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.
| 751. |
The density `("in g mL"^(-1))` of a 3.60 M sulphuric acid solution that is 29 % of acid by mass is : (Given molar mass `= 98 g mol^(-1)`)A. 1.45B. 1.64C. 1.88D. 1.22 |
|
Answer» Correct Answer - D By definition, `3.6 M(3.6xx98=352.8g)` of `H_(2)SO_(4)` are present in 1000 mL of the solution 29.0 g of acid are present in solution = 100 g 352.8 g of acid are present in solution `= ((100g)xx(352.8g))/((29g))` Density of solution `=("Mass of solution")/("Volume of solution")` `= ((1216g))/((1000mL))=1.22"g mL"^(-1)`. |
|
| 752. |
Which weighs the maximum ?A. 2.24 litres of `CO_(2)` at N.T.PB. `6.022 xx 10^(23)` molecules of `CO_(2)`C. `6.022 xx 10^(23)` molecules of `CO_(2)`D. 10 g of `CO_(2)` |
|
Answer» Correct Answer - B `6.022 xx 10^(23)` molecules of `CO_(2)` weigh 44 g (gram molar mass). Weigh maximum. |
|
| 753. |
Cadverine molecule has `58.77% C, 13.81% H` and `27% N` by mass Find Empirical formula of cadverine . |
|
Answer» Correct Answer - `C_(5)H_(14)N_(2)` `|{:("element","massper100gm","mole","simplest ratio",),(C,58.77,58.77//12,5,),(H,13.81,13.81//1,14,),(N,27.42,27.14//2,2,):}|` `E.F = C_(5)H_(14)N_(2)=102=M.E` |
|
| 754. |
When the same amount of zinc is treated separately with excess of sulphric acid and excess of sodium hydroxide, the ratio of volume of hydrogen evolved isA. `1 : 1`B. `1 : 2`C. `2 : 1`D. `9 : 4` |
|
Answer» Correct Answer - A In both cases, the same volume of hydrogen is evolved for the same amount of zinc reacted. `Zn + H_(2)SO_(4) rarr ZnSO_(4) + uarr` `Zn + NaOH rarr Na_(2) ZnO_(2) + H_(2) uarr` |
|
| 755. |
Which of the statement are ture?A. Brass is an elementsB. Dry ice is a mixtureC. Aerated drink, e.g., coca cola, is a mixture.D. Diesel is a mixture |
|
Answer» Correct Answer - C::D (a) and (b) are compound. (c ) and (d) are mixture. |
|
| 756. |
The total number of electrons in one molecular of carbon dioxide isA. 22B. 44C. 66D. 88 |
|
Answer» Correct Answer - C In one molecule of `CO_(2)`, the number of electrons is `6 + 8 + 8 = 22` |
|
| 757. |
A bulb contains `1.6 g` of `O_(2)` contains.A. 0.05 mol of `O_(2)`B. `3.011 xx 10^(22)` molecules of `O_(2)`C. `1.12 L` of `O_(2)` at `STP`D. `1.22 L` of `O_(2)` at `STP` |
|
Answer» Correct Answer - A::B::C::D a. Moles of `O_(2) = (1.6)/(32) = 0.05` b. Molecules of `O_(2) = 0.05 xx 6.023 xx 10^(23) = 3.011 xx 10^(22)` d. Volume of `O_(2)` at `STP = 0.05 xx 22.4 = 1.12 L` d. `V_(O_(2))` at `SATP = 0.05 xx 24.4 = 1.22 L` |
|
| 758. |
When `2.76 g` of silver carbonate is strongly heated, it yields a residue weighingA. `2.16 g`B. `2.48 g`C. `2.32 g`D. `2.64 g` |
|
Answer» Correct Answer - A `Ag_(2) CO_(3) rarr 2 Ag + CO_(2) + (1)/(2) O_(2)` `276 g Ag_(2) CO_(3)` gives `2 xx 108 = 216 g` residue `2.76 g Ag CO_(3)` gives `2.16 g` residue (silver) |
|
| 759. |
Calculate the residue obtained on strongly heating `2.76 Ag_(2) CO_(3)`. |
|
Answer» `underset({:(276 g),((108 xx 2 + 12 + 16 xx 3)):})(Ag_(2)CO_(3)) overset(Delta)rarr underset({:(216 g),((108 xx 2)):})(2 Ag) + CO_(2) + 1//2O_(2)` `276 g of Ag_(2) CO_(3) implies 216 g of Ag` `2.76 g of Ag_(2) CO_(3) implies 2.16 g of Ag` |
|
| 760. |
