Explore topic-wise InterviewSolutions in Class 11.

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.

49101.

A weak acid HX has the dissociation constant 1xx10^(-5)M. It forms a salt NaX on reaction with alkali. The degree of hydrolysis of 0.1 M solution of NaX is

Answer» <html><body><p>`0.0001%`<br/>`0.01%`<br/>`0.1%`<br/>`0.15%`</p>Solution :Hydrolysis reaction is `<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>^(-) + H_(2)O <a href="https://interviewquestions.tuteehub.com/tag/harr-2692945" style="font-weight:bold;" target="_blank" title="Click to know more about HARR">HARR</a> HX+OH^(-)` For a salt of <a href="https://interviewquestions.tuteehub.com/tag/weak-729638" style="font-weight:bold;" target="_blank" title="Click to know more about WEAK">WEAK</a> acid with <a href="https://interviewquestions.tuteehub.com/tag/strong-653928" style="font-weight:bold;" target="_blank" title="Click to know more about STRONG">STRONG</a> base, `K_(h)=(K_(w))/(K_(a))=ch^(2)` <br/> `:. (10^(-14))/(10^(-<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>))=0.1xxh^(2) or h^(2) = 10^(-8) or h = 10^(-4)` <br/> % hydrolysis `= 10^(-4) xx 100 = 10^(-2) = 0.01`</body></html>
Discussion
49102.

A weak acid HX has K_(a) =1 xx 10^(-5)M. If forms a salt NaX onreaction with alkali. The perentage degree of dissociation of 0.1 M solution of NaX is

Answer» <html><body><p>1.0E-5<br/>0.01<br/>0.1<br/>0.15</p>Solution :`underset(("Salt"))(X^(-)) +H_(2)O hArr HX+OH^(-)` <br/> For the salt `K_(h) =(K_(w))/(K_(a)) =Calpha^(2)` <br/> `:. (10^(-4))/(10^(-5)) =0.1 alpha^(2)` <br/> `" or """alpha^(2) =(10^(-14))/(10^(-6)) =10^(-8)` <br/> `alpha =(10^(-8))^(1//2) =10^(-4)` <br/> `% " degree of <a href="https://interviewquestions.tuteehub.com/tag/dissociation-15685" style="font-weight:bold;" target="_blank" title="Click to know more about DISSOCIATION">DISSOCIATION</a> "underset(=10^(-2) =0.01%)(=10^(-4) <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> <a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a>)`</body></html>
Discussion
49103.

A weak acid HX, after mixing with 12 mL of decimolar potash has a pH 5. At the end point of neutralisation 26.6 mL of potash is consumed. What is the dissociation constant of HX ?

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> :`8.2 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(-<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>
Discussion
49104.

A weak acid HX has the dissociation constant 1xx 10 ^(-5)M.It forms a salt NaX on reactions with alkali. The percentage hydrolysis of 0.1 M solution of NaX is :

Answer» <html><body><p>`0.0001 %` <br/>`0.01 %` <br/>` 0.1 %` <br/>` 0.15%` </p>Solution :` h= <a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a> ((K_h)/( <a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>) )=sqrt((K_w)/( K_a.C) )=sqrt( 10 ^(_4)/( 10^(-5)xx 10^(-1))), h = 10 ^(_4) ` <br/> ` % h= 0.01 %`</body></html>
Discussion
49105.

A water sample has ppm level concentration of the following metals: Fe = 0.2 , Mn = 5.0 , Cu = 3.0, Zn = 5.0. The metal that makes the water sample unsuitable for drinking is....

Answer» <html><body><p>Cu<br/>Mn<br/>Zn<br/>Fe</p>Solution :Prescribed <a href="https://interviewquestions.tuteehub.com/tag/level-1072714" style="font-weight:bold;" target="_blank" title="Click to know more about LEVEL">LEVEL</a> of Mn is 0.5 <a href="https://interviewquestions.tuteehub.com/tag/ppm-1162221" style="font-weight:bold;" target="_blank" title="Click to know more about PPM">PPM</a>. So water <a href="https://interviewquestions.tuteehub.com/tag/sample-1194587" style="font-weight:bold;" target="_blank" title="Click to know more about SAMPLE">SAMPLE</a> containing Mn = 5ppm is water unsuitable for <a href="https://interviewquestions.tuteehub.com/tag/drinking-959674" style="font-weight:bold;" target="_blank" title="Click to know more about DRINKING">DRINKING</a>.</body></html>
Discussion
49106.

