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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.
| 101. |
Which among of the following is an element ?A. 22 carat goldB. German silverC. GraphiteD. Dry ice |
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Answer» Correct Answer - C Graphite is allotrope of carbon. |
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| 102. |
Which of the following is not an element.A. DiamondB. SilicaC. TungstenD. Graphite |
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Answer» Correct Answer - B Silica `(SiO_(2))` is a compound |
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| 103. |
Which one of the following properties of an element is not variable ?A. ValencyB. Atomic weightC. Equivalent MassD. All the above |
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Answer» Correct Answer - B At.Wt never vary. |
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| 104. |
If 216g residue is obtained from 0.5 ml silver salt,A. `n = 4`B. `n = 2`C. Molar mass of salt `=718g//mol`D. Molar of mass of sat `=388g mol` |
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Answer» Correct Answer - B::C `0.5 xx n = (216)/(108)`mol of `Ag n = 4` molar mass `=58 +(165) n = 718g//mol` |
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| 105. |
Density of a solution containing `x%` by mass of `H_(2)SO_(4)` is y. The normality isA. `(xy xx 10)/(98)`B. `(xy xx 10)/(98y)xx 2`C. `(xy xx 10)/(98) xx 2`D. `(x xx 10)/(98y)` |
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Answer» Correct Answer - C `N = ("percentage" xx d xx 10)/(GEW)` |
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| 106. |
Normality of 2% of `H_(2)SO_(4)` solution by volume is nearlyA. 2B. 4C. 0.2D. 0.4 |
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Answer» Correct Answer - D `N = (w)/(GEW) xx (1000)/(V("in ml"))` |
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| 107. |
The molarity of pure water isA. `100M`B. `55.6M`C. `50M`D. `18M` |
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Answer» Correct Answer - B `M = (w)/(GMW) xx (1)/(V "in litres")` |
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| 108. |
The reaction `5H_(2)O_(2)+XClO_(2)+2OH^(-) rarr XCl^(-)+YO_(2)+6H_(2)O` is balanced ifA. `x = 5, y = 2`B. `x = 2, y = 5`C. `x = 4, y = 10`D. `x = 5, y = 5` |
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Answer» Correct Answer - B Balancing according to oxidation number method |
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| 109. |
1 L `(M)/(10) Ba(MnO_(4))_(2)` in acidic medium can be oxidised completely with `(1)/(6)` L of x M ferric oxalate. The volume of x is |
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Answer» Correct Answer - 1 (a) `10e^(-) +2MnO_(4)^(-) rarr 2Mn^(2+) (n = 5)` `{:(Eq of MnO_(4)^(-) -=, Eq of Fe_(2) (C_(2)O_(4))_(3)),((n=5),(n=6).):} ` `1L xx (M)/(10) xx 10 -= (1)/(6) L xx 6 xx x x = 1M` |
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| 110. |
If equal volumes of 1 M `KMnO_(4)` and 1M `K_(2)Cr_(2)O_(7)` solutions are allowed to oxidise Fe(II) to Fe(III) in acidic medium, then Fe(II) oxidised will beA. more with `KMnO_(4)`B. more with `K_(2)Cr_(2)O_(7)`C. equal with both oxidizing agentsD. cannot be determined |
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Answer» Correct Answer - B No. of m eqts of `K_(2)Cr_(2)O_(7) = 6` No. of equivalents of `KMnO_(4) = 5` |
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| 111. |
100 mL of x M `KMnO_(4)` is requried oxidise 200 mL of 0.2 M ferric oxalate in acidic medium what is the normality of `KMnO_(4)`?A. `0.0075`B. `0.005`C. `0.01`D. `0.015` |
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Answer» Correct Answer - A `MnO_(4)^(-) +I^(-) rarr MnO_(2) +I_(2)` No: of milli eq of `MnO_(4)^(-) = 0.02 xx 3 xx 250 = 15` No: of milli eq. of `I_(2) = 0.1 xx 2 xx 250 = 25` Here `Mno_(4)^(-)` is limiting reagent. `:.` No of milli eq. of `I_(2)` formed `=15` `:.` No. of milli eq of `I_(2)` formed `=(15)/(1000) = 0.015` `:.` No of milli moles of `I_(2)` formed `=(0.015)/(2) = 0.0075` |