A mixture of `NaCl` and `Na_(2)CO` is given On heating `12 g` of the mixture with dilute `HCl, 2.24 g` of `CO_(2)` is removed. Calculate the amounts of each in the mixture. |
|
Answer» Reactions: `NaCl + HCl rarr` no reaction `Na_(2) CO_(3) + 2 HCl rarr 2 NaCl + CO_(2) + H_(2) O` Let `x g` of `Na_(2)CO_(3)` in the mixture. `:. (x)/(106) mol of Na_(2) CO_(3) = (x)/(106) mol of CO_(2)` `:. (x)/(106) = (2.24)/(44)` or `x = 5.4 g Na_(2) CO_(3)` Weight of `NaCl` in mixture `= 12 - 5.4 = 6.6 g` |
|
| 761. |
Acetylene burns in oxygen to form carbon dioxide and water. Write the skeleton equation for the reaction and balance it. |
|
Answer» The skeleton equation for the reaction is : `C_(2)H_(2)+O_(2)rarrCO_(2)+H_(2)O` The balancing is completed in the following steps : (i) Write the oxygen in the atomic state `C_(2)H_(2)+OrarrCO_(2)+H_(2)O` (ii) Balance the C atoms `C_(2)H_(2)+O rarr2CO_(2)_H_(2)O` (iii) Balance the O atoms `C_(2)H_(2)+5Orarr2CO_(2)+H_(2)O` Upon checking, the H atoms are already balanced (iv) Make the equation molecular or change O into `O_(2)`. For this multiply the equation by the factor 2 `2C_(2)H_(2)+5O_(2)rarr4CO_(2)+2H_(2)O` The final balanced equation is : `2C_(2)H_(2)+5O_(2)rarr4CO_(2)+2H_(2)O`. |
|
| 762. |
If 1 moles of ethanol `(C_(2)H_(5)OH)` completely burns to form carbon dioxide formed is about |
|
Answer» `{:(C_(2)H_(5)OH + 3O_(2) to 2CO_(2) + 3H_(2)O),(1" "3" "2" "2):}` 2 moles of `CO_(2)` are formed =88 g |
|
| 763. |
A gaseous mixture of `H_(2) and NH_(3)` gas contains 68 mass % of `NH_(3)` . The vapour density of the mixture is- |
|
Answer» No. of mole of `NH_(3)` in 100 g mixture `=(68)/(17)=4` No. of moles of `H_(2)` in 100 g mixutre`=(32)/(2)=16` `M_("average")=("Total mass")/("Total moles")=(100)/(4+16)=5` V.d=`(5)/(2)=2.5` |
|
| 764. |
What mass of CaO is formed by heating 50 g `CaCO_(3)` in air? |
|
Answer» `CaCo_(3)(s) to CaO(s) + CO_(2)(g)` 50 gm `=(50)/(100)` mol `=(1)/(2)` mol `" "(1)/(2)` mol `-(1)/(2)xx56=28` gm |
|
| 765. |
By heating `10 g CaCO_(3), 5.6 g CaO` is formed. What is the weight of `CO_(2)` obtained in this reaction |
|
Answer» `underset({:(40 + 12 + 16 xx 3),(= 100 g):})(CaCO_(3)) rarr underset({:(40 + 16),(= 56 g):})(CaO) + underset({:(12 + 16 xx 2),(= 44 g):})(CO_(2))` a. `100 g of CaCO_(3) implies 56 g of CaO` `10 g of CaCO_(3) implies 5.6 g of CaO` b. `100 g of CaCO_(3) implies 44 g of CO_(2)` `10 g of CaCO_(3) implies 4.4 g of CO_(2)` |
|
| 766. |
Many cereals are made with high moisture content so that the product can be formed into various containing 50% `H_(2)O` by mass is produced at the rate of 1000 kg/hr. How much water (in kg) must b evaporated per hour if final product contains only 20% water? (Fill your answer dividing it by 62.5) |
|
Answer» Correct Answer - 6 |
|
| 767. |
Identify the option containing maximum number of atoms:A. 18mg of glucoseB. 2 mg of hydrogen gasC. 10 mg of `H_(2)O`D. 7.8 mg of benzene |
|
Answer» Correct Answer - A |
|
| 768. |
Calculate the mass of ammonia that can be produced from the decomposition of a sample `(NH_(4))_(2)PtCl_(6)` containing 0.100g Pt. `{:("Substance", " Molar mass"), ((NH_(4))_(2)PtCl_(6), 443.9g. mol^(-1)):}`A. 0.0811gB. 0.0766gC. 0.0175gD. 0.00766g |
|
Answer» Correct Answer - C |
|
| 769. |
