InterviewSolution
Saved Bookmarks
InterviewSolution
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.
| 51801. |
1 mole of photons, each of frequency 250 s^(-1) would have approximately a total energy of |
| Answer» <html><body><p>1 <a href="https://interviewquestions.tuteehub.com/tag/erg-446447" style="font-weight:bold;" target="_blank" title="Click to know more about ERG">ERG</a><br/>1 joule<br/>1 eV<br/>1 MeV</p>Solution :`E = N_(A) <a href="https://interviewquestions.tuteehub.com/tag/hv-1033771" style="font-weight:bold;" target="_blank" title="Click to know more about HV">HV</a> = (6.02 xx 10^(<a href="https://interviewquestions.tuteehub.com/tag/23-294845" style="font-weight:bold;" target="_blank" title="Click to know more about 23">23</a>)) (6.6 xx 10^(-34) Js) (250 s^(-1)) = 10^(-<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>) "erg " = 1J`</body></html> | |
| 51802. |
1 mole of photon, each of frequency 2500 s^(-1) , would have approximately a total energy of |
| 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> erg<br/><a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> <a href="https://interviewquestions.tuteehub.com/tag/joule-526299" style="font-weight:bold;" target="_blank" title="Click to know more about JOULE">JOULE</a> <br/>1 eV<br/>1 MeV</p>Solution :`upsilon=2500 s^(-1)` <br/> `therefore` Energy of a photon , E=hv <br/> `=(6.626xx10^(-34)xx2500)`J <br/> `because` Energy of one <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> of photon <br/> `=(6.626xx10^(-34)xx2500xx6.023xx10^23)J` <br/> `=9.977xx10^(-7) J ~~ 10` erg</body></html> | |
| 51803. |
1 mole of NH_(3) gas at 27^(@) C is expanded under reversible adiabatic conditions to make volume 8 times (gamma=1.33). Final temperature and work done respectively are : |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/150-275254" style="font-weight:bold;" target="_blank" title="Click to know more about 150">150</a> <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>, 900 cal<br/>150 K, <a href="https://interviewquestions.tuteehub.com/tag/400-315233" style="font-weight:bold;" target="_blank" title="Click to know more about 400">400</a> cal<br/>250 K, 1000 cal<br/>200 K, 800 cal</p>Answer :A</body></html> | |
| 51804. |
1 mole of PCl_(5) is placed in a closed vessel at 523K. At equilibrium, if it dissociates to an extent of 35%, calculate K_(p)" for "PCl_(5)hArrPCl_(3)+Cl_(2). Equilibrium pressure is found to be 5xx10^(5) Pa. |
| 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>Solution :Initial no. of <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a>: `underset(1)(PCl_(5))hArrunderset(0)(PCl_(3))+underset(0)(Cl_(2))` <br/> `{:("No. of moles present"),("at equilibrium"):}}1-x=0.65" "x""x""(x=0.35)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> number of moles present at equilibrium = 0.65 + 0.35 + 0.35 = 1.35 <br/> We know that partial <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a> = Mole fraction = Mole fraction `xx` total pressure <br/> `thereforeP_(PCl_(5))=(0.65/1.35)xx5xx10^(5)<a href="https://interviewquestions.tuteehub.com/tag/pa-1145246" style="font-weight:bold;" target="_blank" title="Click to know more about PA">PA</a>,P_(PCl_(3))=(0.35/1.35)xx5xx10^(5)Pa,P_(Cl_(2))=(0.35/1.35)xx5xx10^(5)Pa` <br/> `K_(p)=(P_(PCl_(3))xxP_(Cl_(2)))/(P_(PCl_(5)))=((0.35/1.35)xx5xx10^(5)(0.35/1.35)xx5xx10^(5))/((0.65/1.35)xx5xx10^(5))=6.98xx10^(4)Pa`</body></html> | |
| 51805. |
1 mole of N_(2) and 3 mole of H_(2) are mixed in a closed vessel of 1 dm^(3) capacity. At equilibrium if the vessel contains a total of 2.4 moles, calculate equilibrium constant K_(c)" for "N_(2)+3H_(2)hArr2NH_(3). |
| Answer» <html><body><p></p>Solution :Initial no. of moles: `underset(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)(N_(2))+underset(3)(3H_(2))hArrunderset(0)(2NH_(3))` <br/> `{:("No. of moles reacting"),("at equilibrium"):}}x""3x""-` <br/> `{:("No. of moles present"),("at equilibrium"):}}1-x""3-3x""2x` <br/> `therefore` <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> no. of moles present at equilibrium = (1 - x) + (3 - 3x) + 2x = 4 - 2x <br/> Given 4 - 2x = 2.4 `""therefore2x=4-2.4orx=1.6/2=0.8` <br/> Since the volume of the vessel is 1 `<a href="https://interviewquestions.tuteehub.com/tag/dm-432223" style="font-weight:bold;" target="_blank" title="Click to know more about DM">DM</a>^(3)`, <br/> `[N_(2)]=(1-x)/1=(1-0.8)/1=0.2" mol"//dm^(3),[H_(2)]=(3-3x)/1=(3-2.4)/1=0.6"mol"//dm^(3)` <br/> `[NH_(3)]=(2x)/1=1.6/1=1.6"mol"//dm^(3)` <br/> `K_(c)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))=((1.6)^(2))/((0.2)xx(0.6)^(3))=59.26("mol"//dm^(3))^(-2)`.</body></html> | |
| 51806. |
1 mole of methane (CH_4) contains |