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| 112. |
A sample of 1.0g of solid `Fe_(2)O_(3) of 80%` purity is dissolved in a moderately concentrated HCl solution which is reduced by zinc dust. The resulting solution required 16.7mL of a 0.1M solution of the oxidant. Calculate the number of electrons taken up by the oxidant.A. 5B. 2C. 4D. 6 |
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Answer» Correct Answer - D Weight of pure `Fe_(2)O_(3) = (1 xx 80)/(100) =0.8g` `Fe_(2)O_(3) +6HCI rarr 2FeCl_(3) +3H_(2)O` `2FeCl_(3) +H_(2) overset("Zn dust")rarr 2FeCl_(2) +2HCl` `Fe^(++) +` Oxidant `rarr Fe^(+++) +` reductant t Eq. wt of `Fe_(2)O_(3) = ("Mole. wt")/(2) = (160)/(2) = 80g` meq of `Fe_(2)O_(3) =` meq of oxidant `(0.8)/(80) xx 10^(3) = 16.7 xx 0.1(x)` `:. x = 6` |
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| 113. |
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 |
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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` |
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| 114. |
A solution containing `0.2` mole of ferric chloride is allowed to react with `0.24` mole of sodium hydroxide. The correct statement for this reaction isA. Limiting reagent for this reaction is `NaOH`B. `0.08` moles of `Fe(OH_(3))` is formedC. `0.12` mole of `FeCI_(3)` is left unreactedD. All the above |
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Answer» Correct Answer - B `FeCl_(3) +3NaOh rarr Fe(OH)_(3) +3NaCl` |
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| 115. |
The number of moles of `Fe_(2)O_(3)` formed when 0.5 moles of `O_(2)` and `0.5` moles of Fe are allowed to react areA. `0.25`B. `0.5`C. `1//3`D. `0.125` |
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Answer» Correct Answer - A `4Fe +3O_(2) rarr 2Fe_(2)O_(3)` |
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| 116. |
The sulphur dioxide obtained by the combustion of 8 gms of sulphur is passed into Bromine water. The solution is then treated with barium chloride solution. The amount of barium sulphare formed isA. 1 moleB. 0.5 moleC. 0.25 gmsD. 0.25 gm moles |
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Answer» Correct Answer - D 32 gm of sulphur is present in 1 mole of `BaSO_(4)` 8 gm of sulphur is present in `1//4` mole `BaSO_(4)` |
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| 117. |
If `0.7` moles of Barium Chlorine is treated with 0.4 mole of potassium sulphate, number of moles of barium sulphate formed areA. 0.7B. 0.4C. 0.35D. 0.2 |
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Answer» Correct Answer - B `BaCl_(2)+K_(2)SO_(4) rarr BaSO_(4) +2KCl` |
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| 118. |
0.7 moles of potassium sulphate is allowed to react with 0.9 moles of barium chloride in aqueous solutions. The number of moles of the substance precipitated in the reaction isA. 1.4 moles of potassium chlorideB. 0.7 moles of barium sulphateC. 1.6 moles of potassium chlorideD. 1.6 moles of barium sulphate |
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Answer» Correct Answer - B `K_(2)SO_(4) +BaCl_(2) rarr BaSO_(4) + 2KCl 38` `CaCO_(3) rarr CaO +CO_(2)` |
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| 119. |
When V ml of `2.2 M H_(2)SO_(4)` solution is mixed with 10 V ml of water, the volume contraction of 2% take place. Calculate the molarity of diluted solution ?A. `0.2M`B. `0.204M`C. `0.196M`D. `0.224M` |
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Answer» Correct Answer - B Let `V_(1) = V -ml,V_(2) = (10V +V) -((10V+V)2)/(100)` Apply law of dilution, `M_(1)V_(1) = M_(2)V_(2)` |
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| 120. |
If 500mL of a 5M solution is diluted to 1500 mL, what will be the molarity of the solution obtained?A. 1.5MB. 1.66MC. 0.017MD. 1.59M |