If, from 10 moles `NH_(3)` and 5 moles of `H_(2)SO_(4),` all the H-atoms are removed in order to form `H_(2)` gas, then find the number moles of `H_(2)` formed. |
|
Answer» Correct Answer - 20 |
|
| 770. |
2 litres of a mixture of nitrous and nitric oxides at STP have a mean molecular weight of 39.8.What volume of nitrogen measured at STP could be obtained when the mixture has been passed over red hot copper ?A. 1.7 LB. 1.9 LC. 1.5 LD. 1.85 L |
|
Answer» Correct Answer - A |
|
| 771. |
90.8 litres of a mixture of nitrogen and hydrogen measured at STP were passed over a catalyst. After the reaction, the volume of the mixture reduced to 68.1 litres. Ammonia thus formed was dissolved in 101 ml of an aqueous ammonia solution of density of 0.85 g//ml containing `12%` by mass of `NH_(4)OH.` Determine the percent weight strength of the final solution. [Give answer exculding decimal places] |
|
Answer» Correct Answer - 44 |
|
| 772. |
340 g of `NH_(3)(M=17)` when decomposes how many liters of nitrogen gas is produced at STP? |
|
Answer» Correct Answer - 227 |
|
| 773. |
`Br_(2)(l)` reacts with `Cl_(2)(g)` to form `BrCl` and `BrCl_(3),` simultaneously. How many moles of `Cl_(2)(g)` reactas completely with 3 moles of `Br_(2)(i)` to give `BrCl` and `BrCl_(3)` in 5:1 mole ratio? |
|
Answer» Correct Answer - 4 |
|
| 774. |
Find the density of `CO_(2)(g)` with respect to `N_(2)O(g).` |
|
Answer» Correct Answer - 1 |
|
| 775. |
Atomic mass of elements are defined with respect to `(1)/(12)` th of mass of single atom of C-12 [present scale]. If reference is changed to `(1)/(24)` th of mass of single atom of C-12 [new scale] , then select the correct statement(s). Given : Atomic mass of Fe on present scale is 56 Mass of `(1)/(12)` the of mass of single atom of C-12=1 amu New mass of `(1)/(24)` th of mass single atom fo C-12 =1 amuA. Atomic mass of elements will change.B. Mass of an atom (in gm) of an elements remains same on both scale.C. Mass of an atom of Fe will be 112 amu on new scale.D. Atomic mass of Fe on new scale will be 112. |
|
Answer» Correct Answer - A::B::D |
|
| 776. |
Select the correct statement(s)A. Ratio of gm/litres and % (w/v) of a solution is independent fo solute substanceB. Ratio of % (w/v) and molarity of a solution depends on solute substanceC. Ratio of % (w/v) and molarity of a solution depends on solvent substanceD. Ration of % (w/v) and ppm for any solution same |
|
Answer» Correct Answer - A::B |
|
| 777. |
Which of the follewing aqueous solutions of `H_(2)SO_(4)` has 4.9 g of `H_(2)SO_(4)` ?A. 500 mL of 0.1 M `H_(2)SO_(4) (d=1.5 g mL^(-1))`B. 250 mL solution of density 2 g `mL^(-1)` which is 49% (w/w)C. 10 g solution which is 49% (w/w)D. Solution having 500g water with molality 0.1 mol `kg^(-1)` |
|
Answer» Correct Answer - A::C::D |
|
| 778. |
40 ml of 22.7 V `H_(2)O_(2)` solution is mixed with 60 ml of 8.5% (w//v) `H_(2)O_(2)` solution and the mixture is diluted to 230 ml. If 20 ml of diluted solution is callected in an empty beaker, the molarity of collected solution is: |
|
Answer» Correct Answer - 1 |
|
| 779. |
White `P` reacts with caustic soda, the products are `PH_(3)` and `NaH_(2)PO_(2)`. This reaction is an example of:A. OxidationB. ReductionC. DisproportionationD. Neutralisation |
