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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.
| 801. |
Boron occurs in nature in the form of two isotopes having atomic masses 10 and 11. What are the percentage abundances of these isotopes in a simple a boron having average atomic mass of 10.8 are :A. 20,80B. 80,20C. 50,50D. 40,60 |
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Answer» Correct Answer - A Let the percentage abundance of B-10 isotope=x `:.` The percentage abundance of B-11 isotope = (100-x) From the available data : `(x xx10)/(100)+((100-x)xx11)/(100)=10.8` or `10x + 1100 - 11x = 1080or x = 20` Percentage abundance of B-10 isotope = 20 % Percentage abundance of B-11 isotope = 80 %. |
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| 802. |
A chemist decided to determine the molecular formula of an unknown compound. He collects following informations: (P) Compound contains 2:1 H to O atom (number fo atoms). (Q) Compound has 40% C by mass (R) Approximate molecular mass of the compound is 178 g. (S) Compound contains C,H and O only. What is the empirical formula of the compound?A. `CH_(3)O`B. `CH_(2)O`C. `C_(2)H_(2)O`D. `CH_(3)O_(2)` |
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Answer» Correct Answer - B |
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| 803. |
A chemist decided to determine the molecular formula of an unknown compound. He collects following informations: (P) Compound contains 2:1 H to O atom (number fo atoms). (Q) Compound has 40% C by mass (R) Approximate molecular mass of the compound is 178 g. (S) Compound contains C,H and O only. What is the % by mass of oxygen in the compound?A. `53.33%`B. `88.88%`C. `33.33%`D. None of these |
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Answer» Correct Answer - A |
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| 804. |
An impure sample of `CaCO_(3)` contains 38% of Ca. The percentage of impurity present in the sample is :A. `5%`B. `95%`C. `10%`D. `2.5%` |
| Answer» Correct Answer - A | |
| 805. |
Chlorine has two isotopes of atomic mass units `34.97 u and 36.97 u`. The relative abundances of these two isotopes are `0.735 and 0.245` respectively. Find out the average atomic mass of chlorine. |
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Answer» Atomic mass of first isotope `= 34.97 u` Relative abundance of first isotope `= 0.735` Atomic mass of second isotope `= 36.97 u` Relative abundance of second isotope `= 0.245` Average atomic mass of chlorine `= ((34.97u)xx0.735+(36.97u)xx0.245)/(0.735+0.245)` `=((25.703+9.058)u)/(0.980)=35.47u`. |
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| 806. |
Assume isotope of chlorine present on the unknown planet are `.^(34) Cl and .^(38)Cl`. If average molecular weight of `Cl`is found to be 35, what is the sum of moles of proton and neutron in 7 gm sample of chlorine ? |
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Answer» Correct Answer - 7 |
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| 807. |
On analysing an impure sample of sodium chloride, the percentage of chlorine was found to be 45.5. What is the percentage of pure sodium chloride in the sample ? |
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Answer» Molecular mass of pure NaCl = Atomic mass of Na + Atomic mass of Cl `= 23 + 35.5 = 58.5` u Percentage of chlorine in pure NaCl `= ((35.5 "u"))/((58.5 "u"))xx100=60.6` Now, if chlorine is 60.6 parts, NaCl = 100 parts if chlorine is 45.5 parts, NaCl `= ((45.5 "u"))/((60.6 "u")) xx100 = 75` Thus, percentage of pure NaCl = 75 %. |
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| 808. |
The average atomic mass of chlorine is 35.5 u. Do we come across a sample of the element with atomic mass 35.5 u ? |
| Answer» No, it is not possible. The fractional atomic mass of an element is its average mass and not the actual mass. In the case, the element chlorine exists as two isotopes with atomic mass 35 u and 37 u respectively in the ratio 3 : 1. The average come out to be fractional i.e., 35.5 u. | |
| 809. |
Consider the following statements: (1) If all the reactants are not taken in their stoichiometric ratio, then at least one reactant will be left behind. 2 moles of `H_(2)(g)` and 3 moles of `O_(2)(g)` produce 2 moles of water. (3) Equal wt. of carbon and oxygen are taken to produce `CO_(2)` then `O_(2)` is limiting reagent. (Assume 100% yield in all cases) The above statements (1), (2), (3) respectively are (T = True, F=False):A. TTTB. FTFC. FFFD. TFT |
