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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your Class 11 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 |
| Answer» <html><body><p>0.94<br/>1.44<br/>1.5<br/>`3.0` </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 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 |
| Answer» <html><body><p>0.5 <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a><br/>0.8 atm<br/>0.9 atm<br/>1 atm </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 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 |
| Answer» <html><body><p>0.8 <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a><br/>0.9 atm<br/>1 atm<br/>0.5 atm.</p>Solution :0.5 atm.</body></html> | |
| 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? |
| Answer» <html><body><p>6<br/>3<br/>12<br/>8</p>Answer :A</body></html> | |
| 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. |
| Answer» <html><body><p></p>Solution :`P_(A)=3.0" atm", P_(B)=<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.0" atm"` <br/> Hence, `"" P_(C )=10-(3+1)=<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>" atm"` <br/> `P_(C )=(n_(C ))/(n_("<a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a>"))xxP_("total")=(<a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a>)/(M)xx(1)/(n_("total"))xxP_("total")` <br/> `:. "" 6.0=(w)/(2)xx(1)/(10)xx10" or "w=12" g"`</body></html> | |
| 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- |
| Answer» <html><body><p>Only `<a href="https://interviewquestions.tuteehub.com/tag/ar-380980" style="font-weight:bold;" target="_blank" title="Click to know more about AR">AR</a>^(+)`<br/>`Ar^(+) and He^(+)`<br/>`Ar^(+) and <a href="https://interviewquestions.tuteehub.com/tag/ne-1112263" style="font-weight:bold;" target="_blank" title="Click to know more about NE">NE</a>^(+)`<br/>`Ar^(+) and <a href="https://interviewquestions.tuteehub.com/tag/kr-534811" style="font-weight:bold;" target="_blank" title="Click to know more about KR">KR</a>^(+)`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :D</body></html> | |
| 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. |
| Answer» <html><body><p><br/></p>Solution :Let the volume of propane in the mixture = xL <br/> `:.` The volume of butane in the mixture `= (3-x)<a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>` <br/> Now let us calculate the volume of `CO_(2)` evolved with the help of chemical <a href="https://interviewquestions.tuteehub.com/tag/equations-452536" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATIONS">EQUATIONS</a> <br/> Step I. Calculation of volume of `CO_(2)` from xL of propane <br/> The combustion equation for propane is : <br/> `underset(1L)(C_(3)H_(8))+5O_(2)rarrunderset(3L)(3CO_(2))+4H_(2)O` <br/> 1 L of propane `(C_(3)H_(8))` from `CO_(2)=3L` <br/> x L of propane `(C_(3)H_(8))` from `CO_(2)=3xL` <br/> Step <a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>. Calculation of volume of `CO_(2)` from (3-x)L of butane <br/> The combustion equation for butane is : <br/> `underset(1L)(C_(4)H_(10))+(13)/(2)O_(2)rarrunderset(4L)(4CO_(2))+5H_(2)O`<br/> 1L of butance `(C_(4)H_(10))` from `CO_(2)=4L` <br/> (3-x)L of butane `(C_(4)H_(10))` from `CO_(2) = 4 xx (3-x)L` <br/> Step III. Calculation of composition of the mixture <br/> Total volume of `CO_(2)` formed in the step I and step II = [<a href="https://interviewquestions.tuteehub.com/tag/3x-310805" style="font-weight:bold;" target="_blank" title="Click to know more about 3X">3X</a> + 4(3-x)]L <br/>But the volume of `CO_(2)` actually formed = 10 L <br/> `:. 3x +4(3-x)=10 , 3x +12 - 4x =10 - x = -2 or x = 2L` <br/> `:.` Volume of propane = 2L <br/> Volume of butane `= (3-2)=1L`.</body></html> | |
| 50158. |