A wave has a frequency of 3 xx 10^(15) sec^(-1). The energy of that photon is

Answer» <html><body><p>`1.6xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a>)` erg<br/>`3.2xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/11-267621" style="font-weight:bold;" target="_blank" title="Click to know more about 11">11</a>)` erg<br/>`2.0xx10^(-11)` erg <br/>`3xx10^(<a href="https://interviewquestions.tuteehub.com/tag/15-274069" style="font-weight:bold;" target="_blank" title="Click to know more about 15">15</a>)` erg</p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/e-444102" style="font-weight:bold;" target="_blank" title="Click to know more about E">E</a>= <a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a> upsilon`</body></html>
Discussion
49107.

A water sample is said to contain permanent hardness if water contains

Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/sulphates-1234316" style="font-weight:bold;" target="_blank" title="Click to know more about SULPHATES">SULPHATES</a> and chlorides of <a href="https://interviewquestions.tuteehub.com/tag/calcium-412119" style="font-weight:bold;" target="_blank" title="Click to know more about CALCIUM">CALCIUM</a> and magnesium<br/>carbonates of calcium and magnesium<br/>bicarbonates of calcium and magnesium<br/>sulphates and chlorides of sodium and potassium.</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/permanent-590817" style="font-weight:bold;" target="_blank" title="Click to know more about PERMANENT">PERMANENT</a> hardness of <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a> is due to sulphates and chlorides of calcium and magnesium.</body></html>
Discussion
49108.

A water sample has ppm level concentration of following anions F^(-)=10 , SO_4^(2-)=100 , NO_3^(-)=50 The anion/anions that make/makes the water sample unsuitable for drinking is/are

Answer» <html><body><p>only `F^-`<br/>only `SO_4^(2-)`<br/>only `NO_3^-`<br/>both `SO_4^(2-)` and `NO_3^(-)`</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/acceptable-846681" style="font-weight:bold;" target="_blank" title="Click to know more about ACCEPTABLE">ACCEPTABLE</a> <a href="https://interviewquestions.tuteehub.com/tag/level-1072714" style="font-weight:bold;" target="_blank" title="Click to know more about LEVEL">LEVEL</a> of `F^-` is upto 1 ppm, `NO_3^-` is upto 50 ppm and `SO_4^(2-)` is upto 500 ppm `F^-` concentration in water sample is 10 ppm which is higher than the permissible level. <a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>, only `F^-` <a href="https://interviewquestions.tuteehub.com/tag/makes-2811811" style="font-weight:bold;" target="_blank" title="Click to know more about MAKES">MAKES</a> the water sample <a href="https://interviewquestions.tuteehub.com/tag/unsuitable-7379520" style="font-weight:bold;" target="_blank" title="Click to know more about UNSUITABLE">UNSUITABLE</a> for drinking</body></html>
Discussion
49109.

A water sample has ppm level concentration of following anions. F^(-) = 10, SO_(4)^(2-) = 100, NO_(3)^(-) = 50 The anion/anions that make/makes the water sample unsuitable for drinking is/are……..

Answer» <html><body><p>only `NO_(<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/>both `SO_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)^(2-) and NO_(3)^(-)`<br/>only `F^(-)`<br/>only `SO_(4)^(2-)`</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/accepted-366523" style="font-weight:bold;" target="_blank" title="Click to know more about ACCEPTED">ACCEPTED</a> amount of negative ion in sample of <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a> For `F^(-)` up to 1 ppm <br/> `NO_(3)^(-)` up to <a href="https://interviewquestions.tuteehub.com/tag/50-322056" style="font-weight:bold;" target="_blank" title="Click to know more about 50">50</a> ppm `SO_(4)^(2-)` up to 500ppm</body></html>
Discussion
49110.