| Answer» <html><body><p>`6.02 xx 10^23` atoms of H<br/>4 gram atoms of <a href="https://interviewquestions.tuteehub.com/tag/hydrogen-22331" style="font-weight:bold;" target="_blank" title="Click to know more about HYDROGEN">HYDROGEN</a><br/>`1.81 xx 10^23` molecules of methane<br/>3.0 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> of carbon.</p>Answer :B</body></html> | |
| 51807. |
1 mole ofIO_3^(-)ions is heated with excess ofI^(-) ions in the presence of acidic conditions as per the following equationIO_(3)^(-) +I^(-) rarrI_2 .How many moles of acidified hypo solution will be required toreact completley withI_2thus produced ? |
| Answer» <html><body><p></p>Solution :`10e^(-)+2IO_3^(-) rarrI_2,2I^(-) rarrI_2+2e^(-)]xx5` <br/> `2IO_3^(-)+10I^(-) rarr6I_2` <br/>`<a href="https://interviewquestions.tuteehub.com/tag/2mol-1837971" style="font-weight:bold;" target="_blank" title="Click to know more about 2MOL">2MOL</a>""10mol ""6mol`<br/> ii) `I_2+2S_2O_3^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>-)rarrS_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)O_6^(2-)+2I^(-)` <br/> 1 mol 2 mol<br/> 1 mol of `IO_3^(-) =5 ` mol `5I^(-) -= <a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a> ` mol of `I_2 =6` mol of `S_2O_3^(2-)`</body></html> | |
| 51808. |
1 mole of H_(2)SO_(4)will exactly neutralise: |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> mole of ammonia<br/>1 mole of `<a href="https://interviewquestions.tuteehub.com/tag/ba-389206" style="font-weight:bold;" target="_blank" title="Click to know more about BA">BA</a>(<a href="https://interviewquestions.tuteehub.com/tag/oh-585115" style="font-weight:bold;" target="_blank" title="Click to know more about OH">OH</a>)_(2)` <br/>0.5 moles of `Ca(OH)_(2)` <br/>2 mole of KOH</p>Answer :A::B::D</body></html> | |
| 51809. |
1 mole of each of Ca(OH)_(2) and H_(3) PO_(4) are allowed to react under dilute conditions . The maximum number of moles of Ca_(3) (PO_(4))_(2) formed is |
| 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><br/>`1//2`<br/>`1//3`<br/>3</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> | |
| 51810. |
1 mole of chlorine combines with a certain mass of a metal giving 111g of its chloride. The atomic mass of the metal (assuming its valency to be 2) is |
| Answer» <html><body><p>40<br/>20<br/>80<br/>60</p>Solution :`("<a href="https://interviewquestions.tuteehub.com/tag/wt-1462289" style="font-weight:bold;" target="_blank" title="Click to know more about WT">WT</a> of <a href="https://interviewquestions.tuteehub.com/tag/metal-1094457" style="font-weight:bold;" target="_blank" title="Click to know more about METAL">METAL</a> chloride ")/("wt of chlorine ")=(E_("Metal")+E_(<a href="https://interviewquestions.tuteehub.com/tag/cl-408888" style="font-weight:bold;" target="_blank" title="Click to know more about CL">CL</a>^(-)))/(E_(Cl^(-)))`</body></html> | |
| 51811. |
1 mole of an ideal monatomic gas is subjectedtoi the followingreversible charge of stateAunderset("Expansion")overset("Isothermal")toBunderset("Cooling") overset("Isochoric")to (5 atm , 500K)Cunderset("compression")overset("Adiabatic")toA(300K)Then , which of the following are correct?[ Given(0.6)^(2.5) =0.3 , "in"2=0.7](R=0.08 L-atm//"mol"-K=2cal//"mol"-k) |
| Answer» <html><body><p>Pressure of point B is 2.5 atm<br/>Volume at point <a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> is <a href="https://interviewquestions.tuteehub.com/tag/16-276476" style="font-weight:bold;" target="_blank" title="Click to know more about 16">16</a> litre<br/>Magnitude of work <a href="https://interviewquestions.tuteehub.com/tag/involved-7257329" style="font-weight:bold;" target="_blank" title="Click to know more about INVOLVED">INVOLVED</a> in complete process is <a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a> cal<br/>Change in enthalpy of process of process `C-A` is`+1` kcal .</p>Answer :a,b,c,d</body></html> | |
| 51812. |
1 mole of any gas a) Occupies 22.4 lit at STP b) Contains 3.05 xx 10^22 molecules c) Contains 6.023 xx 10^23 molecules d) Contain same number of molecules as in 22 gm of CO_2 |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>, d <br/>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><br/>b, c <br/>a, d</p>Answer :B</body></html> | |
| 51813. |
1 mole of an ideal gas, maintained at 4.1 atm and at a certain temperature, absorbs heat 3710 J and expands to 2 litres. Calculate the entropy change in expansion process. |