A gaseous mixture has oxygen and nitrogen in ratio of 1 : 4 by weight. What is ratio of number of molecules of them ? |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>:8`<br/>`3:16`<br/>`1:<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>`<br/>`7:32`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Oxygen : Nitrogen <br/> 1:4 by weight <br/> `(1)/(32) : (4)/(28)` by <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> `=` by <a href="https://interviewquestions.tuteehub.com/tag/molecule-563028" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULE">MOLECULE</a> `(1)/(32) :(1)/(7)` <br/> `:.7:32`</body></html> | |
| 50159. |
A gaseous mixture contains oxygen and nitrogen in the ratio 1:4 by weight. The ratio of their number of molecules is |
| Answer» <html><body><p>`1:4`<br/>`4:1`<br/>`7:31`<br/>`3:16`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 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 |
| Answer» <html><body><p>19<br/>20<br/>16<br/>18</p>Answer :D</body></html> | |
| 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? |
| Answer» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`p_(<a href="https://interviewquestions.tuteehub.com/tag/co-920124" style="font-weight:bold;" target="_blank" title="Click to know more about CO">CO</a>)=0.495` <a href="https://interviewquestions.tuteehub.com/tag/bar-892478" style="font-weight:bold;" target="_blank" title="Click to know more about BAR">BAR</a> , `p_(CO_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))=1.505` bar</body></html> | |
| 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 |
| Answer» <html><body><p>2.4 <br/>0.6 <br/>1.5 <br/><a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a> </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 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» <html><body><p>Butyne-2 <br/><a href="https://interviewquestions.tuteehub.com/tag/ethene-975852" style="font-weight:bold;" target="_blank" title="Click to know more about ETHENE">ETHENE</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/ethane-975817" style="font-weight:bold;" target="_blank" title="Click to know more about ETHANE">ETHANE</a> <br/>Propyne</p>Answer :D</body></html> | |
| 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. |
| Answer» <html><body><p>`2:2:3`<br/>`6:6:<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(2):sqrt(2):3`<br/>`4:4:3`</p>Solution :`(n'_(He))/(n'_(CH_(4)))=(1)/(2)sqrt((<a href="https://interviewquestions.tuteehub.com/tag/16-276476" style="font-weight:bold;" target="_blank" title="Click to know more about 16">16</a>)/(4))=(1)/(1)`<br/> `(n'_(He))/(n'_(SO_(2)))=(1)/(3)sqrt((64)/(3))=(4)/(3)` <br/> `n'_(He):n'_(CH_(4)):n'_(SO_(2))=4:4:3`</body></html> | |
| 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. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :total `P=1.164` <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a> `, `p_(O_(2))=1.12` <a href="https://interviewquestions.tuteehub.com/tag/mm-1098795" style="font-weight:bold;" target="_blank" title="Click to know more about MM">MM</a> <br/></body></html> | |
| 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 |
| Answer» <html><body><p>1.293atm <br/>1.2315 <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a> <br/>12.315 atm <br/>0.616atm </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 50167. |
A gaseous hydrocarbon 'X' on reaction with bromine in light forms a mixture of two monobromo alkanes and HBr.The hydrocarbon 'X' is : |
| Answer» <html><body><p>`CH_3-CH_3`<br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/RES_CHM_ORM_II_E01_025_O02.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/RES_CHM_ORM_II_E01_025_O03.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/RES_CHM_ORM_II_E01_025_O04.png" width="30%"/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 50168. |
A gaeous hydrocation 'X' on reaction with bromine in light forms a mixture of two monbromo alkanes and HBr. The hydrocarbon 'X' is : |