A water sample has ppm level concentration of following anions F^(-)=10,SO_(4)^(2-)=100,NO_(3)^(-)=50 The anion/anions that make/makes the water sample unsuitable for drinking is/are:

Answer» <html><body><p>only `<a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a>^(-)`<br/>only `SO_4^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>-)`<br/>only `NO_3^(-)` <br/>both `SO_4^(2-) and NO_3^(-)`</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> :only `F^(-)`</body></html>
Discussion
49111.

(A) : Water is a liquid whereas sulphurdi oxide is a gas at room temp (R) : Molecular mass of SO_2 is more than that of H_2 O

Answer» <html><body><p>Both (A) and (R) are true and (R) is the <a href="https://interviewquestions.tuteehub.com/tag/correct-409949" style="font-weight:bold;" target="_blank" title="Click to know more about CORRECT">CORRECT</a> <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 true and (R) is not the correct explanation of (A)<br/>(A) is true but (R) is false <br/>(A) is false but (R) is true </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>
Discussion
49112.

(a) W hat do ou mean by catensation? (b) C Cl_(4) cannot be hydrolysed, but SiCl_(4) can be easily hydrolysed, Why?

Answer» <html><body><p></p>Solution : (a) F or answer, <a href="https://interviewquestions.tuteehub.com/tag/consult-2035487" style="font-weight:bold;" target="_blank" title="Click to know more about CONSULT">CONSULT</a> section 11.6. <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>)For answer, Consult section 11.7.</body></html>
Discussion
49113.

(A): Volume to temperature ratio is constant for a fixed amount of gas at constant pressure. (R) : At constant pressure the volume of a given mass of a gas increases or decreases by 1/273 times of its volume at 0^@C, for every 1^@C change in temp.

Answer» <html><body><p>Both A and R are <a href="https://interviewquestions.tuteehub.com/tag/correct-409949" style="font-weight:bold;" target="_blank" title="Click to know more about CORRECT">CORRECT</a> 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 false. <br/>A is false but R is true </p>Answer :A</body></html>
Discussion
49114.

A vey high cosmic abundance of Fe is due to

Answer» <html><body><p>its ferro-magnetic nature<br/>its high <a href="https://interviewquestions.tuteehub.com/tag/nuclear-2474" style="font-weight:bold;" target="_blank" title="Click to know more about NUCLEAR">NUCLEAR</a> <a href="https://interviewquestions.tuteehub.com/tag/binding-14922" style="font-weight:bold;" target="_blank" title="Click to know more about BINDING">BINDING</a> energy<br/>its high reactivity towards `O_2`<br/><a href="https://interviewquestions.tuteehub.com/tag/none-580659" style="font-weight:bold;" target="_blank" title="Click to know more about NONE">NONE</a> of these.</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>
Discussion
49115.

A vessel with small opening contained equal volumes of oxygen and an unknown gas. Oxygen effused through the opening 1.8 times faster than the unknown gas. If the atomic mass of oxygen is 16, calculate the molecular mass of the unknown gas.

Answer» <html><body><p></p>Solution :Graham.s <a href="https://interviewquestions.tuteehub.com/tag/law-184" style="font-weight:bold;" target="_blank" title="Click to know more about LAW">LAW</a> of diffusion is also applicable to the phenomenon of <a href="https://interviewquestions.tuteehub.com/tag/effusion-966747" style="font-weight:bold;" target="_blank" title="Click to know more about EFFUSION">EFFUSION</a>. According to this law, <br/>`r_1/r_2 = sqrt(M_2/M_1)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/since-644476" style="font-weight:bold;" target="_blank" title="Click to know more about SINCE">SINCE</a>, oxygen <a href="https://interviewquestions.tuteehub.com/tag/effuses-7679868" style="font-weight:bold;" target="_blank" title="Click to know more about EFFUSES">EFFUSES</a> 1.8 times <a href="https://interviewquestions.tuteehub.com/tag/faster-985028" style="font-weight:bold;" target="_blank" title="Click to know more about FASTER">FASTER</a> than the unknown gas, we have `r_1/r_2 = 1.8` <br/> Suppose the molecular mass of the unknown gas is `M_2`. <br/> Hence, `1.8 = sqrt(M_2/32)( :. "Molecular mass of " O_2 = 32)` <br/> or`M_2 = 103.68`</body></html>
Discussion
49116.