| 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> :`n=<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>"mole"` <br/> `P=4.1atm` <br/> `V=2" <a href="https://interviewquestions.tuteehub.com/tag/lit-537267" style="font-weight:bold;" target="_blank" title="Click to know more about LIT">LIT</a>"` <br/> `T=?` <br/> `q=3710J` <br/> `DeltaS=(q)/(T)` <br/> `DeltaS=(q)/(((PV)/(<a href="https://interviewquestions.tuteehub.com/tag/nr-581903" style="font-weight:bold;" target="_blank" title="Click to know more about NR">NR</a>)))` <br/> `DeltaS=(nRq)/(PV)` <br/> `DeltaS=(1xx0.082" lit atm "<a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>^(-1)xx3710J)/(4.1" atm"xx2" lit")` <br/> `DeltaS=37.10JK^(-1)`.</body></html> | |
| 51814. |
1 mole of an ideal gas, maintained at 4.1 atm and at a certain temperature , absorbs heat 3710 J and expands to 2 litres . Calculate the entropy changes in expansion process. |
| 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 href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a> of an ideal gas =`P_i` = 4.1 atm. <br/> <a href="https://interviewquestions.tuteehub.com/tag/expansion-980011" style="font-weight:bold;" target="_blank" title="Click to know more about EXPANSION">EXPANSION</a> in volume `DeltaV` = 2 litres <br/> Heat absorbed =<a href="https://interviewquestions.tuteehub.com/tag/q-609558" style="font-weight:bold;" target="_blank" title="Click to know more about Q">Q</a> = 3710 J <br/>Entropy change = `DeltaS` = ?<br/>For an ideal gas PV = RT for one mole . <br/> `T=(PV)/<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>=(4.1xx2)/0.0830=100^@C` <br/> T=100+273 = 373 K <br/>`DeltaS=q/T_((K))=3710/373` <br/> Entropy change =9.946 `JK^(-1)`</body></html> | |
| 51815. |
1 mole of an ideal gas A(C_(v,m)=3R) and 2 mole of an ideal gas B are (C_(v,m)=(3)/(2)R) taken in a container and expended reversibly and adiabatically from 1 litre to 4 litre starting from initial temperature of 320 K. DeltaE "or" DeltaU for the process is : |
| Answer» <html><body><p>`-<a href="https://interviewquestions.tuteehub.com/tag/240-295938" style="font-weight:bold;" target="_blank" title="Click to know more about 240">240</a> <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>`<br/>`+240 R`<br/><a href="https://interviewquestions.tuteehub.com/tag/480-1879936" style="font-weight:bold;" target="_blank" title="Click to know more about 480">480</a> R<br/>`-<a href="https://interviewquestions.tuteehub.com/tag/960-342633" style="font-weight:bold;" target="_blank" title="Click to know more about 960">960</a> R`</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> | |
| 51816. |
1 mole of an ideal gas A (C_(v,m)=3R) and 2 mole of an ideal gas B are (C_(v.m)= (3)/(2)R) taken in a constainer and expanded reversible and adiabatically from 1 litre of 4 litre starting from initial temperature of 320K. DeltaE or DeltaU for the process is (in Cal) (Give your answer after divide with 240) |
| Answer» <html><body><p>`-240R`<br/>240R<br/>`-<a href="https://interviewquestions.tuteehub.com/tag/1920-1800891" style="font-weight:bold;" target="_blank" title="Click to know more about 1920">1920</a>` Cal<br/>`-960R`</p>Solution :`C_(V_("avg")) = (<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> xx 3R + 2 xx (3)/(2) <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>)/(1 +2) = 2R rArr gamma = 3//2` <br/> `T_(1) V_(1)^(gamma-1) = T_(2) V_(2)^(gamma-1)` <br/> `320 xx (1)^(1//2) = T_(2) (4)^(1//2) rArr T_(2) = (320)/(2) = 160K` <br/> `Delta U= nC_(v) Delta T= 3 xx 2R xx (160-320)` <br/> `= -<a href="https://interviewquestions.tuteehub.com/tag/960-342633" style="font-weight:bold;" target="_blank" title="Click to know more about 960">960</a> R = - 1920` cal</body></html> | |
| 51817. |
1 mole of a diatomic element X_2 contains 34 and 40 moles of electrons and neutrons respectively. The isotopic formula of the element is: |
| Answer» <html><body><p>`""_(<a href="https://interviewquestions.tuteehub.com/tag/34-308171" style="font-weight:bold;" target="_blank" title="Click to know more about 34">34</a>)^(74)<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>`<br/>`""_(<a href="https://interviewquestions.tuteehub.com/tag/17-278001" style="font-weight:bold;" target="_blank" title="Click to know more about 17">17</a>)^(37)X`<br/>`""_(34)^(40)X`<br/>`""_(20)^(40)X` </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> | |
| 51818. |
1 mole of a compound contains 1 mole of C and 2 moles of O. The molecular weight of the compound is |
| Answer» <html><body><p>3<br/>12<br/>32<br/>44</p>Answer :D</body></html> | |
| 51819. |
Calculate the number of protons, neutrons and electrons respectively in ""_(7)^(14) N^(3-) |
| Answer» <html><body><p>`7xx6.023xx10^(23)` <a href="https://interviewquestions.tuteehub.com/tag/electrons-969138" style="font-weight:bold;" target="_blank" title="Click to know more about ELECTRONS">ELECTRONS</a> <br/>`7xx6.023xx10^(23)` protons <br/>`7xx6.023xx10^(23)` neutrons <br/>`14xx6.023xx10^(23)` protons</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> implies 1 mole ions <br/> implies 7 mole protons <br/> 7 mole neutrons <br/> 10 mole `e^(-)`</body></html> | |
