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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.
| 50151. |
A gaseous oxide contains 30.4% of nitrogen, one molecule of which contains one nitrogen atom. The density of the oxide relative to oxygen is |
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Answer» 0.94 |
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| 50152. |
A gaseous mixture was prepared by taking equal mole of CO and N_2 If the total pressure of the mixture of the nitrogen (N_2) in the mixture is |
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Answer» 0.5 ATM |
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| 50153. |
A gaseous mixture was prepared by taking equal moles of CO and N_2. If the total pressure of the mixture was found 1 atmosphere, the partial pressure of the nitrogen (N_2) in the mixture is |
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Answer» 0.8 ATM |
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| 50154. |
A gaseous mixture of three gases A, B and C has a pressure of 10atm. The total number of moles of all the gases is 10. The partial pressure of A and B are 3 and I am respectively. If C has a molecular weight of 2, what is the weight of C in grams present in the mixture? |
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Answer» 6 |
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| 50155. |
A gaseous mixture of three gases A, B and C has a pressure of 10 atm. The total number fo moles of all the gases is 10. If the partial pressures ofA and B are 3.0 and 1.0 atm respectively and if C has a molecular mass of 2.0, then calculate the weight of C in g present in the mixture. |
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Answer» Solution :`P_(A)=3.0" atm", P_(B)=1.0" atm"` Hence, `"" P_(C )=10-(3+1)=6" atm"` `P_(C )=(n_(C ))/(n_("TOTAL"))xxP_("total")=(W)/(M)xx(1)/(n_("total"))xxP_("total")` `:. "" 6.0=(w)/(2)xx(1)/(10)xx10" or "w=12" g"` |
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| 50156. |
A gaseous mixture of He, Ne, Ar and Kr is irradiated with photons of frequency appropriate to ionise Ar. The ion(s) present in the mixuture will be- |
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Answer» Only `AR^(+)` |
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| 50157. |
A gaseous mixture of 3 L of propane (C_(3)H_(8)) and butane (C_(4)H_(10)) on complete combustion at 25^(@)C produced 10 L of CO_(2). Find out the composition of the gaseous mixture. |
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Answer» `:.` The volume of butane in the mixture `= (3-x)L` Now let us calculate the volume of `CO_(2)` evolved with the help of chemical EQUATIONS Step I. Calculation of volume of `CO_(2)` from xL of propane The combustion equation for propane is : `underset(1L)(C_(3)H_(8))+5O_(2)rarrunderset(3L)(3CO_(2))+4H_(2)O` 1 L of propane `(C_(3)H_(8))` from `CO_(2)=3L` x L of propane `(C_(3)H_(8))` from `CO_(2)=3xL` Step II. Calculation of volume of `CO_(2)` from (3-x)L of butane The combustion equation for butane is : `underset(1L)(C_(4)H_(10))+(13)/(2)O_(2)rarrunderset(4L)(4CO_(2))+5H_(2)O` 1L of butance `(C_(4)H_(10))` from `CO_(2)=4L` (3-x)L of butane `(C_(4)H_(10))` from `CO_(2) = 4 xx (3-x)L` Step III. Calculation of composition of the mixture Total volume of `CO_(2)` formed in the step I and step II = [3X + 4(3-x)]L But the volume of `CO_(2)` actually formed = 10 L `:. 3x +4(3-x)=10 , 3x +12 - 4x =10 - x = -2 or x = 2L` `:.` Volume of propane = 2L Volume of butane `= (3-2)=1L`. |
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| 50158. |
A gaseous mixture has oxygen and nitrogen in ratio of 1 : 4 by weight. What is ratio of number of molecules of them ? |
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Answer» SOLUTION :Oxygen : Nitrogen 1:4 by weight `(1)/(32) : (4)/(28)` by MOLES `=` by MOLECULE `(1)/(32) :(1)/(7)` `:.7:32` |
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| 50159. |
A gaseous mixture contains oxygen and nitrogen in the ratio 1:4 by weight. The ratio of their number of molecules is |