|
Answer» Correct Answer - 3 `overset((0))(P_(4))+6NaOHrarr overset((-3))(PH_(3))+3Na_(2)overset((+1))(HPO_(2))` Disproportionation reaction. In this reaction, `P` element present in intermediate oxidation state and `P` undergoes both oxidation and reduction. |
|
| 780. |
100 mL of `PH_(3)` on decomposition produced phosphorus and hydrogen. The change in volume is |
| Answer» Correct Answer - 1 | |
| 781. |
A sugar syrup of weight `214.2 g` contains `34.2 g` of sugar `(C_(12) H_(22) O_(11))`. Calculate a. the molal concentration. b. the mole fraction of the sugar in the syrup. |
|
Answer» Correct Answer - A::B Weight of solvent `= 214. 2 - 34.2 = 180 g` `m = (W_(2) xx 1000)/(Mw_(2) xx W_(1)) = (34.2 xx 1000)/(342 xx 180) = 0.555` b. `chi_(2) = (n_(1))/(n_(1) + n_(2)) = (34.2//342)/((180)/(18) + (34.2)/(342)) = 0.0099` |
|
| 782. |
Given the number : 161 cm, 0.161 cm and 0.0161 cm. The no. of significant figures for the three numbers areA. 3,4,5B. 3,3,3C. 3,3,4D. 3,4,4 |
| Answer» Correct Answer - B | |
| 783. |
Which of the following has maximum number of significant figures ? (i) 0.00453 (ii) 4.8046 (iii) 5.643 |
|
Answer» Correct Answer - 4.8046 (Five) The number 4.8046 (ii) has maximum numberof significant figures i.e., Five. |
|
| 784. |
Add : (i) `93.4 + 4.03` (ii) `77.86, 14.12 and 5.7` |
|
Answer» Correct Answer - (i) 97.4 (ii) 97.7 (i) Add : `{:(93.4),(4.03),(bar(97.43)):}` The correct answer is 97.4 (Digit 3 is to be deleted) (ii) Add : `{:(77.86),(14.12),(" 5.7"),(bar(97.68)):}` The correct answer is 97.7 (Digit 6 has been rounded off to 7) |
|
| 785. |
Answer the following questions: a. Aniline does not undergo Friedel Crafts reactions.b. Although -NH2 is an ortho-para directing group, nitration of aniline produces 47% meta substituted product. c. Aniline cannot be prepared by Gabriel pthalimide synthesis. |
|
Answer» a) Aniline does not undergo Friedel Craft acylation or alkylation. This is due to salt formation between base i.e., aniline and Lewis acid i.e., anhydrous AlCl3 used as octalyst in the reaction. Due to this, nitrogen of aniline acquires positive charge and thus acts as a strong deactivating group. c) Gabriel's synthesis can not prepare aniline. Using this process, aniline can not be prepared, because aryl halides do not undergo nucleophilic substitution with the phthalimide-formed anion. |
|
| 786. |
A gas has a density of `1.25gL^(-1)` at 1 atm and 273 K. Identify it:A. `NO_(2)`B. `O_(2)`C. `N_(2)`D. `SO_(2)` |
|
Answer» Correct Answer - C |
|
| 787. |
Which of the following has same mass?A. 1.0 mole of `O_(2)`B. `3.01 xx 10^(23)` molecular of `SO_(2)`C. 0.5 moles of `CO_(2)`D. 1 g atom of sulphur |
|
Answer» Correct Answer - A::B::D |
|
| 788. |
The density of air of 0.001293 `g//cm^(3)` at 1 atm and 273 K. Identify which of the following statement is correct?A. Vapour density is 14.48.B. Molecular weight is 28.96.C. Vapour density is 0.001293 `g//cm^(3)`D. Vapour density and molecular weight cannot be determined. |
|
Answer» Correct Answer - A::B |
|
| 789. |
Calculate the number of `Na^(+)` ion present in 710 mg of `Na_(2)SO_(4)` in aqueous solution `(N_(A)=6xx10^(23))` If your answer is `x xx 10^(y)` (in scientific notation) then fill `x` in OMR, where `x` is single digit number |