| 51820. |
1 mole each of H_2(g)and I_2(g) are introduced in a 1L evacuated vessel at 523 K and equilibrium H_2(g)+I_2(g)hArr2HI(g) is established. The concentration of HI(g) at equilibrium: |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/changes-913881" style="font-weight:bold;" target="_blank" title="Click to know more about CHANGES">CHANGES</a> on changing pressure. <br/><a href="https://interviewquestions.tuteehub.com/tag/change-913808" style="font-weight:bold;" target="_blank" title="Click to know more about CHANGE">CHANGE</a> on changing temperature. <br/>is same even if only <a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> mol of HI(g) were introduced in the vessel in the <a href="https://interviewquestions.tuteehub.com/tag/begining-894533" style="font-weight:bold;" target="_blank" title="Click to know more about BEGINING">BEGINING</a>. <br/>is same even when a platinum gauze is introduced to catalyse the reaction. </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::B::C::D</body></html> | |
| 51821. |
1 mole each of H_(2)(g) "and" I_(2)(g) are introduced in a 1L evacuated vessel at 523K and equilibriumH_(2)(g)+I_(2)(g)hArr2Hi(g) is established. The concentration of HI(g) at equilibrium: |
| Answer» <html><body><p>Changes on changing <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/>Changes on changing temperature.<br/>Changes on changing volume of the vessel<br/>Is same even if only `2` mol of `HI(g)` were introduced in the vessel in the beginning. <br/> Is same even when a platinum gauze is introduced to catalyst the reaction.</p>Solution :`H_(2)(g)+I_(2)(g)hArr2HI(g)` <br/> (A) For changing pressure volume has to be changed, <a href="https://interviewquestions.tuteehub.com/tag/though-2306877" style="font-weight:bold;" target="_blank" title="Click to know more about THOUGH">THOUGH</a> <a href="https://interviewquestions.tuteehub.com/tag/number-582134" style="font-weight:bold;" target="_blank" title="Click to know more about NUMBER">NUMBER</a> of moles of `HI(g)` do not get changed but its concentration will get changed. <br/> (B) Temperature change will change `K_(P)` and hence concentration. <br/> (C) Volume change will change concentration, not the number of moles. <br/> (D) Same equilibrium will be attained from <a href="https://interviewquestions.tuteehub.com/tag/either-7265046" style="font-weight:bold;" target="_blank" title="Click to know more about EITHER">EITHER</a> direction. <br/> Catalyst does not change equilibrium concentration.</body></html> | |
| 51822. |
1 mole each of FeC_(2)O_4and FeSO_4is oxidised separately by 1M KMnO_4in acid medium. Calculate the volume ratio of KMnO_4used for FeC_(2) O_4 and FeSO_4 |
| Answer» <html><body><p><br/></p>Solution :`Fe^(2+)+C_(2)O_(4)^(2-) rarrFe^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>+)+2CO_(2)^(4-)+3e` <br/> `Fe^(2+) +SO_(4)^(2-) rarrFe^(3+)+SO_(2)^(4-)e` <br/> `Mn^(<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>+)+5erarrMn^(2+)` <br/> Meq.of `F_(e)C_(2)O_4` = Meq.of `KMnO_4` <br/> `1xx3xx1000=1xx5xxV_(1).......(1)` <br/> Meq.of `FeSO_(4)` = Meq.of `KMnO_4` <br/> `1xx1 xx1000=1xx5xxV_2""....(2)` <br/> By <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a>.(1) and (2) `V_1/V_2=3`</body></html> | |
| 51823. |
1 mole each of CaC_(2), Mg_(2)C_(3) reacts with excess water in separate open flasks work done by the gas during the dissolution shows the order: |
| Answer» <html><body><p>`CaC_(2) = Mg_(2)C_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>) lt Al_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)C_(3)`<br/>`CaC_(2) = Mg_(2)C_(3) = Al_(4)C_(3)`<br/>`Mg_(2)C_(3) lt CaC_(-2) lt Al_(40C_(3)`<br/>`Mg_(2)C_(3) lt Al_(4)C_(3) lt CaC_(2)`</p>Solution :`CaC_(2) <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)` <br/> `Mg_(2)C_(3) rarr C_(3)H_(4)` <br/> `Al_(4)C_(3) rarr CH_(4)`</body></html> | |
| 51824. |
1 mole CH_(4) contain.... |
| Answer» <html><body><p>`6.022xx10^(<a href="https://interviewquestions.tuteehub.com/tag/23-294845" style="font-weight:bold;" target="_blank" title="Click to know more about 23">23</a>)` hydrogen <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/>`6.022xx10^(23)` atoms of hydrogen<br/>8g <a href="https://interviewquestions.tuteehub.com/tag/molecules-563030" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULES">MOLECULES</a> of hydrogen<br/>4 g atoms of hydrogen</p>Answer :A::D</body></html> | |
| 51825. |