| Answer» <html><body><p>`C_6H_6`<br/>`C_3H_6`<br/>`C_3H_8`<br/>`C_4H_10`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 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? |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`C_(4)H_(<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>)+6O_(2)rarr4CO_(2)+4H_(2)O` <br/> Ratio of gram <a href="https://interviewquestions.tuteehub.com/tag/atom-887280" style="font-weight:bold;" target="_blank" title="Click to know more about ATOM">ATOM</a> of <a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a> <br/> `<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>=(8//1)/(48//12)=(8)/(4)=2`</body></html> | |
| 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» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Calculation of empirical formula : <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/NTN_HCS_ISC_CHE_XI_P1_C01_SLV_078_S01.png" width="80%"/> <br/> `therefore` The empirical formula of the given hydrocarbon is `C_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)H_(5)`. <br/> Calculation of molecular formula : <br/> Empirical formula mass = `(2 xx 12.01) + (5 xx 1.008) = 29.06` amu. As discussed earlier, the <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> occupied by one <a href="https://interviewquestions.tuteehub.com/tag/gram-1010695" style="font-weight:bold;" target="_blank" title="Click to know more about GRAM">GRAM</a> molecular mass of a substance at S.T.R is 22.4 L (22400 mL) <br/> `therefore` The gram molecular mass of the given hydrocarbon `=0.342/132 xx 22400 = 58.0`<br/> <a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a>, its molecular mass = 58.0 amu `n = ("Molecular mass")/("Empirical formula mass") = 58.0/29.06 =2` <br/> `therefore` Molecular mass = 2 `xx` Empirical formula <br/> `=2 xx C_(2)H_(5) = C_(4)H_(10)` <br/> Hence, the molecular formula of the given hydrocarbon is `C_(4)H_(10)`.</body></html> | |
| 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 |
| Answer» <html><body><p>`C_(7)H_(<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>)`<br/>`C_(2)H_(4)`<br/>`C_(3)H_(4)`<br/>`C_(6)H_(5)`</p>Solution :Let the mass of <a href="https://interviewquestions.tuteehub.com/tag/gaseous-467815" style="font-weight:bold;" target="_blank" title="Click to know more about GASEOUS">GASEOUS</a> <a href="https://interviewquestions.tuteehub.com/tag/hydrocarbon-1034072" style="font-weight:bold;" target="_blank" title="Click to know more about HYDROCARBON">HYDROCARBON</a> taken = w gm <br/> `:. %C = (<a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a>)/(44) xx (3.08)/(w)xx 100 = (84)/(w)` <br/> `%H = (2)/(18)xx (0.72)/(w)xx 100 = (8)/(w)` <br/> Ratio of C : H atoms `= (84)/(w) xx(1)/(12) : (8)/(w) xx (1)/(1) = 7 : 8` <br/> <a href="https://interviewquestions.tuteehub.com/tag/thus-2307358" style="font-weight:bold;" target="_blank" title="Click to know more about THUS">THUS</a>, E.F. = `C_(7)H_(8)`.</body></html> | |
| 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» <html><body><p>`C_(2)H_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)`<br/>`C_(3)H_(4)`<br/>`C_(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)H_(5)`<br/>`C_(7)H_(8)` </p>Answer :D</body></html> | |
| 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? |
| Answer» <html><body><p>1<br/>2<br/>3<br/>4</p>Solution :`C_(2)H_(2):(<a href="https://interviewquestions.tuteehub.com/tag/24-295400" style="font-weight:bold;" target="_blank" title="Click to know more about 24">24</a>)/(26)xx100=92.3, C_(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)H_(6):(72)/(<a href="https://interviewquestions.tuteehub.com/tag/78-335791" style="font-weight:bold;" target="_blank" title="Click to know more about 78">78</a>)xx100=92.3`</body></html> | |
| 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? |