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Answer» Correct Answer - B `M_(1)V = M_(2)V_(2)` |
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| 121. |
The mass of `Na_(2)CO_(3)` required to prepare 500ml of `0.1M` solution isA. `10.6g`B. `5.3g`C. `2.65g`D. `7.95g` |
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Answer» Correct Answer - B `w = (M xx GMW xx V("in ml"))/(1000)` |
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| 122. |
`500ml` of `0.1MKCI, 200ml` of `0.01M NaNO_(3)` and `500ml` of `0.1M AgNO_(3)` was mixed. The molarity of `K^(+), Ag^(+),CI^(-),Na^(+),NO_(3)^(-)` in the solution would beA. `[K^(+)] = 0.04, [Ag^(+)] = 0.04, [Na^(+)] = 0.002`B. `[K^(+)] = 0.04, [Na^(+)] = 0.00166, [NO_(3)^(-)] = 0.04333`C. `[K^(+)] = 0.04, [Ag^(+)] = 0.05, [Na^(+)] = 0.0025`D. `[K^(+)] = 0.05, [Na^(+)] = 0.0025, [Cl^(-)] = 0.05, [NO_(3)^(-)] = 0.0525` |
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Answer» Correct Answer - B `[K^(+)] = (0.1 xx 500)/(500 +200 +500) = (50)/(1200) = 0.04167M` `[Na^(+)] = (0.01 xx 200)/(500 +200 +500) = (2)/(!200) = 0.00167M` `[NO_(3)^(-)] = (200 xx 0.01 xx 500 xx 0.1)/(500 +200 +500) = (52)/(1200) = 0.0433M` Since 50 milli moles of `Cl^(-)` are mixed with 50 milli moles of `Ag^(+)` ions, total `Ag^(+)` & `Cl^(-)` get precipitated as `AgCl`. Hence the solution does not contain `Ag^(+)` & `Cl^(-)` ions |
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| 123. |
10 grams of each `O_(2),N_(2)` and `Cl_(2)` are kept in three bottles. The correct order of arrangement of bottles containing decreasing number of Molecules.A. `O_(2),N_(2),Cl_(2)`B. `Cl_(2),N_(2),O_(2)`C. `Cl_(2),O_(2),N_(2)`D. `N_(2),O_(2),Cl_(2)` |
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Answer» Correct Answer - D No.of molecules `= (W)/(MW) xx N` |
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| 124. |
What is the ratio of number of molecules, if tha mass ratio of `N_(2)` and `O_(2)` is 4:1 ? |
| Answer» Ratio of molecules `N_(2)` and `O_(2) = (4)/(28): (1)/(32) = (1)/(7) : (1)/(32) :7` | |
| 125. |
50.0 kg of `N_(2)(g)` and 10.0 kg of `H_(2)(g)` are mixed to produce `NH_(3)(g)`. Calculate the `NH_(3)(g)` formed. Identify the limiting reagent in the production of `NH_(3)` in this situation.A. `H_(2),56.67kg`B. `H_(2),13.1kg`C. `H_(2),1.56kg`D. `H_(2),36.2kg` |
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Answer» Correct Answer - A `N_(2)+3H_(2) rarr 2NH_(3)` `(H_(2)` is limiting reagent) |
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| 126. |
According to Avogadro, equal volumes of two different gases under same conditions of temperature and pressure contain equal number ofA. AtomsB. MoleculesC. ElectronsD. Protons |
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Answer» Correct Answer - B Under identical conditions equal volumes of gases contain equal number of molecules. |
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| 127. |
The oxiation number of manganese in potassium manganate isA. `+7`B. `+6`C. `+4`D. `+2` |
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Answer» Correct Answer - B `K_(2)overset(+6)(MnO_(4))` |
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| 128. |
One gas bleaches the colour of flowers by reduction and other by oxidation. These gases areA. CO & `CI_(2)`B. `H_(2)S` & `Br_(2)`C. `SO_(2)` & `CI_(2)`D. `NH_(4)` & `SO_(3)` |
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Answer» Correct Answer - C `SO_(2) +2H_(2)O rarr H_(2)SO_(4)+2(H)` `Cl_(2)+H_(2)O rarr 2HCl +(O)` |
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| 129. |
In the reaction between ethylene and alkaline `KMnO_(4)` the oxidation number of Manganese isA. Decreases from `+7` to `+2`B. Decreases from `+7` to `+6`C. Decreases from `+7` to `+4`D. Increases from `+4` to `+7` |
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Answer» Correct Answer - C `overset(+7)(MnO_(4)^(-) rarr overset(+4)(MnO_(2))` |
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| 130. |