A vessel of 120 mL capacity contains a certain mass of a gas at 20^(@)C and 750 mm pressure. Thegas was transferred to a vessel whose volume is 180 mL. Calculate the pressure of the gas at 20^(@)C.

Answer» <html><body><p></p>Solution :Since a gas completely fills the vessel in which it is contained, thesefore, we have : <br/> `V_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)=120 mL,V_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)=180 mL`,`P_(1)=750 mm, P_(2)=?mm` <br/> Since the temperature <a href="https://interviewquestions.tuteehub.com/tag/remains-621920" style="font-weight:bold;" target="_blank" title="Click to know more about REMAINS">REMAINS</a> <a href="https://interviewquestions.tuteehub.com/tag/constant-930172" style="font-weight:bold;" target="_blank" title="Click to know more about CONSTANT">CONSTANT</a>, therefore, by applying Boyle's Law, `P_(1)V_(1)=P_(2)V_(2)`</body></html>
Discussion
49117.

A vessel of 25 litre capacity contains 10 moles of steam under pressure of 50.3 atm. Calculate the temperature of steam using van der Waals equation (if for water a=5.46 bar L^2 mol^(-2) and b=0.031 L mol^(-1))

Answer» <html><body><p>1539.5K<br/>153.95K <br/>15.395K<br/>1.5395K</p>Answer :A</body></html>
Discussion
49118.

A vessel of 120 mL capacity contains a certain amount of gas at temperature 35^(@)C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at temperature 35^(@)C, What would be its pressure ?

Answer» <html><body><p></p>Solution :When <a href="https://interviewquestions.tuteehub.com/tag/fixed-457597" style="font-weight:bold;" target="_blank" title="Click to know more about FIXED">FIXED</a> mass of any gas is <a href="https://interviewquestions.tuteehub.com/tag/poured-7256597" style="font-weight:bold;" target="_blank" title="Click to know more about POURED">POURED</a> to one container to another container then its pressure, volume and temperature is as under.<br/>Container - 1st : Initial volume `(V_(1))=120` <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a><br/>Initial pressure `(p_(1))=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>` bar <br/>Temperature `(T) = 35^(@)C`<br/>Container - 2nd : Final volume `(V_(2))=180` mL<br/>Temperature (T) `= 35^(@)C`<br/>Final pressure `(p_(2))=(?)`<br/>Temperature is constant at `35^(@)C` and mass of gas is constant so according to Boyle.s <a href="https://interviewquestions.tuteehub.com/tag/law-184" style="font-weight:bold;" target="_blank" title="Click to know more about LAW">LAW</a>.<br/>`p_(1)V_(1)=p_(2)V_(2)`<br/>`therefore p_(2)=(p_(1)V_(1))/(V_(2))=(("1.2 bar")xx(120 mL))/(180 mL)=0.8` bar<br/>Pressure of container - 2nd would be 0.8 bar.</body></html>
Discussion
49119.

A vessel of 120 mL capacity contains a certain amount of gas at 35^(@)C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35^(@)C. What would be its pressure ?

Answer» <html><body><p>&lt;<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>&gt;</p>Solution :`V_(1)=<a href="https://interviewquestions.tuteehub.com/tag/120-270396" style="font-weight:bold;" target="_blank" title="Click to know more about 120">120</a> <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a>,""P_(1)=1.2 bar, "" T_(1)=35^(@)C` <br/> `V_(2)=180 mL, ""P_(2)=?, ""T_(2)=35^(@)C` <br/> As temperature remains <a href="https://interviewquestions.tuteehub.com/tag/constant-930172" style="font-weight:bold;" target="_blank" title="Click to know more about CONSTANT">CONSTANT</a>,<a href="https://interviewquestions.tuteehub.com/tag/applying-1982651" style="font-weight:bold;" target="_blank" title="Click to know more about APPLYING">APPLYING</a> Boyle's law, `P_(1)V_(1)=P_(2)V_(2)` <br/> `(1.2"bar")(120 mL)=P_(2)(180 mL) or P_(2)=0.8 "bar"`</body></html>
Discussion
49120.