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Answer» Correct Answer - A |
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| 810. |
The molar ration of `Fe^(++)` to `Fe^(+++)` in a mixture of `FeSO_(4)` and `Fe_(2)(SO_(4))_(3)` having equal number of sulphate ions in both ferrous and ferric sulphate is:A. `1:2`B. `3:2`C. `2:3`D. `25.67` |
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Answer» Correct Answer - B |
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| 811. |
what volume of `CO_2` will be liberated at STP if 12 g of carbon is burnt in excess of oxygen ?A. 11.35 LB. 22.7 LC. 2.27 LD. 1.135 L |
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Answer» Correct Answer - B |
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| 812. |
The density of `0.06 M` solution of `Kl` in water is `1.006 g mL^(-1)`. Determine the molality of this solution `(K = 39, I = 127 "amu")` |
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Answer» `Mw of Kl = 39 + 127 = 166 g` `d = M ((Mw_(2))/(1000) + (1)/(m))` `1.006 = 0.006 ((166)/(1000) + (1)/(m))` Solve for `m`, `m = 0.06024` Alternate method: `M = (% "by Weight" xx 10 xx d)/(Mw_(2))` `0.06 = (%" "by Weight" xx 10 xx 1.006)/(166)` `:. "by weight" = (166 xx 0.06)/(10 xx 1.066) = 0.99%` Weight of solute `= 0.99 g`. Weight of solution `= 100 g` Weight of solvent `= 100 - 0.99 = 99.01` `m = (w_(2) xx 1000)/(Mw_(2) xx w_(1)) = (0.99 xx 1000)/(166 xx 99) = 0.06024` |
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| 813. |
Sulphur trioxide, `SO_(3)` is made by oxidizing sulphur dioxide, `SO_(2)`, according to the equation, `2SO_(2) + O_(2) rightarrow 2SO_(3)`. If a 16.0 g sample of `SO_(2)` yields 18.0g of `SO_(3)`. What is the percent yield?A. 0.7B. 0.8C. 0.9D. 1 |
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Answer» Correct Answer - C |
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| 814. |
An aqueous fo diabasic acid (molecular mass = 118) containing `35.4 g` of acid per litre of the solution has density `1.0077 g mL^(-1)`. Express the concentration in as many ways as you can? |
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Answer» Correct Answer - A::B::C::D `M = (35.4 xx 1000)/(118 xx 1000) = 0.3` `m = (W_(2) xx 1000)/(Mw_(2) xx (V_(sol) xx d_(sol) - W_(2)))` `= (35.4 xx 1000)/(118 (1000 xx 1.0077 - 35.4))` `= (35.4 xx 1000)/(118 xx 972.3) = 0.31` `N = n xx M = 2 xx 0.3 = 0.6` Weight of solvent `(W_(1))` = Weight of solution - Weight of solute `= V_(sol) xx d_(sol) - W_(2)` `= 1000 xx 1.0077 = 35.4 = 972.3` `X_(2) = (n_(2))/(n_(1) + n_(2)) = (W_(2) // Mw_(2))/((W_(1))/(Mw_(1)) + (W_(2))/(Mw_(2)))` `= (35.4 // 118)/((972.3)/(18) + (35.4)/(118))` `= (0.3)/(54 + 0.3) = 0.055` `X_(1) = 1 - X_(2) = 1 - 0.0055 = 0.9945` |
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| 815. |
An iodized salt contains 0.5% of NaI. A person consumes 3 gm of salt everyday. The number of iodide ions going into his body everyday is (I = 127 )A. `10^(-4)`B. `6.02xx10^(-4)`C. `6.02xx 10^(19)`D. `6.02xx 10^(23)` |
| Answer» Correct Answer - C | |
| 816. |
Statement-1 : For the reaction producing Fe and `CO_2` by the raction of `Fe_(2)O_(3)` and C the ratio of stoichiometric coeffecients of `Fe_2O_3`: "Fe is" `1:2`. Statement-2 : During a chemical reaction atoms can neither be created nor be destroyed.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. |
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Answer» Correct Answer - A |
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| 817. |
For the following reaction if equal mass of A and B are taken : A+2B`to` C Which of the following is/are correct ? (`M_A and M_B` are molar mass of A and B respectively )A. If `M_A=2M_B` , then none of the reactant will be left.B. if `M_B lt M_A/2`, then A will be limiting reagentC. If `M_A=M_B` , then A will be limiting reagentD. If `M_B lt M_A/2` , then A will be limiting reagent. |
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Answer» Correct Answer - A::B |
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| 818. |
The percentage by mass of C, H, and Cl in a compound are C 52.2%, H 3.7% and Cl 44.1%. How many carbon atoms are in the simplest formular of the compound?A. 3B. 4C. 6D. 7 |
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Answer» Correct Answer - D |
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| 819. |