1mole CaCO_(3(s)) isheatedin11.2litvesselsothatequilibriumisestablishedat819 K.IfK_pfor CaCO_3 harr CaO + CO_2 at this temperature is 2 atm, equilibrium concentration of CO_2 (in mol-lit^(-1)) |
| Answer» <html><body><p>`1//3`<br/>`1/(11.2)`<br/>`1/(33.6)`<br/>`1/(22.4)`</p>Solution :`CaCO_(3(s)) harr CaO_((s))+CO_(2(g))` <br/> `K_(P)=PCO_(2)=2` <br/> we know that `K_(P)=K_(<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>)(RT)^(Deltan)` <br/> `K_(c)=(K_(P))/((RT)^(<a href="https://interviewquestions.tuteehub.com/tag/delta-947703" style="font-weight:bold;" target="_blank" title="Click to know more about DELTA">DELTA</a> <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>))=(2)/((0.0821 xx 819)^(1))=(1)/(33.6)`</body></html> | |
| 51826. |
1 mole Ba(OH)_(2) will exactly neutralize |
| Answer» <html><body><p>0.5 <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> HCl<br/><a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> mole `H_(2)SO_(4)`<br/>1 mole of `H_(3)PO_(3)`<br/>2 mole of `H_(3)PO_(2)` </p>Solution :1 Mole `<a href="https://interviewquestions.tuteehub.com/tag/ba-389206" style="font-weight:bold;" target="_blank" title="Click to know more about BA">BA</a>(OH)_(2)` = 2 equivalents `Ba(OH)_(2)` <br/> 1 mole HCl = 1 <a href="https://interviewquestions.tuteehub.com/tag/equivalent-446407" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENT">EQUIVALENT</a> HCl <br/> 1 mole `H_(3)PO_(3)` (di-basic acid) <br/> = 2 equivalent `H_(3)PO_(3)` <br/> 1mole `H_(3)PO_(2)` (monobasic acid) <br/> = 1 equivalent `H_(3)PO_(2)`</body></html> | |
| 51827. |
1 molar solution of a non-volatile and non-electrolyte compound will produce an osmotic pressure ….. At 0^(@)C |
| 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> atm<br/>44.8 atm<br/>0.5 atm<br/>0.75 atm</p>Answer :D</body></html> | |
| 51828. |
1 mol PCl_(3) and 1 mol Cl_2 are taken in a 10 lit vessel and heated . Now , if 0.8 mol PCl_(5) is present at equilibrium , then |
| Answer» <html><body><p>`[PCl_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)] = 0.8` M <br/>`[Cl_(2)] = 0.02` M<br/>`[PCl_(5)] = 0.02` M<br/>`[PCl_(3)] = [Cl_(2)]= 0.2 `M</p>Answer :<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> | |
| 51829. |
1 mol of PCl_(5) , kept in a closed container of volume of 1 dm^(3) and was allowed to attain equilibrium at 423 K . Calculate the equilibrium composition of reaction mixture . (The K_(C) value for PCl_(5) dissociation at 423 K is 2 ) |
| Answer» <html><body><p></p>Solution :`PCl_(5) hArr PCl_(3) + Cl_(2)` <br/> Given that `[PCl_(5)]_("initial") =1` mol , `V = 1 dm^(3) , K_(C) = 2` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_CHE_XI_V02_C08_E01_055_S01.png" width="80%"/> <br/> `K_(C) = ([PCl_(3)][Cl_(2)])/([PCl_(5)])` <br/> `2 = ( x xx x)/((1 - x))` <br/> `2 - 2x = x^(2)` <br/> `x^(2) + 2x - 2 = 0` <br/> Solution for a quadratic equation <br/> `ax^(2) + bx + c = 0` are `x = (- <a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a> pm sqrt(b^(2) - 4ac))/(2a)` <br/> `a = 1 . b = 2 . c = -2` <br/> `x = ( - 2 pm sqrt(4 - <a href="https://interviewquestions.tuteehub.com/tag/4xx-1883349" style="font-weight:bold;" target="_blank" title="Click to know more about 4XX">4XX</a> 1 xx (-2)))/(2 xx 1) = (-2 pm sqrt(12))/(2) = (-2 pm sqrt(4 xx 3))/(2)` <br/> `x = (-2 pm 2 sqrt3)/(2) = (- 2 + 2 sqrt3)/(2) , ( - 2 - 2 sqrt3)/(3)` <br/> `x = -1+ sqrt3 , - 1 - sqrt3` (<a href="https://interviewquestions.tuteehub.com/tag/since-644476" style="font-weight:bold;" target="_blank" title="Click to know more about SINCE">SINCE</a> x is +ve ) , `-1 - sqrt3` not possible <br/> `= - 1 + 1.732 = 0.732` <br/> `therefore` Equilibrium concentration of `[PCl_(5)]_(<a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a>) = (1- x)/(1) = 1 - 0.732 = 0. <a href="https://interviewquestions.tuteehub.com/tag/268-1832096" style="font-weight:bold;" target="_blank" title="Click to know more about 268">268</a>` M <br/> `[PCl_(3)]_(eq) = (x)/(1) = (0.732)/(1) = 0.732` <br/> `[Cl_(2)]_(eq) = (x)/(1) = (0.732)/(1) = 0.732`</body></html> | |
| 51830. |
1 mol of NH_(3) gas( gamma = 1.33) at 27^(@)Cis allowed to expand adiabatically so that the final volume becomes 8 times. Work done will be |
| Answer» <html><body><p>450 cal<br/> 1800 cal<br/>900 cal<br/>300 cal</p>Solution :Work done in <a href="https://interviewquestions.tuteehub.com/tag/adiabatic-849856" style="font-weight:bold;" target="_blank" title="Click to know more about ADIABATIC">ADIABATIC</a> expansion` = - nC_(v) (T_(2) -T_(1))`. To calculate <a href="https://interviewquestions.tuteehub.com/tag/final-461168" style="font-weight:bold;" target="_blank" title="Click to know more about FINAL">FINAL</a> tem. `T_(2)`, proceeding as in the above case, `T_(2) = 150 <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>`.<br/> `C_(v)` for `NH_(3)= 3R` <br/> `( NH_(3)` is a non-linear molecule with`N=4` .<a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a>, translational degrees`=3`, Rotational degrees`=3`. Total `U = ( 3)/(2) RT + (3)/(2)RT =3RT`) <br/> `:. w=-1 xx3 xx2(150-300) = 900cal`</body></html> | |