| Answer» <html><body><p>7.2 gm <br/><a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>.6 gm<br/>14.4 gm <br/>1.8 gm</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/let-11597" style="font-weight:bold;" target="_blank" title="Click to know more about LET">LET</a> <a href="https://interviewquestions.tuteehub.com/tag/atomic-2477" style="font-weight:bold;" target="_blank" title="Click to know more about ATOMIC">ATOMIC</a> weight of metal = a <br/> `(<a href="https://interviewquestions.tuteehub.com/tag/64-330663" style="font-weight:bold;" target="_blank" title="Click to know more about 64">64</a>)/(3a+64)xx100=27.6impliesa=55.96~~56` <br/> `M_(2)O_(3)(48)/(112+48)xx100=(48)/(160)xx100=30`</body></html> | |
| 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 |
| Answer» <html><body><p>`C_(2) H_(2) N`<br/>`C_(2) H_4 N_2`<br/>`C_2 H_2 N_2`<br/>`C_4 H_2 N_2`</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/ratio-13379" style="font-weight:bold;" target="_blank" title="Click to know more about RATIO">RATIO</a> of number of atoms in the molecule = Ratio of gram atoms of elements in he <a href="https://interviewquestions.tuteehub.com/tag/compound-926863" style="font-weight:bold;" target="_blank" title="Click to know more about COMPOUND">COMPOUND</a> <br/> `<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> : N : H= (44.4)/(12) : (51.9)/(14) : (3.7)/(1)` <br/> `= 3.7 : 3.7 : 3.7 = 1 : 1 :1 ` <br/> `therefore ` empirical <a href="https://interviewquestions.tuteehub.com/tag/formula-464310" style="font-weight:bold;" target="_blank" title="Click to know more about FORMULA">FORMULA</a> of `X = CNH`<br/> Ratio of <a href="https://interviewquestions.tuteehub.com/tag/rate-1177476" style="font-weight:bold;" target="_blank" title="Click to know more about RATE">RATE</a> of diffusion of X and that of `H_(2)` is given by <br/> `(r_(x))/(r_(H_(2))) = (30 xx 4.81 )/(25 xx 30) = sqrt((2)/(M))` <br/> On solving , M = 54 <br/> `therefore` molecular formula of x is `C_(2) H_(2) N_(2)`</body></html> | |
| 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 |
| Answer» <html><body><p>`C_(2)H_(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)`<br/>`C_(4)H_(10)`<br/>`C_(3)H_(<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>)`<br/>`CH_(4)`</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 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 |
| Answer» <html><body><p>`NH_(2)`<br/>`NH_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)`<br/>`NH_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)`<br/>`N_(2)H_(4)`</p>Solution :`%H 12.5 (12.5)/(1) = 12.5` <br/> `%N 87.5 (87.5)/(14) = 6.25 :. E.F = NH_(2)` <br/> V.D of compound =<a href="https://interviewquestions.tuteehub.com/tag/16-276476" style="font-weight:bold;" target="_blank" title="Click to know more about 16">16</a>` <br/> `( :'` given in question) <br/> `M.F. Wt = 16 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 2 = 32`</body></html> | |
| 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? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :RMS velocity of (C) is given as <br/> `C= sqrt((3RT)/(M))` <br/> Given temperature `T_1= 308K ` <br/> Elevated temperature `T_2= 616K` <br/> ` C_1= sqrt((3RT_1)/(M_1)) andC_2= sqrt((3RT_2)/(M_2))` <br/> The ratio of RMS <a href="https://interviewquestions.tuteehub.com/tag/velocities-1444510" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITIES">VELOCITIES</a> <br/> ` (C_2)/(C_1) = sqrt((T_2)/(T_1) (M_1)/(M_2))= sqrt((<a href="https://interviewquestions.tuteehub.com/tag/161-277444" style="font-weight:bold;" target="_blank" title="Click to know more about 161">161</a> )/( 308) xx ( 2M_2)/(M_2))=2`<br/> RMS speed of `XY_2`molecules = `C_2` <br/> `2 xx C_1= 2 xx 12 = 24ms^(-1)`</body></html> | |
| 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' |