A compound with 69.41% C, 4.16% H and 26.42%O has a molar mass of 23--250g `mol^(-1)`. What is its molecualr formula?A. `C_(13)H_(9)O_(4)`B. `C_(14)H_(10)O_(4)`C. `C_(13)H_(6)O_(4)`D. `C_(15)H_(14)O_(3)` |
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Answer» Correct Answer - B |
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| 820. |
When burnt in air, a 12.0 g mixture of carbon and sulphur yields a mixture of `CO_2` and `SO_2` , in which the number of moles of `SO_2` is half that of `CO_2`.The mass of the carbon the mixture contains is : (At . Wt. S=32)A. 4.08 gB. 5.14 gC. 8.74 gD. 1.54 g |
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Answer» Correct Answer - B |
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| 821. |
The mass percentage of O in a potassium salt, `K_(2)S_(2)O_(x)` , is 36.0%. What is the formular of the polyatomic ion?A. `S_(2)O_(3)^(2-)`B. `S_(2)O_(5)^(2-)`C. `S_(2)O_(7)^(2-)`D. `S_(2)O_(8)^(2-)` |
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Answer» Correct Answer - B |
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| 822. |
For which compound are the empirical and molecualr formular the same?A. `C_(6)H_(5)COOH`B. `C_(6)H_(4)(COOH)_(2)`C. HOOCCOOHD. `CH_(3)COOH` |
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Answer» Correct Answer - A |
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| 823. |
The molecular mass of iodide of tin `(Sn)` is 626.5 amu. What is the empirical formula of the substance? `(I = 127, Sn = 118.5)` |
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Answer» Let the formula iodide of `SN = SnI_(x)` Since the valency of Iodile `= -1` `:. SnI_(x) = 118.5 + 127 xx x = 625.5` Solve for `x`: `:.x = 4` Formula `= SnI_(4)` |
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| 824. |
An aqueous solution of NaOH has a molarity of `0.05` M. Select the option having incorrect representation of concentration of solution. `[d_("solution")=1.002gm//ml]`A. `%w//v=0.2%`B. `X_(NaOH)`(Mole fraction)=`(0.05)/(0.05+55056)`C. `%w//w=(2)/(10.02)%`D. `"ppm"=5"ppm NaOH`"` solution |
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Answer» Correct Answer - D |
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| 825. |
A metal oxide has the formular `M_(2)O_(3)`. It can be reduced by hydrogen to give free metal and water 0.1596g of the metal oxide required 6 mg hydrogen for complete reduction. The atomic weight of the metal is:A. 27.9B. 159.6C. 0.486gmD. 55.80gm |
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Answer» Correct Answer - D |
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| 826. |
An 18.5 g sample of tin (M = 118.7) combines with 10.0g of sulphur (M=32.07) to form a compound. What is the empirical formular of this compound?A. SnSB. `SnS_(2)`C. `Sn_(2)S`D. `Sn_(2)S_(3)` |
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Answer» Correct Answer - B |
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| 827. |
Molarity and molality of pure `CH_(3)COOH` are respectively : `(d_(CH_(3)COOH)=1.5g//ml)`A. `16.67,25`B. `25, 16.67`C. `50, 33.3`D. 25 both |
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Answer» Correct Answer - B |
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| 828. |
A mineral water sample was analysed and found to contain `1xx10^(-3)%` ammonia (w/w). The mole of dissolved ammonia gas in one litre water bottle is `(d_("water")~~1gm//ml)` :A. `5.8xx10^(-4)mol`B. `1xx10^(-2)mol`C. `0.58xx10^(-2)mol`D. same as w/w |
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Answer» Correct Answer - A |
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| 829. |
An oxide of element A was analysed and found to have mass ratio of A to oxygen equal to `7:3`. Then formular of oxide can be : [Atomic mass of A = 56]A. `A_(2)O_(2)`B. `A_(2)O_(3)`C. AOD. `A_(2)O` |
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Answer» Correct Answer - B |
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| 830. |
Disilane, `Si_(2)H_(X)` is analysed and found to contain 90.32% silicon by mass. What is the value of X? [Si = 28]A. 3B. 4C. 6D. 8 |
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Answer» Correct Answer - C |
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| 831. |
A pure gas that is `14.3%` hydrogen and `85.7%` carbon by mass has a density of `2.5g L^(-1)` at `0^(@)C` and 1 atm pressure. What is the molecular formula of the gas :A. `CH_(2)`B. `C_(2)H_(4)`C. `C_(4)H_(8)`D. `C_(6)H_(12)` |