| 51831. |
1 mol of equimolar mixture of ferric oxalate and ferrous oxalate will require x mol of KMnO_(4) in acidic medium for complete oxidation. X is |
| Answer» <html><body><p>0.8 mol<br/>0.9 mol<br/>1.6 mol <br/>1.8 mol </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> :Equimolar implies 0.5 <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> each <br/> eq. `FeC_(2)O_(4)+eq. Fe_(2)(C_(2)O_(4))_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)=eq. KMnO_(4)` <br/> `0.5xx3+0.5xx6=x xx5impliesx=0.9`</body></html> | |
| 51832. |
1 mol of CH_4, 1 mole of CS_2 and 2 mole of H_2S are 2 mol of H_2 are mixed in a 500 mL flask. The equilibrium constant for the reaction K_C = 4 xx 10^(-2) mol^2 lit^(-2). In which direaction will the reaction proceed to reach equilibrium ? |
| Answer» <html><body><p></p>Solution :`CH_4 (g)+ 2H_2S (g) hArr CS_2(g) + 4H_2(g)`<br/> `K_C =<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a> xx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) mol^2 L^(-2)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a>`= 500 mL= 1/2 L` <br/> `[CH_4]= (1 mol)/(1/2L) = 2mol L^(-1)`<br/> `[CS_2] = (1 mol)/(1/2L) = 2mol L^(-1)`<br/> `[H_2S] = (2 mol)/(1/2L) = 4mol L^(-1)`<br/> `[H_2]= (2mol)/(1/2L) = 4 mol L^(-1)` <br/> `Q = ([CS_2][H_2]^4)/([CH_4][H_2S]^2)` <br/> `:. Q = (2 xx (4)^4)/(2 xx (4)^2) = 16` <br/> `Q gt K_C` <br/> `:.` The reaction will <a href="https://interviewquestions.tuteehub.com/tag/proceed-592932" style="font-weight:bold;" target="_blank" title="Click to know more about PROCEED">PROCEED</a> in the reverse direction to <a href="https://interviewquestions.tuteehub.com/tag/reach-1178062" style="font-weight:bold;" target="_blank" title="Click to know more about REACH">REACH</a> the equilibrium.</body></html> | |
| 51833. |
1 mol of CH_(4), 1 mole of CS_(2) and 2 mol of H_(2)S are 2 mol of H_(2) are mixed in a 500 ml flask The equilibrium constant for the reaction K_(C)=4xx10^(-2)"mol"^(2)" lit"^(-2). In which direcition will the reaction proceed to reach equilibrium ? |
| Answer» <html><body><p></p>Solution :`CH_(4)(g)+2H_(2)S(g)hArrCS_(2)(g)+4H_(2)(g)` <br/> `K_(C)=4xx10^(-2)" <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> <a href="https://interviewquestions.tuteehub.com/tag/lit-537267" style="font-weight:bold;" target="_blank" title="Click to know more about LIT">LIT</a>"^(-2)` <br/> `"Volume = 500 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a>"=1/2L` <br/> `{:([CH_(4)]_("in")=("1 mol")/(1)""[CS_(2)]_("in")=("1 mol")/(1/2L)),("= 2 mol L"^(-1)"= 2 mol L"^(-1)),([H_(2)S]_("in")=("2 mol")/(1/2L)""[H_(2)]=("2 mol")/(1/2L)),("= 4 mol L"^(-1)"= 4 mol L"^(-1)):}`<br/> `Q=([CS_(2)][H_(2)]^(4))/([CH_(4)][H_(2)S]^(2))` <br/> `:.Q=(2xx(4)^(4))/((2)xx(2)^(2))=64` <br/> `QgtKC` <br/> The reaction will proceed in the reverse direation to reach the equilibrium.</body></html> | |
| 51834. |
1 mol N_2 and 3 mol H_2 taken in 4L definite temperature of closed vessel. Reaction N_(2(g)) + 3H_(2(g)) hArr2NH_(3(g)) according to 0.25% N_2 convert into ammonia. Calculate K_c and how much reaction of K_c? 1/2N_(2(g)) + 3/2 H_(2(g)) hArr NH_(3(g)) |
| 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> :`K_c=1.49xx10^(-5) ("mol L"^(-1) )^(-2)` and `K_c=<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>.86xx10^(-3) ("mol L"^(-1))`</body></html> | |
| 51835. |
1 mol N_2 and 3 mol H_2 heated at 473 K and 100 atm pressure. At equilibrium moles of NH_3 is 0.5 mol. Than calculate the equilibrium constant of the given reaction N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g)) |
| 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> :`K_p=7.5xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)`</body></html> | |
| 51836. |
1 mol A_((g)) is heated in1 lit closed vessel and equilibrium is reached at 300^(@) C in A_((g)) hArr B_((g)) . If K_(C) =4 , concentration of B_((g)) at equilibrium is (in mol/lit) |
| Answer» <html><body><p>`0.2`<br/>`0.6`<br/>`0.8`<br/>`0.1`</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> | |
| 51837. |
1 ml of 13.6 M HCl is diluted with water to give 1 litre of the solution. Calculate pH of the resulting solution. |
| Answer» <html><body><p><br/></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> :`1.67`</body></html> | |
| 51838. |
1 ml of 1 M solution is mixed with 999 ml of pure water. |