| Answer» <html><body><p>`X = CO_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>), Y = Cl_(2)`<br/>`X = Cl_(2), Y = CO_(2)`<br/>`X = Cl_(2), Y = H_(2)`<br/>`X = H_(2), Y = Cl_(2)`</p>Solution :`underset((X))Cl_(2) + H_(2)O rarr HOCl + HCl` <br/> `AgNO_(3) + HCl rarr underset(("White ppt"))(AgCl) + HNO_(3)` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a> + underset((X))(Cl_(2)) rarr MgCl_(2) + underset((Y))(H_(2))`</body></html> | |
| 50180. |
A gas X causes heating effect when allowed to expand. This is because |
| Answer» <html><body><p>the <a href="https://interviewquestions.tuteehub.com/tag/gas-1003521" style="font-weight:bold;" target="_blank" title="Click to know more about GAS">GAS</a> is a <a href="https://interviewquestions.tuteehub.com/tag/noble-2197707" style="font-weight:bold;" target="_blank" title="Click to know more about NOBLE">NOBLE</a> gas<br/>the inversion <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a> of the gas is very <a href="https://interviewquestions.tuteehub.com/tag/low-537644" style="font-weight:bold;" target="_blank" title="Click to know more about LOW">LOW</a><br/>the gas is ideal gas<br/>the boiling point of the gas is very low</p>Solution :The inversion temperature of the gas is very low.</body></html> | |
| 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 |
| Answer» <html><body><p>Y will <a href="https://interviewquestions.tuteehub.com/tag/oxidise-1144503" style="font-weight:bold;" target="_blank" title="Click to know more about OXIDISE">OXIDISE</a> X and not <a href="https://interviewquestions.tuteehub.com/tag/z-750254" style="font-weight:bold;" target="_blank" title="Click to know more about Z">Z</a> <br/>Y will oxidise Z and not X <br/>Y will Oxidise both Z and X <br/>Y will <a href="https://interviewquestions.tuteehub.com/tag/reduce-1181332" style="font-weight:bold;" target="_blank" title="Click to know more about REDUCE">REDUCE</a> both X and Z </p>Solution :is the correct <a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a></body></html> | |
| 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 |
| Answer» <html><body><p>y will oxidize x but not z <br/>y will oxidise <a href="https://interviewquestions.tuteehub.com/tag/bothx-2481332" style="font-weight:bold;" target="_blank" title="Click to know more about BOTHX">BOTHX</a> andz <br/>y will oxidise z but not x <br/>y will reduce both x and z </p>Solution :Higher the reduction potential <a href="https://interviewquestions.tuteehub.com/tag/stronger-1229999" style="font-weight:bold;" target="_blank" title="Click to know more about STRONGER">STRONGER</a> the oxidising agent since reduction potential <a href="https://interviewquestions.tuteehub.com/tag/decrease-946104" style="font-weight:bold;" target="_blank" title="Click to know more about DECREASE">DECREASE</a> in the order `ZgtYgtY` therefore their oxidising power <a href="https://interviewquestions.tuteehub.com/tag/also-373387" style="font-weight:bold;" target="_blank" title="Click to know more about ALSO">ALSO</a> decrease in the same order <a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a> y is a stronger oxidising agent than x but weaker than z therefore y can oxidise x but not z</body></html> | |
| 50183. |
A gas behave most like an Ideal gas under conditions of |
| Answer» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>><a href="https://interviewquestions.tuteehub.com/tag/low-537644" style="font-weight:bold;" target="_blank" title="Click to know more about LOW">LOW</a> T and <a href="https://interviewquestions.tuteehub.com/tag/high-479925" style="font-weight:bold;" target="_blank" title="Click to know more about HIGH">HIGH</a> P<br/>High T and high P <br/>Low T and low P <br/>High T and low P</p>Answer :D</body></html> | |
| 50184. |
A real gas behaves like an ideal gas if its |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/low-537644" style="font-weight:bold;" target="_blank" title="Click to know more about LOW">LOW</a> <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a> and low <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a><br/>low temperature and <a href="https://interviewquestions.tuteehub.com/tag/high-479925" style="font-weight:bold;" target="_blank" title="Click to know more about HIGH">HIGH</a> pressure<br/>high temperature and low pressure<br/>high temperature and high pressure.</p>Solution :high temperature and low pressure</body></html> | |
