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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.
| 201. |
Number of gram atoms of oxygen present in `0.3` mole of `(COOH)_(2).2H_(2)O` isA. 9B. 18C. 0.9D. 1.8 |
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Answer» Correct Answer - D No. of gram atoms =No: of moles `xx 6` |
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| 202. |
What is the oxidation state of Fe in the product formed when acidified Potasssium ferrocyanide `(K_(4)[Fe(CN)_(6)])` is treated with hydrogen peroxide?A. `+2`B. `+3`C. `+1`D. `+6` |
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Answer» Correct Answer - B `K_(4) [Fe(CN)_(6)] rarr K_(3)[Fe(CN)_(6)]` `Fe^(+2) rarr Fe^(+3)` |
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| 203. |
No . of gram atoms in 1kg of Helium |
| Answer» `=(1000g)/(4g) = 250` | |
| 204. |
Assertion(A) Significant figures for 0.200 is 3 where as for 200 it is 1. Reason(R) Zero at the end or right of a number are significatn provided they are not on the right side of the decimal point.A. iB. iiC. iiiD. iv |
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Answer» Correct Answer - C `0.200` contains 3 while 200 contains only one significan figure because zero at the end or right of a number are significant provided they are on the right side of the decimal point. |
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| 205. |
No of gram atoms in `12.044 xx 10^(23)` Na atoms |
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Answer» No of gram atom `= ("No. of atoms")/("Avogadro Number")` `=(12.044 xx 10^(23))/(6.023 xx 10^(23)) = 2` |
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| 206. |
The number of significant figures in `0.0045` areA. 2B. 3C. 4D. 5 |
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Answer» Correct Answer - A Zeros to the left of the first non-zero digit are not significant |
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| 207. |
What is the concentration of sugar `(C_(12)H_(22)O_(11))` in `mol L^(-1)` if its `20 g` are dissolved in enough water to make a final volume up to `2L`?A. `0.092`B. `0.029`C. `0.059`D. `0.069` |
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Answer» Correct Answer - B `M = (w)/(GMW) xx (1)/(V "in litres")` |
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| 208. |
On analysis of a compund, it was found that it contains (i) 4.08 grams Na (ii) 0.264 mole oxygen atoms (ii) `5.3 xx 10^(22)` atoms of carbon. Determine its E.F. |
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Answer» Determine the mole ratio `Na : C : O = (4.08)/(23) : (5.3 xx 10^(22))/(6.23 xx 10^(23)) : 0.264` or Simple mole ratio `= 2:1:3` or `E.F. = Na_(2)CO_(3)` |
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| 209. |
24 grams of metal reacts with an acid and liberated 22.4 lit of hydrogen at S.T.P. The eq.mass of metal is |
| Answer» eq. mass of metal = `("weight of metal")/("volume of hydrogen") xx 11.2 = (24)/(22.4) xx 11.2 = 12` | |
| 210. |
If the mass of an electron is `9 xx 10^(-28)` grams weight of one mole of electron isA. `9 xx 10^(-28)gm`B. `6 xx 10^(-28)`C. `1.008 gm`D. `0.00054 gm` |
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Answer» Correct Answer - D `9 xx 10^(-28) xx 6.023 xx 10^(23)` |
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| 211. |
Molecular weight of a gas is 44. The volume occupied by 2.2 grams of the gas under STP isA. 1.12 litB. 1.14 litC. 2.24 litD. 5.6 lit |
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Answer» Correct Answer - A Under STP 1 mole `= 22.4L` `(2.2)/(44) = ?` |
