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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.
| 401. |
Which of the following pairs have the same number of atoms ?A. 16 g of `O_(2)(g)` and 4 g of `H_(2)(g)`B. 16 g of `O_(2)` and 44 g of `CO_(2)`C. 28 g of `N_(2)` and 32 g of `O_(2)`D. 12 g of C(s) and 23 g of Na(s) |
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Answer» Correct Answer - C::D 28g of `N_(2)` and 32 g `O_(2)` both represent 1 mole and have same no. of atoms. `= 2 xx N_(A)` 12 g of C(s) and 23 g of Na (s)k represent 1 gram atom and have same no. of atoms `(N_(A))` |
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| 402. |
`10 L` of hard water requires `0.28 g` of line `(CaO)` for removing hardness. Calculate the temporary hardness in ppm of `CaCO_(3)`. |
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Answer» Temporary hardness is due to `HCO_(3)^(ɵ)` of `Ca^(2+)` and `Mg^(2+)` `Ca(HCO_(3))_(2) + underset(56 g)(CaO) rarr underset(2 xx 100 g)(2CaCO_(3) + H_(2)O)` `56 g CaO = 200 g CaCO_(3)` in `10 L of H_(2) O` `0.28 g CaO = (200 xx 0.28)/(56)` `1 g CaCO` in `10 L of H_(2)O` `= 1 g CaCO_(3)` in `10 xx 1000 mL of H_(2) O` `= 1 g CaCO_(3)` in `10^(4) mL of H_(2) O` `100 g CaCO_(3)` in `10^(6) mL of H_(2) O = 100 ppm` Hence, temporary hardness of `CaCO_(3) = 100 ppm` |
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| 403. |
`10 L` of hard water required `0.56 g` of lime `(CaO)` for removing hardness. Hence, temporary hardness in ppm (part per million `10^(6)`) of `CaCO_(3)` is:A. 100B. 200C. 10D. 20 |
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Answer» Correct Answer - B Temporary hardness is due to `HCO_(3)^(ɵ)` of `Ca^(2+)` and `Mg^(2+)` `Ca (HCO_(3))_(2) + (CaO)/(56 g) rarr (2 CaCO_(3))/((2 xx 100) g) + H_(2) O` `0.56 g CaO -= 2 g CaCO_(3) "in" 10 L H_(2) O` `= 2 g CaCO_(3) "in" 10^(4) mL H_(2) O` `= 200 g CaCO_(3) "in" 10^(6) mL H_(2) O` |
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| 404. |
Accoridng to EPA guidelines the permissible level for lead in drinking water is 15 parts per billion (ppb). What is the maximum allowable mass of lead that could be present in 1.00L of `H_2`?A. 0.015 ngB. ` 0.015 mug`C. 0.015mgD. 0.015 g |
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Answer» Correct Answer - C |
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| 405. |
Four one litre flaske are separately filled with the gases `H_(2), He, O_(2) "and" O_(3)` at the same temperature and pressure. The ratio of total number of atoms of these gases present in different flask would be:A. `1:1:1:1`B. `1:2:2:3`C. `2:1:2:3`D. `3:2:2:1` |
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Answer» Correct Answer - C `{:(,H_(2),:,He,:,O_(2),:,O_(3),),("no.of atoms",2N_(A),:,1N_(A),:,2N_(A),:,2N_(A),):}` `{:(=2,:,1,:,2,:,3):}` |
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| 406. |
Are the atomic masses of the elements their actual masses ? |
| Answer» No, atomic masses of the elements are not actual masses. These are only relative masses because the actual masses are very small. | |
| 407. |
Select the correct statement(s) for `(NH_(4))_(3)PO_(4)`A. Ratio of number of oxygen atoms to number of hydrogen atoms is 1:3B. Ratio of number of cations to number of anions is 3:1C. Ratio of number of gm-atoms of nitrogen to gm-atoms of oxygen is 3:2D. Total number of atoms in one mole of `(MH_(4))_(3)PO_(4)` is 20. |
| Answer» Correct Answer - A::B | |
| 408. |