26 cc of `CO_(2)` are passed over red hot coke. The volume of `CO evolved isA. 15 ccB. 10ccC. 32ccD. 52cc |
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Answer» Correct Answer - D `CO_(2) +C rarr 2CO` |
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| 131. |
Metal X forms wo oxides. Formula of the first oxide is `XO_(2)`. The first oxide contains `50%` of oxygen. If the second oxide contains `60%` of oxygen, the formula of the second oxide isA. `X_(2)O`B. `XO_(3)`C. `X_(2)O_(3)`D. `X_(3)O_(2)` |
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Answer» Correct Answer - B `X: O = 1:2 = (50)/("Atomic weight of metal"): (50)/(16)` `X : O = (40)/(32): (60)/(16)` |
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| 132. |
When equal masses of methane and sulphur dioxide are taken, then the ratio of their molecule isA. `1:1`B. `1:2`C. `2:1`D. `4:1` |
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Answer» Correct Answer - D Ratio of molecules is equal to inverse ratio of their mol. Wts .if the wts are equal. |
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| 133. |
Atomic weight of elements is 40 and its valency is 2. what is its equivalent weigtA. 20B. 40C. 80D. 10 |
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Answer» Correct Answer - A `E.W = ("At. Wt")/("Valency")` |
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| 134. |
15 cc of gaseous hydrocarbon required 45 cc of oxygen for complete combusion if 30 cc of `CO_(2)` is formed, the formula of the gaseous compound isA. `C_(3)H_(6)`B. `C_(2)H_(2)`C. `C_(4)H_(10)`D. `C_(2)H_(2)` |
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Answer» Correct Answer - D `C_(x)H_(y)+ (x+(y)/(4)) O_(2) rarr xCO_(2) +(y)/(2) H_(2)O` |
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| 135. |
X and Y are two different elements having their atomic masses in 1:2 ratio. The compound formed by the combination of X and Y contains 50% of X by weight. The empirical formula of the compound isA. `X_(2)Y`B. `XY_(2)`C. `XY`D. `X_(4)Y` |
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Answer» Correct Answer - A `(%X)/(1) (%y)/(2)` |
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| 136. |
Which of the following sequence is correct with reference to the oxidation number of iodine?A. `I_(2) lt ICI lt HI lt HIO_(4)`B. `HIO_(4) lt ICI lt I_(2) lt HI`C. `I_(2) lt HI lt ICI lt HIO_(4)`D. `HI lt I_(2) lt ICI lt HIO_(4)` |
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Answer» Correct Answer - D `HI lt I_(2) lt ICI lt HIO_(4)` |
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| 137. |
Oxidation state of carbon is not zero inA. `CH_(2)O`B. `C_(6)H_(12)O_(6)`C. `CH_(2)Cl_(2)`D. `CHCl_(3)` |
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Answer» Correct Answer - D `overset(+2)(CH) Cl_(3)` |
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| 138. |
Oxidation number of carbon in carbon suboxide `(C_(3)O_(2))` is :A. `(+2)/(3)`B. `(+4)/(3)`C. `+4`D. `(-4)/(3)` |
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Answer» Correct Answer - B `C_(3)O_(2)` `3 xx +2 (-2) = 0` `3 xx +4 rArr x = +(4)/(3)` |
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| 139. |
10 mL of `N//20 NaOH` solution is mixed with 20 mL of `N//20 HCI` solution. The resulting solution will :A. `40mL`B. `20mL`C. `30mL`D. `5mL` |
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Answer» Correct Answer - A `HCl +AgNO_(3) rarr AgCl +HNO_(3)` m equivalents of `HCl` left `=0.5` m. equivalents of `HNO_(3)` formed `= 1.5` `V_(a) N_(a) = V_(b)N_(b), 2 = V_(b) xx 0.05` Volume of `NaOH = 40ml` |
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| 140. |
The number of atoms present in 10 gms of `CaCO_(3)` areA. `5N^(3)`B. 0.5NC. 5D. N |
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Answer» Correct Answer - B `100......5N` |
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| 141. |
The mass of `K_(2)Cr_(2)O_(7)` required to produce 5.0 L `CO_(2)` at `77^(@)C` and 0.82 atm pressure from excess of oxalic acid and volume of 0.1 N `NaOH` required to neutralise the `CO_(2)` evolved respectively areA. `7g, 2.86L`B. `5g, 1.86L`C. `4g,0.86L`D. `14g,2.86L` |