A vessel of 120 mL capacity contains a certain amount of gas at 35^(@)C and 1.2 bar pressure. The gas 1p transferred to another vessel of volume 180 mL at 35^(@) C. What would be its pressure ?

Answer» <html><body><p></p>Solution :`P_(1)=1*2` bar `P_(2) =?, v_(1)=<a href="https://interviewquestions.tuteehub.com/tag/120-270396" style="font-weight:bold;" target="_blank" title="Click to know more about 120">120</a> <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a>, v_(2)=<a href="https://interviewquestions.tuteehub.com/tag/180-279527" style="font-weight:bold;" target="_blank" title="Click to know more about 180">180</a> mL` <br/> <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a> contant.<br/> According contant. <br/> According to Boyle.s law `P_(1)V_(1)=P_(2)V_(2) `or <br/> `P_(2) =(P_(1)V_(1))/(V_(2))=(1*2 "bar" xx12 mL)/(180mL)=0*8 "bar"`.</body></html>
Discussion
49121.

A vessel of 120 mL capacity contains a certain amount of gas at 35 ^(@)C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35 ^(@)C. What would be its pressure?

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> :Since temperature <a href="https://interviewquestions.tuteehub.com/tag/remains-621920" style="font-weight:bold;" target="_blank" title="Click to know more about REMAINS">REMAINS</a> constant, we have according to Boyle.s <a href="https://interviewquestions.tuteehub.com/tag/law-184" style="font-weight:bold;" target="_blank" title="Click to know more about LAW">LAW</a>, <br/> `P_1 V_1 = P_2 V_2` <br/> or `1.2 xx 120 = P_2 xx <a href="https://interviewquestions.tuteehub.com/tag/180-279527" style="font-weight:bold;" target="_blank" title="Click to know more about 180">180</a>` <br/>or `P_2 =(1.2xx120)/180 = 0.8`bar</body></html>
Discussion
49122.

A vessel of 10L was filled with 6 mole of Sb_(2)S_(3) and 6 mole of H_(2) to attain the equilibrium at 440^(@)C as: Sb_(2)S_(3)(s)+3H_(2)(g)hArr2Sb(s)+3H_(2)S(g) After equilibrium the H_(2)S formed was analysed by dissolving it in water and treating with excess of Pb^(2+) to give 708 g "of" PbS as precipitate. What is value of K_(c) of the reaction at 440^(@)C?(At. weight of Pb=206).

Answer» <html><body><p>`0.08`<br/>`0.8`<br/>`0.4`<br/>`0.04`</p>Solution :<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 `PbS=708//236="mole of" H_(2)S` <br/> `Sb_(2)S_(3)(s)+3H_(2)(g)hArr2Sb(s)+3H_(2)S(g)` <br/> `{:("Initial",6,6,0,0),("at <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a>.",5,3,2,3):}` <br/> `K_(c)=((3//10)^(3)xx(2//10)^(2))/((5//10)xx(3//10)^(3))=(4)/(<a href="https://interviewquestions.tuteehub.com/tag/50-322056" style="font-weight:bold;" target="_blank" title="Click to know more about 50">50</a>)=0.08`</body></html>
Discussion
49123.

A vessel is filled with a mixture of oxygen and nitrogen . At what ratio of partial pressures will the mass of the gases be indentical ?

Answer» <html><body><p></p>Solution :`PV=nRT=(<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)RT` <br/> `P_(O_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))V=(w_(O_(2)))/(32)xxRT` <br/> `P_(N_(2))V=(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` <br/> `:. "" (P_(O_(2)))/(P_(N_(2)))=(w_(O_(2)))/(32)xx(28)/(w_(N_(2)))=(28)/(32)""(because w_(O_(2))=w_(N_(2)))` <br/> `=(<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>)/(8)=7:8`</body></html>
Discussion
49124.