Matching list type: A. `{:(P,Q,R,S),(3,4,1,2):}`B. `{:(P,Q,R,S),(2,4,1,3):}`C. `{:(P,Q,R,S),(4,3,1,2):}`D. `{:(P,Q,R,S),(2,3,1,4):}` |
| Answer» Correct Answer - A | |
| 409. |
NaBr , used to produce AgBr for use in photography can be self prepared as follows: `Fe+Br_(2) to FeBr_(2) " "…(i) ` `FeBr_(2) + Br_(2) to Fe_(3) Br_(8)" "…(ii) `(not balanced) `Fe_(3)Br_(8) + Na_(2)CO_(3) to NaBr+CO_(2) + Fe_(3)O_(4)" "...(iii)" "` (not balanced) (At mass: Fe=56, Br=80) Mass of iron required to produce `2.06xx10^(3)` kg NaBrA. 420 gmB. 420 kgC. `4.2 xx 10^(5)` kgD. `4.2 xx 10^(8)` gm |
| Answer» Correct Answer - B | |
| 410. |
NaBr , used to produce AgBr for use in photography can be self prepared as follows: `Fe+Br_(2) to FeBr_(2) " "…(i) ` `FeBr_(2) + Br_(2) to Fe_(3) Br_(8)" "…(ii) `(not balanced) `Fe_(3)Br_(8) + Na_(2)CO_(3) to NaBr+CO_(2) + Fe_(3)O_(4)" "...(iii)" "` (not balanced) (At mass: Fe=56, Br=80) If yield of (iii) reaction is 90% then mole of `CO_(2)` formed when `2.06xx10^(3)` gm NaBr is formedA. 20B. 10C. 9D. 440 |
| Answer» Correct Answer - B | |
| 411. |
NaBr , used to produce AgBr for use in photography can be self prepared as follows: `Fe+Br_(2) to FeBr_(2) " "…(i) ` `FeBr_(2) + Br_(2) to Fe_(3) Br_(8)" "…(ii) `(not balanced) `Fe_(3)Br_(8) + Na_(2)CO_(3) to NaBr+CO_(2) + Fe_(3)O_(4)" "...(iii)" "` (not balanced) (At mass: Fe=56, Br=80) If the yield of (ii) is 60% & reaction is 70% then mass of iron required to produce `2.06xx10^(3)` kg NaBrA. `10^(5)` kgB. `10^(5)` gmC. `10^(3)` kgD. None |
| Answer» Correct Answer - C | |
| 412. |
Calculate the speed of a train `ms^(-1)` which covers 972 miles in 27 hours (1 mile = 1.60 km) |
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Answer» Correct Answer - `16.0 ms^(-1)` `("972 miles")/("27 hrs")=(("972 miles")xx("conversion factor"))/(("27 hrs")xx("conversion factor"))` `= (("972 miles")xx(1.6xx10^(3)m))/(("1 mile"))xx(1)/(("27 hrs"))xx(("1 hr"))/((3.6 xx10^(3)s))=16.0 "m s"^(-1)`. |
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| 413. |
`3 g` of ethane `C_(2)H_(6)` on complete combustion gave `8.8 g` of `CO_(2)` and `5.4 g` of water. Show that the results are in accordance with the law of conservation of mass. |
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Answer» `underset({:(12 xx 2 + 6),(= 30 g):})(C_(2) H_(6)) + underset(underset(= 122g)(32 xx (7)/(2)))((7)/(2) O_(2)) rarr underset({:(44 xx 2),(= 88 g):})(2CO_(2)) + underset({:(18 xx 3),(= 54 g):})(3H_(2) O)` a. `30 g of C_(2) H_(6) = 88 g of CO_(2)` `3 g of C_(2) H_(6) = 8.8 g of CO_(2)` b. `30 g of C_(2) H_(6) = 54 g of H_(2) O` `3 g of C_(2) H_(6) = 5.4 of H_(2) O` c. `30 g of C_(6) H_(6) = 112 g of O_(2)` `3 g of C_(2) H_(6) = 11.2 g of O_(2)` weight of reactant `= 3 + 11.2 = 14.2 g` Weight of product `= 8.8 + 5.4 = 14.2 g` Hence, the law of conversation of mass is verified. Second method: Weight of `C` and `H` contained in `8.8 g`of `CO_(2)` and `5.4 g` of `H_(2) O` are Weight of `C` in `8.8 g of CO_(2) = (12 xx 8.8)/(44) = 2.4 g` Weight of `H` in `H_(2) O = (2 xx 5.4)/(18) = 0.45 g` Total weight of `C` and `H` in the hydrocarbon after combustion `= 2.4 + 0.6 = 3 g` Since the weight of `C` and `H` after combustion is the same as the weight of hydrocarbon before burining, the result are in accordance with the law of conservation of mass. |
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| 414. |
A 4 : 1 molar mixture of `He` and `CH_(4)` is contained in vessel at 20 per pressure. Due to a hole in the vessel the gas mixture leakes out. What is the compostion of mixture effusing out initially.A. `33.3% He, 66.7% CH_(4)`B. `66.7% He, 33.3% CH_(4)`C. `40% He, 60% CH_(4)`D. `60% He, 40% CH_(4)` |