|
Answer» Correct Answer - 6 |
|
| 790. |
13.5 gmof aluminium when changes to `Al^(+3)` ion in solution, will lose: `[Ai = 27, N_(A) = 6 xx 10^(23)]`A. `18.0 xx 10^(23)` electronsB. `6.0 xx 10^(23)` electronsC. `3.0 xx 10^(23)` electronsD. `9.0 xx 10^(23)` electrons |
|
Answer» Correct Answer - D |
|
| 791. |
A compound contains 7 carbon atoms, 2 oxygen atoms and `1.0 xx 10^(-23)` gmof other elements. The molecular mass of compound is : `(N_(A) = 6 xx 10^(23))`A. 122B. 116C. 148D. 154 |
|
Answer» Correct Answer - A |
|
| 792. |
20 gmof a mixture of NaCl and NaOH exactly requires 7.3gm HCL for complete reaction. The mass percent of NaCl in the original mixture is:A. 0.4B. 0.6C. 0.5D. 0.8 |
|
Answer» Correct Answer - B |
|
| 793. |
400 ml of `0.2` M-HCl is mixed with 600 ml of `0.1` M-NaOH solution. The maximum mass of NaCl fromed is :A. `4.68` gmB. `2.34` gmC. `7.02` gmD. `3.51` gm |
|
Answer» Correct Answer - D |
|
| 794. |
The number of hydrogen atoms in 0.9 gm glucose, `C_(6)H_(12)O_(6)` is same as:A. 0.048 gm hydrazine, `N_(2)H_(4)`B. 0.17gm ammonia, `NH_(3)`C. 0.30gm ethane, `C_(2)H_(6)`D. 0.03 gm hydrogen, `H_(2)` |
|
Answer» Correct Answer - C |
|
| 795. |
90 gm glucose is dissolved in 410 gm water to get a solution. The concentration of solution isA. `(900)/(41)%`B. `1.8%`(w/w)C. `(50)/(41)`D. `1.0` m |
|
Answer» Correct Answer - C |
|
| 796. |
100 gm water is saturated with glucose to form a solution of density X gm//ml and contains 50% gulcose, by mass. If the volume of solution formed is 100 ml, the value of X is: |
|
Answer» Correct Answer - 2 |
|
| 797. |
A mixture of sodium chloride(NaCl) and anhydrous sodium carbonate `(Na_(2)CO_(3))` has a mole ratio, `2:1`. It is dissolved in water and treated with `BaCl_(2)` solution. The mass of `BaCO_(3)` precipitated is 197 gm. Calculate the mass (in gm) of NaCl in the mixture. |
|
Answer» Correct Answer - 117 |
|
| 798. |
In order to remove `Mg^(2+)` and `Ca^(+2)` from `H_(2)O`, impure water is treated with sodium tripolyphosephate `Na_(5)P_(3)O_(10)+Mg^(+2) rarr Na_(3)MgP_(3)O_(10)+2Na^(+)` `Na_(3)MgP_(3)O_(10)+Ca^(+2)rarr NaCaMgP_(3)O_(10)+2Na^(+)` Select the correc statement about treatment of 10 L `H_(2)O` having 48 ppm of `Mg^(+2)` and 40 ppm of `Ca^(+2)`A. In order of remove all `Mg^(+2)` from `H_(2)O` at least 7.36 gm of `Na_(5)P_(3)O_(10)` is requiredB. In order of remove all `Mg^(+2)` from `H_(2)O` at least 3.68 gm of `Na_(5)P_(3)O_(10)` is requiredC. In order to remove all `mg^(+2)` and `Ca^(+2)` ,7.36 gm of `Na_(5)P_(3)O_(10)` is requiredD. In order to remove all `Mg^(+2)` and `Ca^(+2)` , at least 11.04 gm of `Na_(5)P_(3)O_(10)` is required |
|
Answer» Correct Answer - A::C |
|
| 799. |
A chemist has sythesized a greenish yellow gaseous compound of chlorine and oxygen and oxygen and finds that its density is 7.71 g/L at `36^@ C` and 2.88 atm. Then the molcular formula of the compound will be :A. `CIO_3`B. `CIO_2`C. `CIO`D. `CI_2O_2` |
|
Answer» Correct Answer - B |
|
| 800. |
An impure sample of `CaCO_(3)` contains 38 % of Ca. The percentage of impurity present in the sample is : |
|
Answer» Correct Answer - 5 |
|