| Answer» <html><body><p>`10^(-3)` M solution is formed <br/>The mass of <a href="https://interviewquestions.tuteehub.com/tag/solute-1217068" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTE">SOLUTE</a> per ml <a href="https://interviewquestions.tuteehub.com/tag/decreases-946143" style="font-weight:bold;" target="_blank" title="Click to know more about DECREASES">DECREASES</a> by <a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a> times<br/>The quantity of solute <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 solution<br/>10 ml of resultant solutoin contains `10^(-5)` moles of solute </p>Solution :`M_(1)V_(1)=M_(2)V_(2)` <br/> `1xx1=M_(2)xx1000=M_(2)=10^(-3)M` <br/> 10 ml contains `(10)/(1000)xx10^(-3)` moles `=10^(-5)` moles Quantity of solute in solution remains same but mass of solute/ml decreases by `10^(3)` times</body></html> | |
| 51839. |
1-Methylcyclopentene can be converted into the given compound by the use of which of the followign reagetns? |
| Answer» <html><body><p>`BD_(<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/followed-2079285" style="font-weight:bold;" target="_blank" title="Click to know more about FOLLOWED">FOLLOWED</a> by `HCOOH`<br/>`BH_(3)` followed HCOOD<br/>`BD_(3)` followed by HCOOD<br/>`BH_(3)` followed by `CH_(3)-overset(O)overset(||)(C)-O-D`</p>Answer :B::D</body></html> | |
| 51840. |
1 litres of oxygen is allowed to react with three litres of carbon monoxide at N.T.P. Calculate the volume of each gas found after the reaction. |
| Answer» <html><body><p><br/></p>Solution :The reaction takes place <a href="https://interviewquestions.tuteehub.com/tag/according-366619" style="font-weight:bold;" target="_blank" title="Click to know more about ACCORDING">ACCORDING</a> to the equation : <br/> `underset(<a href="https://interviewquestions.tuteehub.com/tag/2l-300409" style="font-weight:bold;" target="_blank" title="Click to know more about 2L">2L</a>)(2CO)+underset(1L)(O_(2))rarrunderset(2L)(2CO_(2))`<br/> 1L of `O_(2)` <a href="https://interviewquestions.tuteehub.com/tag/reacts-1178303" style="font-weight:bold;" target="_blank" title="Click to know more about REACTS">REACTS</a> with CO = 2L <br/> `:.` <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> of CO left unreacted `= 3-2 = 1L` <br/> Volume of `CO_(2)` formed = 2L.</body></html> | |
| 51841. |
1 litre of N_(2) and 7/8 litre of O_(2) at the same temperature and pressure were mixed together. What is the relationship between the masses of the two gases in the mixture ? |
| Answer» <html><body><p></p>Solution :`PV=nRT=(w)/(M)RT` <br/> For`N_(2),"" Pxx1=(w_(N_(2)))/(<a href="https://interviewquestions.tuteehub.com/tag/28-299271" style="font-weight:bold;" target="_blank" title="Click to know more about 28">28</a>)xxRT ""`……(i)<br/> For`O_(2), "" Pxx(7)/(<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>)=(w_(N_(2)))/(32)xxRT ""`......(<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>) <br/> Dividing eqn.(ii) by eqn. (i)<br/> `(7)/(8)=(w_(O_(2)))/(32)<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>(28)/(w_(N_(2)))`. This give `w_(N_(2))=w_(O_(2))` <br/> Thus, the <a href="https://interviewquestions.tuteehub.com/tag/mixture-1098735" style="font-weight:bold;" target="_blank" title="Click to know more about MIXTURE">MIXTURE</a> contains equal masses of `N_(2)` and `O_(2)`.</body></html> | |
| 51842. |
1 lit of oxygen and 3 lit of SO_(2) at STP are reacted to produce sulphur trioxide . Then the ratio of between volume of sulphur trioxide and that of sulphur dioxide after reaction and weight of SO_3 formed (in grams) respectively are |
| Answer» <html><body><p>`1:<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> , 7:14 gms` <br/>`2:1 , 14.28 gms` <br/>`2:1 , 7.14 gms`<br/>`1:1 , 14.28 gms`</p>Solution :`2SO_(2)+O_(2)rarr2SO_(3)` <br/> 2 lit `<a href="https://interviewquestions.tuteehub.com/tag/larr-536544" style="font-weight:bold;" target="_blank" title="Click to know more about LARR">LARR</a>` 1 lit `larr` 2 lit <br/> <a href="https://interviewquestions.tuteehub.com/tag/implies-1037962" style="font-weight:bold;" target="_blank" title="Click to know more about IMPLIES">IMPLIES</a> 1 lit `SO_(2)` left <br/> `therefore (SO_(3))/(SO_(2))=(2)/(1)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/weight-1451304" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHT">WEIGHT</a> of `SO_(3)=(2)/(22.4)xx80=7.14g`</body></html> | |
| 51843. |
1 lit of M/10 Ba(MnO_(4))_(2), in acidic medium can be oxidised completely with 1/6 lit of x M ferricoxalate. The value of x is |