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Answer» Correct Answer - B `(r_(He))/(r_(CH_(4))) = sqrt((M_(CH_(4)))/(M_(He))) xx (P_(He))/(P_(CH_(4)))` `P prop` moles `(r_(He))/(r_(CH_(4))) = sqrt((16)/(4)) xx (4)/(1) = 2 xx 4 = 8.1` Molar composition effusing = 8:1 Weight composition effustiing `= 8 xx 4 : 1 xx 16 = 32.:16` % of `He = (32)/(48) xx 100 = 66.7%` % `CH_(4) = 100 = 66.7 = 33.3%` |
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| 415. |
Uranium is isolated from its ore by dissolving it as `UO_(2)(NO_(3))_(2)` and separating it as solid `UO_(2)(C_(2)O_(4)).xH_(2)O " "A 1.0 g` sample of ore on treatment with nitric acid yielded `1.48 g UO_(2)(NO_(3))_(2)` which on further treatment with `0.4 g Na_(2)C_(2)O_(4)` yielded `1.23 g " "UO_(2)" "(C_(2)O_(4)).xH_(2)O`. Determine weight percentage of uranium in the original sample and `x` : |
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Answer» Correct Answer - `89.4.3` Mass of uranium in the sample `=(1.48)/(394)xx238=0.894g` Mass `%` of uranium in the sample `=89.4` `UO_(2)(NO_(3))_(2)+Na_(2)C_(2)O_(4)+xH_(2)OrarrUO_(2)(C_(2)O_(4))xH_(2)Odarr` `m" "mol 3.756" "2.985" "+2NaNO_(3)` Here `Na_(2)C_(2)O_(4)` is the limiting reagent, therefore, `m` mol of `UO_(2)(C_(2)O_(4)).xH_(2)O` formed is `2.985` `implies M(UO_(2))(C_(2) O_(4)).xH_(2)O=(1.23)/(2.985)xx1000=412` `238+32+88+18x` `implies x =(54)/(18) = 3` |
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| 416. |
Equal moles of `H_(2)O` and a solute (of negligible molar mass) are present in a solution. Hence, molarity of solution is :A. `0.55`B. `55.5`C. `1.00`D. `0.18` |
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Answer» Correct Answer - B |
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| 417. |
Convert 2-6 minutes in seconds |
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Answer» We know that `1 min = 60 s` `:.` Conversion factor `= ((60s))/((1min))` `2.6min=(2.6min)xx"conversion factor"=(2.6 min)xx((60s))/((1min))=156s`. |
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| 418. |
Convert 6 cubic metres into cubic centimetres. |
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Answer» `1m=100cm and 1m^(3)=(100cm)^(3)=10^(6)cm^(3)` `:.` Conversion factor `= ((10^(6)cm^(3)))/((1m^(3)))` `6m^(3)=6m^(3)xx`conversion factor `=(6m^(3))xx((10^(6)cm^(3)))/((1m^(3)))=6xx10^(6)cm^(3)`. |
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| 419. |
When do zeros present in a number become insignificant ? |
| Answer» The zeroes written to the left of the first non-zero digit in a number are insignificant. For example, in the number 0.014, both the zeros are insignigicant. | |
| 420. |
Is velocity a basic or derived quantity according to S.I system ? |
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Answer» Velocity is a derived quantity and it depends upon two basic quantities i.e. distance and time. The `SI` units of velocity are : Velocity = Distance/Time = `ms^(-1)` |
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| 421. |
Why do we regard the gaseous state of water as vapours while that of ammonia as gas ? |
| Answer» Only the gaseous states of those substances are regarded as vapours which are liquid at room temperature. Since water is a liquid, its gaseous state is called vapours. However, the gaseous state of ammonia is called gas because it is not a liquid at room temperature. | |