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Answer» `1:4` |
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| 50160. |
A gaseous mixture contains hydrogen atoms in the 4^(th) excited state, He^(+) ions in 3^(rd) excited state and Li^(2+) in 2^(nd) excited state. The number of spectral lines obtained in the emission spectrum of this sample when all these atoms/ions return to the ground state is |
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Answer» 19 |
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| 50161. |
A gaseous mixture contains 5.6 g of carbon (II) oxide and rest carbon (IV) oxide. When it is enclosed in a vessel of 10 dm^(3) at 293 K, it recorded a pressure of 2.0 bar. What is the partical pressure of each oxide of carbon? |
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Answer» <P> SOLUTION :`p_(CO)=0.495` BAR , `p_(CO_(2))=1.505` bar |
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| 50162. |
A gaseous mixture contains 160 grams of neon and 64 grams of oxygen in 82.1 litre vessel at 27^@C, then the partical pressure of Neon may be - atm |
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Answer» 2.4 |
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| 50163. |
A gaseous mixture containing two hydro carbonsX & Y of volume 44.8 lit (STP) when passed through ammonical Cu_(2)Cl_2has suffered a reduction in volume of about 11.2 lit. If remaining volume is due to Y, X is |
| Answer» Answer :D | |
| 50164. |
A gaseous mixture containing He,CH_(4) and SO_(2) was allowedwas allowedto effuse through a fine hole then find what molar ratio of gases coming out initially ? If mixture contain He, CH_(4) and SO _(2) in 1:2:3 mole ratio. |
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Answer» `2:2:3` `(n'_(He))/(n'_(SO_(2)))=(1)/(3)sqrt((64)/(3))=(4)/(3)` `n'_(He):n'_(CH_(4)):n'_(SO_(2))=4:4:3` |
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| 50165. |
A gaseous mixturecontaining 8g of O_(2) and 227 mL of N_(2) at STPis enclosed in flask of 5 L capacity at 0^(@)C. Find the partial pressure of each gas and calculate the total pressure in the vessel. |
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Answer» SOLUTION :total `P=1.164` ATM `, `p_(O_(2))=1.12` MM |
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| 50166. |
A gaseous mixture containing 0.35g of N_(2)and 5600 ml of O_(2)at STP is kept in a 5 litres flask at 300K. The total pressure of the gaseous mixture is |
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Answer» 1.293atm |
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| 50167. |
A gaseous hydrocarbon 'X' on reaction with bromine in light forms a mixture of two monobromo alkanes and HBr.The hydrocarbon 'X' is : |
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Answer» `CH_3-CH_3` |
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| 50168. |
A gaeous hydrocation 'X' on reaction with bromine in light forms a mixture of two monbromo alkanes and HBr. The hydrocarbon 'X' is : |
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Answer» `C_6H_6` |
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| 50169. |
A gaseous hydrocarbon on combustion produces four times its volume of CO_(2) by consuming six times its volume of oxygen. What is the ratio of atoms of Hydrogen and carbon in that hydrocarbon? |
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Answer» Ratio of gram ATOM of H `C=(8//1)/(48//12)=(8)/(4)=2` |
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| 50170. |
A gaseous hydrocarbon on analysis gave the following data : (i) It contains C = 82.7% and H = 17.3% (ii) The mass of 132 mL (measured at S.T.R) of it is 0.342 g.Find the molecular formula of the hydrocarbon. |
Answer» SOLUTION :Calculation of empirical formula :  `therefore` The empirical formula of the given hydrocarbon is `C_(2)H_(5)`. Calculation of molecular formula : Empirical formula mass = `(2 xx 12.01) + (5 xx 1.008) = 29.06` amu. As discussed earlier, the VOLUME occupied by one GRAM molecular mass of a substance at S.T.R is 22.4 L (22400 mL) `therefore` The gram molecular mass of the given hydrocarbon `=0.342/132 xx 22400 = 58.0` HENCE, its molecular mass = 58.0 amu `n = ("Molecular mass")/("Empirical formula mass") = 58.0/29.06 =2` `therefore` Molecular mass = 2 `xx` Empirical formula `=2 xx C_(2)H_(5) = C_(4)H_(10)` Hence, the molecular formula of the given hydrocarbon is `C_(4)H_(10)`. |