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Answer» Correct Answer - C `% "mol Simples ratio"` `C" " 85.7 " "85.7//12 = 7.14" "7.14// 7.14 = 1 " "1` `H " "14.3" " 14.3//1 = 14.3" " 14.3//7.14 = 2 " "2` `:.` Emperical formal `=CH_(2)` `:. PMw = DRT` `Mw = (DRT)/(P) = (2.5 xx 0821 xx 273)/(1) =56` `n = ("molecular wt")/("Ewt") = (56)/(14) = 4` Molecular formula `= n xx E.F` `= 4 xx CH_(2)` `= C_(4) H_(8)` . |
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| 832. |
Calculate the weight of lime (CaO) obtained by heating 200 kg of 95 % pure lime stone `(CaCO_(3))` |
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Answer» Step I : Calculate of the weight of pure lime stone 100 kg of lime stone `(CaCO_(3))` contain pure substance = 95 kg 200 kg of lime stone `(CaCO_(3))` contain pure substance `= (95)/(100) xx 200 = 190` kg Step II. Calculation of the weight of lime (CaO) formed. The chemical equation for the reaction is : `{:(CaCO_(3),overset(Heat)(rarr),CaO+CO_(2)),(40+12+48,,40+16),(=100g(=100kg),,=56g(=56 kg)):}` 100 kg of lime stone form lime = 56 kg 1.6 kg of lime stone forms lime `= (56)/(100)` kg 190 kg of lime stone form lime `= (56)/(100)xx190` kg = 106.4 kg. |
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| 833. |
A sample of protein was analysed for metal content and analysis revealed that it contained magnesium and titanium in equal amounts, by mass. If these are the only metallic species present in the protein and it contains 0.008% metal by mass, the minimum possible molar mass of the protein is : [Mg = 24, Ti = 48]A. `1.2 xx 10^(22)`B. `1.2 xx 10^(25)`C. `7.2 xx 10^(21)`D. `1.08 xx 10^(22)` |
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Answer» Correct Answer - D |
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| 834. |
Analysis of a compound known to contain only Mg, P, and O gives this analysis. 21.8% Mg, 27.7% P , 50.3% O by mass What is its empirical formula?A. `MgPO_(2)`B. `MgPO_(3)`C. `Mg_(2)P_(2)O_(7)`D. `Mg_(3)P_(2)O_(8)` |
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Answer» Correct Answer - C |
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| 835. |
How many kg of `CaCo_(3)` [Mol. Wt = 100g `mol e^(-1)]` is needed to produce 336 kg of CaO [Mol wt = 56g `mol^(-1)]` if % yield of the reaction given is 60%. `CaCO_(3)(s) rightarrow CaO(s) + CO_(2)(g)`A. `10^(3)`B. `10^(2)`C. 900D. 800 |
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Answer» Correct Answer - A |
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| 836. |
1.44 gram of Titanium (Ti) reacted with excess of `O_(2)` and produced x gram of non-stoichiometric compound Ti `._(0.44)O`. The value of x will be :[Ti = 48]A. 2.77gB. 3.77gC. 1.77gD. 3.0g |
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Answer» Correct Answer - C |
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| 837. |
What will be the percentage loss in mass when `NaHCO_(3)` is heated at `300^(@)C`?A. 0.6B. 0.455C. 0.369D. 0.7 |
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Answer» Correct Answer - C |
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| 838. |
A sample of a mixture of `Na_(2)CO_(3) and NaHCO_(3)` is subjected to heating till there is no further loss in weight. Assuming, that the loss in weight of the sample is `22%` of the initial weight of the mixture due to the evolution of `CO_(2)`, find out the relative percentages of the two components in the mixture. |
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Answer» (i) action of heating on the compounds `Na_(2)CO_(3) and NaHCO_(3)`. (ii) determination of the mass of `Na_(2)CO_(3) and NaHCO_(3)` from the mass of `CO_(2)` produced (iii) `84% NaHCO_(3)` `16% Na_(2)CO_(3)` |
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| 839. |
In a cartain operation 358 g of `TiCl_(4)` is reacted with 96 g of Mg. Calculate % yield of Ti if 32 g of Ti is actually obtained [At. Wt. Ti=48, Mg=24][Hint: `358/190=1.88`]A. 0.3546B. 0.666C. 1D. 0.6 |
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Answer» Correct Answer - A |
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| 840. |
In the formation reaction of `NH_(3)` from `N_(2)` and `H_(2)` 140g of `N_(2)` and 40g `H_(2)` were mixed. Select the option which is correct.A. Maximum mass of `NH_(3)` which can be formed is 180gmB. If % yield of reaction is 80%, then `H_(2)` consumed will be 32gm.C. Some `N_(2)(g)` will be left after the reaction.D. If `NH_(3)` formed is 85gm then % yield will be 50%. |
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Answer» Correct Answer - D |