| 422. |
Which is more precise measurement of the length of a thread (i) 10.0 cm (ii) 10.00 cm ? |
| Answer» The measurement 10.0 cm represents a length between 9.9 cm and 10.1 cm while the measurement 10.00 cm is between 9.99 cm and 10.010 cm. This means that the measurement 10.00 cm is the most precise representation. | |
| 423. |
A student wishes to determine the thickness of a reactangualr piece of aluminuim foil but cannot measure it directly. She can measure its density (d), length (l), mass (m) and width (w). Which relationship will give the thickness?A. `(m)/(d.l.w)`B. `(m.l.w)/(d)`C. `(d.l.w)/(m)`D. `(d.m)/(l.w)` |
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Answer» Correct Answer - A |
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| 424. |
The largest number of molecules is in:A. 34 g of waterB. 28g of `CO_(2)`C. 46 g of `CH_(3)OH`D. 54gm of `N_(2)O_(5)` |
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Answer» Correct Answer - A |
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| 425. |
A hydrocarbon contain `80%` C . The vapour density of compound in 30 . Molecular formula of compound is :-A. `CH_(3)`B. `C_(2)H_(6)`C. `C_(4)H_(12)`D. All of these |
| Answer» Correct Answer - 3 | |
| 426. |
Which of the following solution contains approximately equal hydrogen ion concentrationA. `100 mL of 0.1 M HCl + 50 mL H_(2) O`B. `75 mL of 0.1 M HCl + 75 mL H_(2) O`C. `50 mL of 0.1 M H_(2) SO_(4) + 100 mL H_(2)`D. `100 mL of 0.1 NH_(2)SO_(4) + 50 mL H_(2) O` |
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Answer» Correct Answer - C::D a. `100 xx 0.1 = 10 "mmol" H^(o+)//150 mL` b. `75 xx 0.1 = 7.5 "mmol" H^(o+)//150 mL` c. `50 xx 0.1 xx 2` (`n` factor) `= 10 "mmol"//150 mL` d. `100 xx 0.1 = 100 "mmol"//150 mL` |
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| 427. |
Calculate the number of atoms present in one drop of water having mass 1.8 g .A. `0.3N_(A)`B. `0.2N_(A)`C. `0.1N_(A)`D. `0.01N_(A)` |
| Answer» Correct Answer - 1 | |
| 428. |
1 mol of `CH_(4)` containsA. `6.02xx10^(23)` atoms of HB. 4 g atom of hydrogenC. `1.81xx10^(23)` molecules of `CH_(4)`D. 3.0 g of carbon. |
| Answer» Correct Answer - 2 | |
| 429. |
One mole of `CO_(2)` contains:A. 3 atomB. `18.1xx10^(23)` molecules of `CO_(2)`C. `6.02xx10^(23)`atom of oxygenD. `6.02xx10^(23)` atom of carbon |
| Answer» Correct Answer - 4 | |
| 430. |
Which of the following samples contains the largest number of atoms ? .A. 1g of (Ni) sB. 1g of Ca(s)C. 1g of `N_(2)(g)`D. 1g of B(s) |
| Answer» Correct Answer - D | |
| 431. |
The empirical formula of a compound of molecular mass `120` is `CH_(2)O`. The molecular formula of the compound is :A. `C_(2)H_(4)O_(2)`B. `C_(4)H_(8)O_(4)`C. `C_(3)H_(6)O_(3)`D. all of these |
| Answer» Correct Answer - B | |
| 432. |
Under the same conditions, two gases have the same number of molecules. They mustA. be noble gasesB. have equal volumesC. have a volume of `22.4 dm^(3)` eachD. have an equal number of atoms |
| Answer» Correct Answer - B | |
| 433. |
What weight of a metal of equivalent weight 12 will give `0.475 g` of its chloide?A. `0.12 g`B. `0.24 g`C. `0.36 g`D. `0.48 g` |
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Answer» Correct Answer - A `Ew` of metal chloride `= Ew "of" M + Ew "of" Cl` `= 12 + 35.5 = 47.5` `47.5 g` of metal chlofide `implies` weight of metal `implies 12 g` `0.475 g` of metal chloride `implies (12 xx 0.475)/(47.5)` `implies 0.12 g` |
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| 434. |