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| 50171. |
A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO_(2). The empirical formula of the hydrocarbon is |
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Answer» `C_(7)H_(8)` `:. %C = (12)/(44) xx (3.08)/(w)xx 100 = (84)/(w)` `%H = (2)/(18)xx (0.72)/(w)xx 100 = (8)/(w)` Ratio of C : H atoms `= (84)/(w) xx(1)/(12) : (8)/(w) xx (1)/(1) = 7 : 8` THUS, E.F. = `C_(7)H_(8)`. |
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| 50172. |
A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO_(2). The empirical formula of the hydrocarbon is: |
| Answer» Answer :D | |
| 50173. |
A gaseous hydrocarbon consumed 5 times its volume of oxygen as for combustion. The volume of CO_(2) produced in the reaction is thrice the volume of hydrocarbon under the same conditions What is the ratio of molecular weight to emperical formula weight of the hydrocarbon? |
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Answer» 1 |
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| 50174. |
A gaseous hydrocarbon consumed 5 times its volume of oxygen as for combustion. The volume of CO_(2) produced in the reaction is thrice the volume of hydrocarbon under the same conditions How many grams of water is produced by combustion of 0.1 mol of the givenn hydrocarbon? |
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Answer» 7.2 gm `(64)/(3a+64)xx100=27.6impliesa=55.96~~56` `M_(2)O_(3)(48)/(112+48)xx100=(48)/(160)xx100=30` |
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| 50175. |
A gaseous compound X contained 44.4 %C, 51.9 % N and 3.7% H . Under like conditions 30 cm^(3) of X diffused through a pinhole in 25 sec and the same volume of H_(2) diffused in 4.81 sec . The molecular formula ofX is |
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Answer» `C_(2) H_(2) N` `C : N : H= (44.4)/(12) : (51.9)/(14) : (3.7)/(1)` `= 3.7 : 3.7 : 3.7 = 1 : 1 :1 ` `therefore ` empirical FORMULA of `X = CNH` Ratio of RATE of diffusion of X and that of `H_(2)` is given by `(r_(x))/(r_(H_(2))) = (30 xx 4.81 )/(25 xx 30) = sqrt((2)/(M))` On solving , M = 54 `therefore` molecular formula of x is `C_(2) H_(2) N_(2)` |
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| 50176. |
A gaseous alkane requires five times its volume of oxygen under the same conditions for complete combustion. The molecular formula of the alkane is |
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Answer» `C_(2)H_(6)` |
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| 50177. |
A Gasaeous compound of Nitrogen and Hydrogen contains 12.5% by weight of Hydrogen. The density of the compound relative to Hydrogen is 16, the molecular formula of the compound is |
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Answer» `NH_(2)` `%N 87.5 (87.5)/(14) = 6.25 :. E.F = NH_(2)` V.D of compound =16` `( :'` given in question) `M.F. Wt = 16 XX 2 = 32` |
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| 50178. |
A gas X,Y, at 35^@C has RMS speed 12ms^(-1). On heating the gas twice to the original absolute temperature, the dimer totally dissociated to give monomer. What is the RMS speed of XY_2 molecules at the given elevated temperature? |
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Answer» SOLUTION :RMS velocity of (C) is given as `C= sqrt((3RT)/(M))` Given temperature `T_1= 308K ` Elevated temperature `T_2= 616K` ` C_1= sqrt((3RT_1)/(M_1)) andC_2= sqrt((3RT_2)/(M_2))` The ratio of RMS VELOCITIES ` (C_2)/(C_1) = sqrt((T_2)/(T_1) (M_1)/(M_2))= sqrt((161 )/( 308) xx ( 2M_2)/(M_2))=2` RMS speed of `XY_2`molecules = `C_2` `2 xx C_1= 2 xx 12 = 24ms^(-1)` |
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| 50179. |