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| 212. |
What is the volume (in lit) of carbon dioxide liberated at STP, when 2.12 grams of sodium carbonate (mol. wt = 106) is treated with excess dilute HCl?A. `2.28`B. `0.448`C. `44.8`D. `22.4` |
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Answer» Correct Answer - B `Na_(2)CO_(3) +3HCI rarr 2NaCI +CO_(2) +H_(2)O` |
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| 213. |
The density of a gas at S.T.P is 1.50 grams per litre. The molecular weight of the gas isA. 28B. 30C. 3.14D. 35 |
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Answer» Correct Answer - C At S.T.P mol wt of a gas = density `xx 22.4` |
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| 214. |
At STP one litre of a gas weight `1.25` grams. The gas contains `85.71%` of carbon and `14.29%` of hydrogen. The formula of the compound isA. `CH_(4)`B. `C_(2)H_(6)`C. `C_(3)H_(8)`D. `C_(2)H_(4)` |
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Answer» Correct Answer - D `1L rarr 1.25 gm` `22.4L` gives Molecular Wt. `M.F = E.F xx (MW)/(EF Wt.)` |
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| 215. |
A granulated sample of aircraft alloy `(Al, Mg, Cu)` weighing `8.72 g` was first treated with alkali and then with very dilute `HCl`, leaving a residue. The residue after alkali boiling weighed `2.10 g` and the acid insoluble residue weighed `0.69 g`. What is the composition of the alloy?A. Weight % of `AI = 75.9%`B. Weigth % of `Cu = 7.9%`C. Weight % of `Mg = 16.2%`D. All the above |
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Answer» Correct Answer - D Let `Al,Mg` and Cu be a,b and c g respectively. `2AI + 2NaOH rarr 2NaAIO_(2) +3H_(2)` `Mg +2HCI rarr MgCI_(2) +H_(2)` `Cu +HCI rarr` No reaction i.e., only AI reacts with `NaOH` and then only Mg reacts with `HCI :. a + b +c = 8.72` `b +c = 2.10` (Residue left afte alkali treatment) `c = 0.69` (Residue left after acid treatment) `:. b = 2.1 - 0.69 = 1.41 g` `a = 8.72 - 2.1 = 6.62g` `:. %` of `AI = (6.62)/(8.72) xx 100 = 75.9` `%` of `Mg = (1.41)/(8.72) xx 100 = 16.2` `%` of `Cu = (0.69)/(8.72) xx 100 = 7.9` |
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| 216. |
The number of moles of `H_(2)` in 0.224 L of hydrogen gas a STP (273 K, 1 atm) assuming ideal gas behaviour isA. 1B. 0.1C. 0.01D. 0.001 |
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Answer» Correct Answer - C 1 mole `= 22.4L` `? = 0.224L` |
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| 217. |
0.01 mole of iodoform `(CHI_(3))` reacts with Ag to produce a gas whose volume at NTP is `2CHI_(3)+6Ag to6Agl(s)+C_(2)H_(2)(g)`A. 224 mlB. 112 mlC. 336 mlD. 448 ml |
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Answer» Correct Answer - B `2CHI_(3) +6Ag rarr 6AGI +C_(2)H_(2)` |
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| 218. |
The amount of zinc required to produce 224 ml of `H_(2)` at NTP on treatement with dilute `H_(2)SO_(4)` solution will beA. `0.65 g`B. `0.065g`C. `65g`D. `6.5g` |
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Answer» Correct Answer - A `Zn +H_(2)SO_(4) rarr ZnSO_(4)+H_(2)` |
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| 219. |
`0.132g` of an organic compound gave 50 ml of `N_(2)` at NTP. The weight percentage of nitrogen in the compound is close toA. 15B. 20C. 48.9D. 47.34 |
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Answer» Correct Answer - D `%N = (28 xx V_(N_(2)))/(22400) ("under" STP (ml))/(W_("org.comp")) xx 100` |
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| 220. |