Equal masses of `SO_(2)` and `O_(2)` are placed in a flask at STP . Choose the correct statement .A. The number of molecules of `O_(2)` are more than `SO_(2)` .B. Volume occupied at STP is more for `O_(2)` than `SO_(2)`C. The ratio of number of atoms of `SO_(2)` and `O_(2)` is 3:4.D. Moles of `SO_(2)` is greater than the moles of `O_(2)` |
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Answer» Correct Answer - A::B::C |
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| 435. |
Two flask of equal capacity contain `NH_(3)` and `SO_(2)` gases respectively, are kept under similar conditions of temperature and pressure. Select the correct option on the basis of above information.A. More moles are present in flask contain `NH_(3)` .B. Flask of `SO_(2)` has more mass.C. Flask of `NH_(3)` has more number of atoms.D. Both flask contain same number of molecules of `NH_(3)` and `SO_(2)` respectively |
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Answer» Correct Answer - B::C::D |
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| 436. |
The volume of water is required to make `0.20` M solution from 1 mL of `0.5` M solution is :A. 40 mLB. 16 mLC. 50 mLD. 24 mL |
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Answer» Correct Answer - D |
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| 437. |
A 10.0 g sample of an oxide of copper forms metallic copper and `1.26` g of water when heated in a stream of hydrogen. What is the mass percent of copper in this oxide?A. `11.2%`B. `66.6%`C. `79.9%`D. `88.8%` |
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Answer» Correct Answer - D |
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| 438. |
How many ozone molecules are iin 3.20g of `O_(3)`?A. `4.0 xx 10^(22)`B. `6.0 xx 10^(22)`C. `1.2 xx 10^(23)`D. `6.0 xx 10^(23)` |
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Answer» Correct Answer - A |
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| 439. |
A solution of `FeCl_(3)` is `(M)/(30)`. Its molarity for `Cl^(-)` ion will be :A. `(M)/(90)`B. `(M)/(30)`C. `(M)/(10)`D. `(M)/(5)` |
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Answer» Correct Answer - C |
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| 440. |
A student is asked to analyze a water sample from a stream for total solids (TS), dissolved solids (DS) and suspended solids (SS). She carries out the experiments below. (P) A 25-mL portion of the water sample is evaporated to dryness in a pre-weighed evaporating dish to give mass 1. (Q) A seperate 25-mL portion is filtred into a second pre-weighed evaporating dish and evaporated to dryness to give mass 2. How are the values for TS, SS and DS (per 25 mL water) determined?A. TS = mass 1, SS = mass 1 - mass 2, DS = mass 2B. TS = mass 1, SS = mass 2, DS = mass 1 - mass 2C. TS = mass 1 + mass 2, SS = mass 1 , DS = mass 2D. TS = mass 1 + Mass 2, SS = Mass 2, DS = mass 1 |
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Answer» Correct Answer - A |
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| 441. |
Equal masses of oxygen, hydrogen and methane are taken in identical conditions. What is the ratio of the volume of the gases under identical conditions?A. `16:1:8`B. `1:16:2`C. `1:16:8`D. `2:16:1` |
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Answer» Correct Answer - B |
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| 442. |