A gas 'X' is passed through water to form a solution. The aqeous solution on treatment with AgNO_(3) solution gives a white precipitate. The sturated aqueous solution also dissolves magnesium ribbon with the evolution of colourless gas 'Y'. Identify 'X' and 'Y' |
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Answer» `X = CO_(2), Y = Cl_(2)` `AgNO_(3) + HCl rarr underset(("White ppt"))(AgCl) + HNO_(3)` `MG + underset((X))(Cl_(2)) rarr MgCl_(2) + underset((Y))(H_(2))` |
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| 50180. |
A gas X causes heating effect when allowed to expand. This is because |
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Answer» the GAS is a NOBLE gas |
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| 50181. |
A gas X at 1 atm pressure is bubbled through a solution containing a mixture of 1 MY^(-) and 1 MZ^(-) at 25^(@)C.If the reduction potential of ZgtYgtX, then |
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Answer» Y will OXIDISE X and not Z |
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| 50182. |
A gas X at 1 atm is bubbled through a solutioncontaining a mixture of 1M Y^(-) nad 1M Z^(-)ionsat 25^(@)c if the reduction potentials of ZgtYgtX then |
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Answer» y will oxidize x but not z |
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| 50183. |
A gas behave most like an Ideal gas under conditions of |
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Answer» <P>LOW T and HIGH P |
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| 50184. |
A real gas behaves like an ideal gas if its |
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Answer» LOW TEMPERATURE and low PRESSURE |
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| 50185. |
(A): Gas with lower molecular weight will effuse or diffuse faster than the gas with higher molecular weight. (R) : Kinetic energy of any gas is independent of temperature. |
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Answer» Both A and R are correct and R is the correct EXPLANATION of A. |
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| 50186. |
A gas that follows Boyl's law, Charles' law and Avogardo's law is called an ideal gas. Under what conditions a real gas would behave ideally ? |
| Answer» SOLUTION :A real gas BEHAVES as an IDEAL gas when the pressure is low and temperature is HIGH. | |
| 50187. |
A gas underoes change from state A to state B . In this process , the heat absorbed and work done by the gas is 5 J and 8J respectively . Now gas is brought back to A by another process during which 3 J of heat is evolved . In this reverse process of B to A |
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Answer» 10 J of the WORK will be done by the surrounding on GAS . |
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| 50188. |
A gas that follows Boyle.s law, Charle.s law and Avogadro.s law is called an ideal gas. Under what conditions a real gas would behave ideally ? |
| Answer» Solution :At low pressure and high temperature, a REAL GAS behaves as an IDEAL gas. Almost all gases are real gas. | |
| 50189. |
A gas such as carbon monoxide would be most likely to obey the ideal gas law at |
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Answer» low TEMPERATURE and HIGH PRESSURES |
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| 50190. |
A gas such as carbon monoxide would be most likely to obey the ideal gas law at : |
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Answer» high temperatures and high PRESSURES |
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| 50191. |
A gas produced by dropping water over calcium carbide is bubbled through dilute H_(2)SO_(4) containing HgSO_(4). Which reagent can convert the product of above reaction into ethyledenedichloride ? |
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Answer» `Cl_(2)` `CH -= CH + H_(2)O underset(H_(2)SO_(4) // HgSO_(4))rarr CH_(3)CHO` `CH_(3) - overset(H)overset(|)(C) = O + SOCl_(2) rarr underset("Ethylendene dichloride")(CH_(3) - overset(H)overset(|)(C Cl_(2)))` |
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| 50192. |