`0.132g` of an organic compound gave `50ml` of `N_(2)` at NTP. The weight percentage of nitrogen in the compound is close toA. 15B. 20C. 48.9D. 47.43 |
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Answer» Correct Answer - D wt% of `N_(2) = (28)/(22400) xx ("vol of" N_(2)(ml) "at stp")/("wt. of compound") xx 100` |
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| 221. |
1.0 g of a metal oxide gave 0.2 g of metal. Calculate the equivalent weight of the metal. |
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Answer» Weight of oxygen `= 1.0-0.2 = 0.8g` let the equivalent weight of metal (M) be E,EW (oxygen) = 8g Equivalent of metl oxide = Equivalent of metal `("Weight of Metal oxide")/(Ew(M) +Ew(O)) = ("Weight of oxygen")/(Ew(O))` `= (1.0)/(E+8) = (0.8)/(8) = 0.1` `0.1 (E+8) = 1.0 E = 2` |
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| 222. |
A gaseous mixture of propane and butane of volume 3 litre on complete combustion produces 11.0 litre `CO_(2)` under standard conditions of temperature and pressure. The ration of volume of butane to propane is:A. `1:2`B. `2:1`C. `3:2`D. `3:1` |
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Answer» Correct Answer - B `C_(3)H_(8) +5O_(2) rarr 3CO_(2)+4H_(2)O` x litres of propane produce `3x` litre of `CO_(2)` `C_(4)H_(10) +6.5O_(2) rarr 4CO_(2) +5H_(2)O` (3-x) litres of butane produce `4(3-x)` lit of `CO_(2)` `3x +4 (3-x) = 11, 3x + 12 - 4x = 11` `12 - x = 11, x = 1` litre volume of butane: propane `= 2:1` |
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| 223. |
The equivalent weight of phosphoric acid `(H_(3)PO_(4))` in the reaction `NaOH+H_(3)PO_(4) rarr NaH_(2)PO_(4)+H_(2)O` isA. 49B. 98C. 32.6D. 40 |
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Answer» Correct Answer - B No of `H^(+)` ions given by `H_(3)PO_(4)` to `NaOH` is one |
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| 224. |
For which of the following compound equivalent weight is equal to molecular weightA. `H_(2)SO_(4)`B. `H_(3)PO_(2)`C. `H_(3)PO_(4)`D. `H_(3)PO_(3)` |
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Answer» Correct Answer - B If basicity of acid `=1, MW = EW` |
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| 225. |
How much volume of `CO_(2)` at S.T.P. is liberated by the combustion of `100 cm^(3)` of propane `(C_(3)H_(8))`?A. `100 cm^(3)`B. `200 cm^(3)`C. `300 cm^(3)`D. `400 cm^(3)` |
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Answer» Correct Answer - C `C_(3)H_(8) +5O_(2) rarr 3CO_(2) +4H_(2)O` |
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| 226. |
The number of molecules of `CO_(2)` liberated by the complete combustion of `0.1g` atoms of graphite in air isA. `3.01 xx 10^(22)`B. `6.02 xx 10^(23)`C. `6.02 xx 10^(22)`D. `3.01 xx 10^(23)` |
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Answer» Correct Answer - C `C_((s)) +O_(2(g)) rarr CO_(2(g))` 1mole `-6.023 xx 10^(23)` molecules `CO_(2)` `0.1` mole `-? = 6.023 xx 10^(22)` |
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| 227. |
If 4 g of NaOH dissovles in 36g of `H_(2)O`, calculate the mole fraction of each component in the solution. (specific gravity of solution is `1g mL^(-1)`).A. 2.5MB. 3MC. 2MD. 1.5M |
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Answer» Correct Answer - A Mass of `NaOH = 4g` Number of moles of `NaOH = (4g)/(40g) = 0.1 mol` Mass of `H_(2)O = 36g` Number of moles of `H_(2)O = (36g)/(18g) = 2 mol` Mass of solution = mass of water `+` mass of `NaOH = 36g + 4g = 40g` Volume of solution` = 40 xx 1 = 40 mL` (Since specific gravity of solution is `1g mL^(-1))` Molarity of solution `=("Number of moles of solute")/("Volume of solution in litre")` `= (0.1 mol NaOH)/(0.04L) = 2.5M` |