| 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. |
| Answer» <html><body><p>Both A and R are correct and R is the correct <a href="https://interviewquestions.tuteehub.com/tag/explanation-455162" style="font-weight:bold;" target="_blank" title="Click to know more about EXPLANATION">EXPLANATION</a> of A. <br/>Both A and R are correct but R is not the correct explanation of A. <br/>A is <a href="https://interviewquestions.tuteehub.com/tag/true-713260" style="font-weight:bold;" target="_blank" title="Click to know more about TRUE">TRUE</a> but R is <a href="https://interviewquestions.tuteehub.com/tag/false-459184" style="font-weight:bold;" target="_blank" title="Click to know more about FALSE">FALSE</a>. <br/>A is false but R is true. </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 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» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :A real gas <a href="https://interviewquestions.tuteehub.com/tag/behaves-894569" style="font-weight:bold;" target="_blank" title="Click to know more about BEHAVES">BEHAVES</a> as an <a href="https://interviewquestions.tuteehub.com/tag/ideal-1035490" style="font-weight:bold;" target="_blank" title="Click to know more about IDEAL">IDEAL</a> gas when the pressure is low and temperature is <a href="https://interviewquestions.tuteehub.com/tag/high-479925" style="font-weight:bold;" target="_blank" title="Click to know more about HIGH">HIGH</a>.</body></html> | |
| 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 |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> J of the <a href="https://interviewquestions.tuteehub.com/tag/work-20377" style="font-weight:bold;" target="_blank" title="Click to know more about WORK">WORK</a> will be done by the surrounding on <a href="https://interviewquestions.tuteehub.com/tag/gas-1003521" style="font-weight:bold;" target="_blank" title="Click to know more about GAS">GAS</a> . <br/>10 J of the work will be done by the gas . <br/>6 J of the work will be done by the surrounding on gas . <br/>6 J of the work will be done by the gas . </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 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» <html><body><p></p>Solution :At low pressure and high temperature, a <a href="https://interviewquestions.tuteehub.com/tag/real-1178490" style="font-weight:bold;" target="_blank" title="Click to know more about REAL">REAL</a> <a href="https://interviewquestions.tuteehub.com/tag/gas-1003521" style="font-weight:bold;" target="_blank" title="Click to know more about GAS">GAS</a> behaves as an <a href="https://interviewquestions.tuteehub.com/tag/ideal-1035490" style="font-weight:bold;" target="_blank" title="Click to know more about IDEAL">IDEAL</a> gas. Almost all gases are real gas.</body></html> | |
| 50189. |
A gas such as carbon monoxide would be most likely to obey the ideal gas law at |
| Answer» <html><body><p> low <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a> and <a href="https://interviewquestions.tuteehub.com/tag/high-479925" style="font-weight:bold;" target="_blank" title="Click to know more about HIGH">HIGH</a> <a href="https://interviewquestions.tuteehub.com/tag/pressures-1164423" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURES">PRESSURES</a><br/>high temperature and high pressures <br/>low temperature and low pressures<br/>high temperatures and low pressures </p>Answer :D</body></html> | |
| 50190. |
A gas such as carbon monoxide would be most likely to obey the ideal gas law at : |