A gas of volume 2000ml is kept in a vessel at a pressure of 10^3 pascals at a temperature of 27^@C. If the pressure is increased to 10^5 pascals at the same temperature, the volume of the gas becomes |
| Answer» ANSWER :B | |
| 50193. |
A gas of molecular mass 71 g mol^(-1) is enclosed in a vessel at a temperature of 30^@C. If the pressure exerted by the gas is 1065 mm of Hg, calculate the density of the gas. |
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Answer» Solution :GIVEN that, `P = 1065 mm = (1065)/(760)= 1.4` atm `T=30^@C = 30 + 273 = 303 K, M = 71 g mol^(-1)` and `R=0.0821 L " atm " K^(-1) mol^(-1)`. `:."" d=(MP)/(RT)` `:. "" d=(71 xx 1.4)/(0.0821 xx 303) = 4.0 g L^(-1)` HENCE, the DENSITY of the given gas under given conditions is 4.0 grams per litre. |
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| 50194. |
A gas of identical H-like atom has someatoms in the lowest (ground) energy level A and some atoms in a particular upper (excited) energy level B and there are no atoms in any other energy level. The atoms of the gas make transition to a higher energy level by aborbing monochromatic light of photon energy 2.7eV. Subsequently, the atoms emit radiation of only six different photons energies. Some of the emitted photons have energy 2.7eV. Some have more and some have less than 2.7eV. (a) Find theprincipal quantum number of initially excitied level B. (b) Find the ionisation energy for the gas atoms. (c ) Find the maximum and the minimum energies of the emitted photons. |
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Answer» `E_(4)-E_(2)=2.7eV,""E_(4)-E_(3)lt2.7eV`, `E_(4)-E_(1)gt2.7eV` (a) `n=2` `(E_(4)-E_(2))^("atom")=(E_(4)-E_(2))^(H)xxZ^(2)` `2.7=2.55xxZ^(2)=1.029` (B) `IP=13.6Z^(2)=13.6xx(1.029)^(2)=14.4eV` (C ) Maximum energy emitted `=E_(4)-E_(1)=(E_(4)-E_(1))^(H)xxZ^(2)` `=12.75xx(1.029)^(2)` `13.5eV` Minimum energy emitted `=E_(4)-E_(3)=(E_(4)-E_(3))^(H)xxZ^(2)` `=66xx(1.029)^(2)=0.7eV` |
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| 50195. |
A gas occupies a volume of 300 cc at 27^@C and 620 mm pressure. The volume of the gas at 47^@C and 640 mm pressure is |
| Answer» ANSWER :C | |
| 50196. |
A gas occupies a volume of 2.5L at 9 xx 10^(5) Nm^(-2). The additional pressure required to decrease the volume of the gas to 1.5L keeping the temperature constant is …………. xx 10^(5) Nm^(-2) ? |
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Answer» Solution :`P_1V_1 = P_2V_2` `V_2 = (9 XX 10^5 xx 2.5)/(115) = 16 xx 10^5 N.m^(-2)`. |
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| 50197. |
A gas occupies a volume of 2.5 L at 9xx10^(5)N m^(-2). Calculate the additional pressure required to decrease the volume of the gas to 1.5 L, keeping the temperature constant. |
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Answer» <P> Apply `P_(1)V_(1)=P_(2)V_(2)`. Calculate `P_(2)`. We get `P_(2)=15xx10^(6) N m^(-2)` Additional PRESSURE required `=15xx10^(6)N m^(-2)-9xx10^(5)N m^(-2)=6xx10^(5)N m^(-2)`. |
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| 50198. |
A gas occupies 300.0 mL at 27^@C and 730 mm pressure. What would be its volume at standard temperature and pressure (S.T.P.) ? |
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Answer» In the present case, `P_1 = 730 " mm HG, " V_1 = 300.0 mL`, `T_1 = 27 + 273 = 300 K` `P_2 = 760 " mm Hg, " V_2 = ?, T_2 = 273 K` (at S.T.P.) SUBSTITUTING the VALUES, we have `(730 xx 300.0)/300 = (760 xxV_2)/273` `:. "" V_2 = 262.2 mL`. |
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| 50199. |
A gas occupies 300 ml at 27^(@)C and 730 mm pressure what would be its volume at STP. |
| Answer» SOLUTION :`262.2 L` | |
| 50200. |
A gas obeys P(V-b) = RT. Then which of the following are correct |
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Answer» Isochoric curves have the slope = R/V-B `P(V- b) = RT , P = ((R )/(V-b))^(T)` , slope = `R/(V-b)` b) Isobaric = P constant `(Vb) = (R/P)^T , V = (R/P)^T + b` slope= `R/P`, constant = b c) `P (V- b) = RT, PV e= RT + Pb` `(PV)/(RT) =1+ (b/(RT))P` ` Z > 1` = repulsive forces dominate. |
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