| Answer» <html><body><p><br/></p>Solution :(a) `<a href="https://interviewquestions.tuteehub.com/tag/10e-1774335" style="font-weight:bold;" target="_blank" title="Click to know more about 10E">10E</a>^(-) +2MnO_4^(-) rarr2Mn^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>+)(n=5)` <br/> Eq of `MnO_4^(-) -= Eq ` of `Fe_(2)(C_2O_4)_3`<br/> (n=5) `""` (n=6)<br/> `1LxxM/(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)xx10 -=1/6 L xx6 xx "x" , x = 1M `</body></html> | |
| 51844. |
1 kg of a sample of water contained 222 mg of CaCl_2and 219 mg of Mg(HCO_3)_2 . The permanent and temporary hardness are .... ppm and .... ppm |
| Answer» <html><body><p> <a href="https://interviewquestions.tuteehub.com/tag/200-288914" style="font-weight:bold;" target="_blank" title="Click to know more about 200">200</a>, 200 <br/>200, 150<br/>200, 300<br/>150, 220</p>Answer :<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> | |
| 51845. |
1 kg of 2m urea solution is mixed with 2 kg of 4m urea solution. The molality of the resulting solution is |
| Answer» <html><body><p>`3.33m`<br/><a href="https://interviewquestions.tuteehub.com/tag/10m-267115" style="font-weight:bold;" target="_blank" title="Click to know more about 10M">10M</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.67m<br/>5m</p>Solution :`(m_(1)w_(1)+m_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)w_(2))/(w_(1)+w_(2))=m` (<a href="https://interviewquestions.tuteehub.com/tag/approx-882876" style="font-weight:bold;" target="_blank" title="Click to know more about APPROX">APPROX</a>)</body></html> | |
| 51846. |
1-Hydroxypropane can be obtained from propene by treating with |
| Answer» <html><body><p>`B_(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>)+NaOH`<br/>`B_(2)H_(6)+H_(2)O //OH^(-)`<br/>`LiAlH_(4) +H_(2)O //H^(+)`<br/>`HCl+H_(2)O //H^(+)`.</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> :`CH_(3)CH=CH_(2) underset((ii)H_(2)O_(2)//OH^(-))<a href="https://interviewquestions.tuteehub.com/tag/overset-2905731" style="font-weight:bold;" target="_blank" title="Click to know more about OVERSET">OVERSET</a>((i)B_(2)H_(6))rarrCH_(3)CH_(2)CH_(2)OH`</body></html> | |
| 51847. |
1 grams of a carbonate (M_(2)CO_(3))on tratment with excess HCl produces 0.01186 mole of CO_(2). The molar mass of M_(2)CO_(3)in gmol^(-1) is ..... |
| Answer» <html><body><p>1186<br/>84.3<br/>118.6<br/>11.86</p>Solution :`M_(2)CO_(3) + 2HCl rarr MCl_(2) + H_(2)O + CO_(2)` <br/> `(1)/(M_(<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>))` <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> `rarr ""0.01186 ` mole <br/> where , `M_(0) = ` <a href="https://interviewquestions.tuteehub.com/tag/molecular-562994" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULAR">MOLECULAR</a> mass of `M_(2)CO_(3)` <br/> `(1)/(M_(0))= 0.01186` <br/> `:.M_(0)=84.3` gram/mole</body></html> | |
| 51848. |
1 gram of hydrogen contains6xx10^(23) atoms. Then 4 grams of He contains |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/6xx10-1913073" style="font-weight:bold;" target="_blank" title="Click to know more about 6XX10">6XX10</a>^(23)` <a href="https://interviewquestions.tuteehub.com/tag/atoms-887421" style="font-weight:bold;" target="_blank" title="Click to know more about ATOMS">ATOMS</a><br/>`<a href="https://interviewquestions.tuteehub.com/tag/12xx10-1784290" style="font-weight:bold;" target="_blank" title="Click to know more about 12XX10">12XX10</a>^(23)` atoms<br/>`24xx10^(23)` atoms <br/>`1.5xx10^(23)` atoms</p>Solution :1 <a href="https://interviewquestions.tuteehub.com/tag/gm-1008640" style="font-weight:bold;" target="_blank" title="Click to know more about GM">GM</a> atom `H=N_(<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>)` atoms <br/> 1 gm atom `He=N_(0)` atoms</body></html> | |
| 51849. |
1 gram hydrogen is present in 0.0833 mole carbohydrate. The empirical formula of carbohydrate is CH_(2)O. What will be the molecular formula ? |
| Answer» <html><body><p>`C_(5)H_(10)O_(5)`<br/>`C_(3)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>)O_(3)`<br/>`C_(12)H_(22)O_(<a href="https://interviewquestions.tuteehub.com/tag/11-267621" style="font-weight:bold;" target="_blank" title="Click to know more about 11">11</a>)`<br/>`C_(6)H_(12)O_(6)`</p>Solution :In 0.0833 mole carbohydrate1 gm hydrogen is present . <br/> `:.` In 1 mole `=(1xx1)/(0.0833)=12` gm hydrogen <br/> The number 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> atoms `= 12/1 = 12` <br/> <a href="https://interviewquestions.tuteehub.com/tag/molecular-562994" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULAR">MOLECULAR</a> formula `=C_(6)H_(12)O_(6)`</body></html> | |
| 51850. |
1 gram of a carbonate (M_(2)CO_(3)) o treatment with excess HCl produces 0.01186 mole of CO_(2). The molar mass of M_(2)CO_(3) in g mol^(-1) is:- |
| Answer» <html><body><p>118.6<br/>11.86<br/>1186<br/>84.3</p>Answer :D</body></html> | |