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| 228. |
`7.36g` of a mixture of `KCI` and KI was dissolved in `H_(2)O` to prepare 1 litre solution 25 ml of this required `8.45 ml` of `0.2N AgNO_(3)`, what is % of KI in mixture ?A. `57.28`B. `47.28`C. `5.72`D. `149.12` |
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Answer» Correct Answer - A geq of `KI +KCl` mixture =geq of `AgNO_(3)` `xg (7.36 -x)g` `((x)/(166)+(7.36-x)/(74.5)) = (8.45 xx 0.2 xx 40)/(1000)` `:.` On solving `x = 4.216g` `:. %` of `KI 57.28%` |
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| 229. |
2 mole , equimplar mixture of `Na_(2)C_(2)O_(4) and H_(2)C_(2)O_(4) "required" V_(1) L of 0.1 M KMnO_(4)` in acidic medium for complete oxidation. The same amount of the mixture required ` V_(2) L of 0.2 M NaOH` for neutralisaation. The raation of ` V_(1) and V_(2)` is: |
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Answer» Correct Answer - 7 No. of moles of `Na_(2)C_(2)O_(4) = H_(2)C_(2)O_(4) = 1` `4 = v_(1)xx 0.5, v_(1) = 8L` `2 = v_(2) xx 0.1 rArr v_(2) = 20L, v_(1): v_(2) = 8:20` `= 2:5 = x:y, x+y = 7` |
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| 230. |
A mixture of `H_(2)C_(2)O_(4)` and `K_(2)C_(2)O_(4)` required 0.2 N, 25mL `KMnO_(4)` solution for complete oxidation. Same mixture needs 0.2 M,20 mL NaOH solution for its complete neutralisation. Calculate mole percentage of `H_(2)C_(2)O_(4)` in the given mixture.A. `90%`B. `26.8%`C. `40%`D. `50%` |
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Answer» Correct Answer - B No. of milli equivalents of `H_(2)C_(2)O_(4) =`No. of milli equivalents of `NaOH = 4` No. of milli equivalents of `H_(2)C_(2)O_(4) + Na_(2)C_(2)O_(4) =` No. of milli equivalents of `KMnO_(4) = 6 xx 0.2 xx 5 = 6` milli equivalents `Na_(2)C_(2)O_(4) = 2` weight of `Na_(2)C_(2)O_(4)` `= 2xx 10^(-3) xx 67 = 0.134g` `% Na_(2)C_(2)O_(4) = (0.134)/(0.5) xx 100 = 26.8` |
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| 231. |
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. Statement -1 is True, Statement -2 is True, Statement -2 is a correctB. Statement -1 is True, Statement -2 is True, statement -2 is NOT a correct explanation for Statement-9C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
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Answer» Correct Answer - C `H_(3)BO_(3) +NaOH rarr Na [B(OH)_(4)]` `3C_(2)H_(5)OH + H_(3)BO_(3) rarr (C_(2)H_(5))_(3) BO_(3) +3H_(2)O` `(C_(2)H_(5))_(3) BO_(3) +3NaOH rarr Na_(2)BO_(3) +3C_(2)H_(5)OH` `(C_(2)H_(5))_(3)BO_(3) +3NaOH rarr Na_(3)BO_(3)+3C_(2)H_(5)OH` |
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| 232. |
Which of the following units represent the largest amount of energy?A. CalorieB. ErgC. JouleD. Electron-volt |
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Answer» Correct Answer - A Calorie `gt` joule `gt` erg `gt` eV |
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| 233. |
The M.K.S system was first introduced byA. ArchimedesB. GalileoC. NewtonD. Giorgi |
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Answer» Correct Answer - D Giorgi introduced MKS system. |
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| 234. |
An organic compound contains `C = 40%, H = 13.33%`, and `N = 46.67%`. Its empirical formula will beA. `C_(3)H_(13)N_(3)`B. `CH_(2)N`C. `CH_(4)N`D. `CH_(6)N` |
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Answer» Correct Answer - C `CH_(4)N` has `40%C, 13.3% H` and `46.7%N` |
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