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| 841. |
Statement-1 :`3.4g` of `NH_(3)(g)` on complete decomposition into `N_2` and `H_2(g)` produces 0.6g of `H_2(g)`. Statement- 2 : Law of conservartion of mass is followed by the chemical reaction.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 True, Statement-2 : is False.D. Statement-1 : is False, Statement-2 : is True. |
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Answer» Correct Answer - A |
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| 842. |
One mole mixture of `CH_4` and air (containing 80% `N_2` 20 % `O_2` by volume ) of a composition such that when underwent combustion gave maximum heat (assume combustion of only `CH_4`).Then which of the statements are correct , regarding composition of initial mixture ? (X presents mole fraction )A. `X_(CH_4)=1/11, X_(O_2)=2/11,X_(N_2)=8/11`B. `X_(CH_4)=3/8, X_(O_2)=1/8,X_(N_2)=1/2`C. `X_(CH_4)=1/6, X_(O_2)=1/6,X_(N_2)=2/3`D. Data insufficient |
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Answer» Correct Answer - A |
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| 843. |
A sample of `NaHCO_(3)(s)` on heating undergoes `1.845 gm` loss of mass. Approximate mass of `NaHCO_(3)` (in nearest integer) in gm is : |
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Answer» Correct Answer - 5 |
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| 844. |
Assume 0.10 L of `N_2` and 0.18 L of `H_2`, both at 50 atm and `450^@C`, are reacted to form `NH_3` Assuming the reation goes to completion , identify the reagent that is in excess and determine the volume of that remains at the same temperature and pressure.A. `H_2` 0.02 LB. `H_2` 0.08 LC. `N_2` 0.01 LD. `N_2` 0.04 L |
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Answer» Correct Answer - D |
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| 845. |
Ethanol burns in excess oxygen to form `CO_2(g)` and `H_2O(g)` according to this balanced equation. `C_2H_5OH(g)+3O_2(g) to2CO_2(g)+3H_2O(g)` What value is closest to the volume of `CO_2(g)` , measured at 200 K and 1 atm produced from the combustion of 0.25 mol of `C_2H_5OH(g)`?A. 5 LB. 8 LC. 10 LD. 15 L |
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Answer» Correct Answer - B |
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| 846. |
Relationship between volume strength of `H_2O_2` aqueous solution and molarity depeds on pressre and temperature of `O_2` (g) collected Example : `"At1 atm ,273K," "Volume strength =11.2xxM"` `"At STP, " "Volume strength =11.35 x M"`[Note : By default STP is taken for volume strength and calculationl. Find the relationship between volume strength and molarity at 300 K and 1 atm.A. `12.308xx`MolarityB. `24.63xx`MolarityC. `11.35xx`MolarityD. `22.7xx`Molarity |
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Answer» Correct Answer - A |
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| 847. |
A 12 gm sample of `CH_4` and `C_(2)H_(4)` on complete oxidation with ` O_2` forms 35.2gm`CO_2` . Find the mean molar mass of original sample:A. 20B. 22C. 14.7D. 23 |
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Answer» Correct Answer - A |
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| 848. |
Calculate the amount of `ZnO` produced (in gm) when 195 gm of ZnS reacts with `89.6 L O_(2)` at 1 atm and 274 K. Write nearest integral value `ZnS+O_(2)overset(2)(rarr) ZnO+SO_(2)` |
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Answer» Correct Answer - 163 |
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| 849. |
when a mixture of aluminium powder and iron (III) oxide is ignited, it produces molten iron and aluminium oxide. In an experiment, 5.4gm of aluminium was mixed with 18.5 gm of iron oxide. At the end of the reaction, the mixture contained 11.2 gm of iron, 10.2 gm of aluminium oxide and an undetermined amount of unreacted iron (III) oxide. No aluminium was left. What is the mass of the iron (III) oxide left?A. 2.5 gmB. 7.3gmC. 8.3gmD. 2.9gm |
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Answer» Correct Answer - A |
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| 850. |
`64g` of an organic compound contains `24 g ` of carbon, `8gm` of hydrogen and the rest oxygen. The empirical formula of the compound isA. `CH_(4)O`B. `CH_(2)O`C. `C_(2)H_(4)O`D. none |
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Answer» Correct Answer - A |
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