. (c )If assertion is true but reason is false.D. (d)If assertion is false but reason is true. |
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Answer» Correct Answer - C `B(CH)_(3)+NaOH rarr Na[B(OH)_(4)]` |
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| 978. |
Assertion: In the titration of `Na_(2)CO_(3)` with `HCl` using methyl orange indicator, the volume of acid required is twice that of the acid required using phenolphthalein as indicaton. Reason: Two moles of `HCl` are required for the complete neutralisation of one mole of `Na_(2)CO_(3)`.A. (a)If both assertion and reason are true and the reason is the correct explanation of the assertion.B. (b)If both assertion and reason are true and the reason is not the correct explanation of the assertion.C. (c )If assertion is true but reason is false.D. (d)If assertion is false but reason is true. |
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Answer» Correct Answer - B `Na_(2)O_(3)+2 HCl rarr 2Na Cl+HO+CO_(2)`, phenolphthalein is used for `Na_(2)O_(3)+2 HCl rarr NaHCO_(3)+NaCl+CO_(2)` |
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| 979. |
Equivalent mass of a metal is 12 g `"mol"^(-1)`. Hence, equivalent mass of its oxide isA. 24 g `"mol"^(-1)`B. 28 g `"mol"^(-1)`C. 20 g `"mol"^(-1)`D. 34 g `"mol"^(-1)` |
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Answer» Correct Answer - C Equivalent mass of metal =12 Equivalent mass of oxygen = 8 g `"mol"^(-1)` `therefore` Equivalent mass of metal oxide = 20 g `"mol"^(-1)` |
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| 980. |
What is the molecular mass of a compound `X`, if its `3.0115 xx 10^(9)` molucules weigh `1.0 xx 10^(-12) g?`A. 150 gB. 200 gC. 630 gD. 500 g |
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Answer» Correct Answer - B `3.0115 xx 10^(9) " molecules of X"=10^(-12)g` `6.023xx 10^(23)" molecules of X"=(10^(-12)xx6.023xx10^(23))/(3.0115xx10^(9))` = 200 g = 200 amu |
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| 981. |
`3.92 g` of ferrous ammonium sulphate crystals are dissolved in `100 ml` of water, `20 ml` of this solution requires `18 ml` of `KMnO_(4)` during titration for complete oxidation. The weight of `KMnO_(4)` present in one litre of the solution isA. (a)` 3.476 g`B. (b)` 12.38 g`C. (c )` 34.76 g`D. (d)` 1.238 g` |
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Answer» Correct Answer - A `KMnO_(4)`= Mohr salt `(M_(1)V_(1))/1=(M_(2)V_(2))/5` `=[W/(MxxV)xx1000]xxV_(2)/5` `[(Wxx1000)/(58xx1000)]xx18= (3.92xx1000)/(392xx1000)xx20/5 W=3.476 g//L`. |
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| 982. |
How many significant figures are present in the following? a. 0.0025 b. 208 c. 5005 d. 126000 e. 500.0 f. 2.0034 |
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Answer» Correct Answer - (i) 2 (ii)3 (iii) 4 (iv) 3 (v) 4 (vi) 5 (i) 0.0025 there are 2 significant figures (ii) 208 there are 3 sigificant figures . (iii) 5005 (iv) 126,000 there are 3 significant figures . (v) 5000.0 there arae 4 significant figures. (vi) 2.0034 there are 5 sinnificant figures. |
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