Consider the ionisation of `H_(2) SO_(4)` as follow" `H_(2) SO_(4) + 2H_(2)P rarr 2H_(3) O^(o+) SO_(4)^(2-)` The total number of ions furnised by `100 mL` of `0.1 M` `H_(2)SO_(4)` will beA. `1.2 xx 10^(23)`B. `0.12 xx 10^(23)`C. `0.18 xx 10^(23)`D. `1.8 xx 10^(23)` |
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Answer» Correct Answer - C 1 mol `H_(2) SO_(4) -= 2 "mol" H_(3)O^(O+) -= 1 "mol" SO_(4)^(2-)` `((100)/(1000) xx 0.1 = 0.01) -= 2 xx 0.01 + 0.01 = 0.03 "mo"` `= 0.18069 xx 10^(23)` |
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| 443. |
0.05 mole of `LiAlH_(4)` in ether solution was placed in a flask containing 74g (1 mole) of t-butyl alcohol. The product `LiAlHC_(12)H_(27)O_(3)` weighed 12.6 g. If Li atoms are conserved, the percentage yield is: [Li = 7, Al = 27, H=1, C=12, O=16]A. 0.25B. 0.75C. 1D. 0.15 |
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Answer» Correct Answer - C |
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| 444. |
The molarity of `Cl^(-)` in an aqueous solution which was `(w//w)2%NaCl,4%CaCl_(2)` and `6%NH_(4)Cl` will be :A. `0.342`B. `0.721`C. `1.12`D. `2.18` |
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Answer» Correct Answer - D |
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| 445. |
How many millimoles of methane, `CH_(4)` are present in 6.4 g of this gas?A. 0.4B. 4C. 40D. `4.0 xx 10^(2)` |
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Answer» Correct Answer - D |
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| 446. |
An alloy of gold and silver contains `38.5%` silver by mass and has a density of `14.6 g.mL^(-1)`. What is the molar concentration of silver in this alloy :A. `52.1 mol.L^(-1)`B. `45.6 mol.L^(-1)`C. `3.57 mol.L^(-1)`D. `2.64 mol.L^(-1)` |
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Answer» Correct Answer - A `38.5%((w)/(w))` Ag i.e `38.5 g` Ag contain in `100 g` solution Molarity `=("moles of solute")/("Vol.of solution")` `=(38.5xx146)/(108xx1)=52.1"mol" L^(-1)` |
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| 447. |
What volume of a `0.8M` solution contains 100 millimoles of ther solute .A. `100 mL`B. `125mL`C. `500 mL`D. `62.5mL` |
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Answer» Correct Answer - B Molarity `=("moles of solute")/("vol of solution")` mol of solution `=(100xx10^(-3))/(8) = 125 mL` |
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| 448. |
What volume of 36 M and 1 M sulphuric acid must be mixed to get 1L of 6 M sulphuric acid ? |
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Answer» Suppose the volume of 36 M `H_(2)SO_(4)` solution `=xL` `:.` the volume of `1 M H_(2)SO_(4)` solution `= (1-x)L` Both these acid solutions upon mixing are to form 1 L of `6 M H_(2)SO_(4)` solution The value of x can be calculated as follows : `M_(2)V_(1)+M_(2)V_(2)=M_(3)V_(3),36xxx+1xx(1-x)=6xx1` or `36x+1-x=6or 35x=5` `x=(5)/(35)=0.1428L=142.8mL` `:.` Volume of `36 M H_(2)SO_(4)` solution `= 142.8 mL` Volume of `1 M H_(2)SO_(4)` solution `= 1000 - 142.8 = 857.2 mL`. |
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| 449. |
Which of the following have same significant figures?A. 0.07B. 0.7C. 7D. 70 |
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Answer» Correct Answer - A::B::C::D Two |
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| 450. |
Which of the following have same significant figures?A. `6.02 xx 10^(23)`B. `7.70 xx 10^(-20)`C. 7.50D. 0.75 |
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Answer» Correct Answer - A::B::C Three |
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