| Answer» <html><body><p>high temperatures and high <a href="https://interviewquestions.tuteehub.com/tag/pressures-1164423" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURES">PRESSURES</a> <br/>low temperatures and low pressures <br/>high temperatures and low pressures <br/>low temperatures and high pressures </p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/real-1178490" style="font-weight:bold;" target="_blank" title="Click to know more about REAL">REAL</a> <a href="https://interviewquestions.tuteehub.com/tag/gases-13668" style="font-weight:bold;" target="_blank" title="Click to know more about GASES">GASES</a> show <a href="https://interviewquestions.tuteehub.com/tag/ideal-1035490" style="font-weight:bold;" target="_blank" title="Click to know more about IDEAL">IDEAL</a> gas <a href="https://interviewquestions.tuteehub.com/tag/behaviour-894636" style="font-weight:bold;" target="_blank" title="Click to know more about BEHAVIOUR">BEHAVIOUR</a> at high temperatures and low pressures.</body></html> | |
| 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 ? |
| Answer» <html><body><p>`Cl_(2)`<br/>`SOCl_(2) //Pt`<br/>`SbCl_(5)`<br/>Chloroform</p>Solution :`CaC_(2) + H_(2)O <a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> C_(2)H_(2) uarr + Ca(OH)_(2)` <br/> `CH -= CH + H_(2)O underset(H_(2)SO_(4) // HgSO_(4))rarr CH_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)<a href="https://interviewquestions.tuteehub.com/tag/cho-408392" style="font-weight:bold;" target="_blank" title="Click to know more about CHO">CHO</a>`<br/> `CH_(3) - overset(<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>)overset(|)(C) = O + SOCl_(2) rarr underset("Ethylendene dichloride")(CH_(3) - overset(H)overset(|)(C Cl_(2)))`</body></html> | |
| 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» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/1000ml-1771304" style="font-weight:bold;" target="_blank" title="Click to know more about 1000ML">1000ML</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/20ml-293069" style="font-weight:bold;" target="_blank" title="Click to know more about 20ML">20ML</a> <br/>2ml <br/>200ml </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :B</body></html> | |
| 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. |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> that, `P = 1065 mm = (1065)/(760)= 1.4` atm <br/>`T=30^@C = 30 + 273 = 303 K, M = <a href="https://interviewquestions.tuteehub.com/tag/71-334184" style="font-weight:bold;" target="_blank" title="Click to know more about 71">71</a> g mol^(-1)` <br/>and `R=0.0821 L " atm " K^(-1) mol^(-1)`. <br/> `:."" d=(MP)/(RT)`<br/> `:. "" d=(71 xx 1.4)/(0.0821 xx 303) = 4.0 g L^(-1)` <br/><a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a>, the <a href="https://interviewquestions.tuteehub.com/tag/density-17451" style="font-weight:bold;" target="_blank" title="Click to know more about DENSITY">DENSITY</a> of the given gas under given conditions is 4.0 grams per litre.</body></html> | |
| 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. |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Since we obtain 6 emission lines, it means <a href="https://interviewquestions.tuteehub.com/tag/electron-968715" style="font-weight:bold;" target="_blank" title="Click to know more about ELECTRON">ELECTRON</a> comes from <a href="https://interviewquestions.tuteehub.com/tag/4th-319068" style="font-weight:bold;" target="_blank" title="Click to know more about 4TH">4TH</a> orbit energy emitted is equal to less than and more than `2.7eV`. So it can be like this<br/> `E_(4)-E_(2)=2.7eV,""E_(4)-E_(3)lt2.7eV`, <br/> `E_(4)-E_(1)gt2.7eV` <br/> (a) `n=2` <br/> `(E_(4)-E_(2))^("atom")=(E_(4)-E_(2))^(H)xxZ^(2)`<br/> `2.7=2.55xxZ^(2)=1.029`<br/> (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>) `IP=13.6Z^(2)=13.6xx(1.029)^(2)=14.4eV` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> ) Maximum energy emitted `=E_(4)-E_(1)=(E_(4)-E_(1))^(H)xxZ^(2)`<br/> `=12.75xx(1.029)^(2)`<br/>`13.5eV` <br/> Minimum energy emitted `=E_(4)-E_(3)=(E_(4)-E_(3))^(H)xxZ^(2)`<br/> `=66xx(1.029)^(2)=0.7eV`</body></html> | |