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Answer» Correct Answer - D `H_(2)C_(2)O_(4) rarr 2CO_(2) +2H^(+) +2e^(-)` Moles of `CO_(2) =(PV)/(RT) = (0.82 xx 5)/(0.082 xx 350) = (1)/(7)` `rArr (1)/(7) (W_(K_(2)Cr_(2)O_(7)))/(49) rArr W_(K_(2)Cr_(2)O_(7)) = 7g` and `(1)/(7)mol CO_(2) -= (1)/(7) mol H_(2)CO_(3) = (2)/(7)` Eqts `(2)/(7) = 0.1 xx V = 2.86L` Volume of `NaOH,V = 2.86` lit |
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| 142. |
In the chemical reaction, `K_(2)Cr_(2)O_(7)+xH_(2)SO_(4)+ySO_(2)rarrK_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+zH_(2)O` `x,y, and z` areA. 1,3,1B. 4,1,4C. 3,2,3D. 2,1,2 |
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Answer» Correct Answer - A Balancing according to oxidation number method |
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| 143. |
`4.9 g` of `K_(2)Cr_(2)O_(7)` is taken to prepare `0.1 L` of the solutio. `10 mL` of this solution is further taken to oxidise `Sn^(2+)` ion into `Sn^(4+) ion` so produced is used in second reaction to prepare `Fe^(3+)` ion then the millimoles of `Fe^(3+)` ion formed will be (assume all other components are in sufficient amount)[Molar mass of `K_(2)Cr_(2)O_(7)=294 g`].A. 5B. 20C. 10D. 15 |
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Answer» Correct Answer - C N of `K_(2)Cr_(2)O_(7) = (4.9 xx 1)/(49xx 0.1) = 1N` No. of m.equts orf `Fe^(+3) =` No. of m.eqts of `Sn^(4+)` =No. of m.equts of `K_(2)Cr_(2)O_(7) = 10 xx 1 = 10` |
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| 144. |
The volume of water to be added to 400 ml of `N//8 HCI` to make it exactly `N//12` isA. 400 mlB. 300mlC. 200 mlD. 100 ml |
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Answer» Correct Answer - C `N_(1)V_(1) = N_(2)V_(2)` Volume of water is added `=V_(2) -V_(1)` |
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| 145. |
The vapour density of a gas is 11.2. The volume occupied by 11.2 gms of the gas at S.T.P isA. 11.2 litB. 4 litC. 2 litD. 22.4 lit |
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Answer» Correct Answer - A `M.W = 2 xx VD` |
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| 146. |
Substance having more number of moles isA. 90 gms of waterB. 112 litre of hydrogen at S.T.PC. 24 gm of helium gasD. `3.01 xx 10^(24)` molecules of `CO_(2)` |
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Answer» Correct Answer - C `n = (W)/(M)` or `("Volume in lit")/(22.4)` |
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| 147. |
The following data are available. (i) % of Mg in `MgO` and in `MgCl_(2)` (ii) % of C in CO & `CO_(2)` (ii) % of Cr in `K_(2)Cr_(2)O_(7)` and `K_(2)CrO_(4)` (iv) % of Cu isotopes in Cu metal. The law of multiple proportions may be illustrated by data.A. i & iiB. Only iiC. i,ii & iiiD. Only iii |
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Answer» Correct Answer - B `CO` & `CO_(2)` only illustrates laws of multiple proportions |
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| 148. |
`Mg(OH)_(2)+HCI rarr Mg(OH)CI +H_(2)O` In this reaction, the equivalent weight of `Mg(OH)_(2)` |
| Answer» `=` Formula weight`//1 = 58//1 = 58`. | |
| 149. |
From 320 mg. of `O_(2), 6.023 xx 10^(20)` molecules are removed, the no. of moles remained areA. `9 xx 10^(-3)` molesB. `9 xx 10^(-2)` molesC. ZeroD. `3 xx 10^(-3)` moles |
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Answer» Correct Answer - A No. of moles of `(Wt)/(M.Wt)` or `("No. of molecules")/(N)` |
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| 150. |
Given the number: 161 cm, 0.161 cm, 0.0161 cm. The number of significant figures for the three numbers areA. 3,4 and 5B. 3,3 and 3C. 3,3 and 4D. 3,4 and 4 |
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Answer» Correct Answer - B Based on signification figures concept. |
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