| 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» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/260-298496" style="font-weight:bold;" target="_blank" title="Click to know more about 260">260</a> <a href="https://interviewquestions.tuteehub.com/tag/cc-911065" style="font-weight:bold;" target="_blank" title="Click to know more about CC">CC</a><br/>310 cc<br/>390 cc <br/><a href="https://interviewquestions.tuteehub.com/tag/450-317308" style="font-weight:bold;" target="_blank" title="Click to know more about 450">450</a> cc </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 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) ? |
| Answer» <html><body><p></p>Solution :`P_1V_1 = P_2V_2` <br/> `V_2 = (9 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10^5 xx <a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>.5)/(115) = <a href="https://interviewquestions.tuteehub.com/tag/16-276476" style="font-weight:bold;" target="_blank" title="Click to know more about 16">16</a> xx 10^5 N.m^(-2)`.</body></html> | |
| 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. |
| Answer» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`V_(1)=2.5 L, P_(1)=9xx10^(5) N m^(-2) , V_(2)=1.5 L, P_(2)`=?<br/> Apply `P_(1)V_(1)=P_(2)V_(2)`. Calculate `P_(2)`. We get `P_(2)=15xx10^(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>) N m^(-2)` <br/> Additional <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a> required `=15xx10^(6)N m^(-2)-9xx10^(5)N m^(-2)=6xx10^(5)N m^(-2)`.</body></html> | |
| 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.) ? |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :According to the gas equation, `(P_1V_1)/T_1 = (P_2V_2)/T_2` <br/> In the present case, <br/> `P_1 = 730 " mm <a href="https://interviewquestions.tuteehub.com/tag/hg-485049" style="font-weight:bold;" target="_blank" title="Click to know more about HG">HG</a>, " V_1 = 300.0 mL`, <br/>`T_1 = 27 + 273 = 300 K` <br/> `P_2 = 760 " mm Hg, " V_2 = ?, T_2 = 273 K` (at S.T.P.) <br/> <a href="https://interviewquestions.tuteehub.com/tag/substituting-1231652" style="font-weight:bold;" target="_blank" title="Click to know more about SUBSTITUTING">SUBSTITUTING</a> the <a href="https://interviewquestions.tuteehub.com/tag/values-25920" style="font-weight:bold;" target="_blank" title="Click to know more about VALUES">VALUES</a>, we have <br/> `(730 xx 300.0)/300 = (760 xxV_2)/273` <br/> `:. "" V_2 = 262.2 mL`.</body></html> | |
| 50199. |
A gas occupies 300 ml at 27^(@)C and 730 mm pressure what would be its volume at STP. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`262.2 <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>`</body></html> | |
| 50200. |
A gas obeys P(V-b) = RT. Then which of the following are correct |
| Answer» <html><body><p>Isochoric curves have the slope = R/V-<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a><br/>Isobaric curves have the slope =R/P with <a href="https://interviewquestions.tuteehub.com/tag/intercept-1047794" style="font-weight:bold;" target="_blank" title="Click to know more about INTERCEPT">INTERCEPT</a> = b<br/>For the gas compressibility factor = 1+ Pb/RT<br/>Attractive forces predominate <a href="https://interviewquestions.tuteehub.com/tag/repulsive-613883" style="font-weight:bold;" target="_blank" title="Click to know more about REPULSIVE">REPULSIVE</a> forces<br/></p>Solution :a) Isochoric = V constant <br/> `P(V- b) = RT , P = ((R )/(V-b))^(T)` , slope = `R/(V-b)` <br/> b) Isobaric = P constant <br/>`(Vb) = (R/P)^T , V = (R/P)^T + b` <br/>slope= `R/P`, constant = b <br/> c) `P (V- b) = RT, PV e= RT + Pb` <br/> `(PV)/(RT) =1+ (b/(RT))P` <br/> ` Z > 1` = repulsive forces dominate.</body></html> | |