A vessel of irregular shape has a volume 'V'. It is first evacuated and coupled with a vessel of 4L capacity at 35^@C and 10 atm pressure. If the final pressure in both the vessels is 3atm, calculate the volume V.

Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/unknown-1438723" style="font-weight:bold;" target="_blank" title="Click to know more about UNKNOWN">UNKNOWN</a> vesselKnown vessel<br/> `V_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>) = VL ""V_(2) = 4L` <br/> `P_(1) = 0 "atm" P_(2) = 10` atm <br/> The average pressure (P) of the mixture of the <a href="https://interviewquestions.tuteehub.com/tag/gases-13668" style="font-weight:bold;" target="_blank" title="Click to know more about GASES">GASES</a> from two vessels ` = P = (P_(1)V_(1)+ P_(2)V_(2))/(V_(1) + V_(2))` <br/> ` 3 = (0 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> V + 10 xx 4)/(V + 4) = (<a href="https://interviewquestions.tuteehub.com/tag/40-314386" style="font-weight:bold;" target="_blank" title="Click to know more about 40">40</a>)/(V + 4)` <br/> Solving, V = 9.3 L <br/> The volume of the given vessel of irregular shape is 9.3 L</body></html>
Discussion
49125.

A vessel has N_2 gas saturated with water vapour at a total pressure of 1 atm. The partial pressure of water vapour is 0.3 atm. The contents of this vessel are completely transferred to another vessel having one third of the capacity of the original volume, at the same temperature. The total pressure of this system in the new vessel is

Answer» <html><body><p>3.0 <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/>1 atm <br/>3.33 atm <br/>2.4 atm<br/></p>Solution :`P_(N_2) = P_("dry <a href="https://interviewquestions.tuteehub.com/tag/gas-1003521" style="font-weight:bold;" target="_blank" title="Click to know more about GAS">GAS</a>") + P_(H_2O)` <br/> `1 = P_1 + 0.3 implies P_1 = 0.7 atm` <br/> on making volume = `(V_1)/3`, <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/> `P_2 = 3P_1 = 3 xx 0.7 = 2.1 atm`. <br/> `P_(H_2O)` does not change <br/> `:.` <a href="https://interviewquestions.tuteehub.com/tag/new-1114486" style="font-weight:bold;" target="_blank" title="Click to know more about NEW">NEW</a> `P_(N_2) = 2.1 + 0.3 = 2.4 atm.`</body></html>
Discussion
49126.

A vessel contains N_(2)O_(4) & NO_(2) in 2:3 molar ratio at 10 atm under equilibrium. Now, K_(P) for N_(2)O_(4) harr 2NO_(2) is

Answer» <html><body><p>9 <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/>9 `atm^(-1)`<br/>4.5 `atm^(2)`<br/>10 atm</p>Solution :`N_(2)O_(4) <a href="https://interviewquestions.tuteehub.com/tag/harr-2692945" style="font-weight:bold;" target="_blank" title="Click to know more about HARR">HARR</a> 2NO_(2)` <br/> `K_(P)=(P_(NO_(2))^(2))/(P_(N_(2)O_(4)))=(((3)/(5) xx 10)^(2))/((2)/(5) xx 10)=9` atm</body></html>
Discussion
49127.

A vessel contains N_(2) O_(4) at certain temperature & 2 atm . If the degree of dissociation of N_(2) O_(4) at equilibrium is 0.2 , final pressure of system (in atm) ?

Answer» <html><body><p>`1.2`<br/>`2.4`<br/>`3.6`<br/>`0.8`</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>
Discussion
49128.

A vessel contains methane and oxygen in the mass ratio 2:1. The fraction of the partial pressure of oxygen in the total pressure is

Answer» <html><body><p>`1//3`<br/>`2//3`<br/>`1//4`<br/>`1//2`</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>
Discussion
49129.

A vessel contains helium and methane in 4:1 molar ratio at 20 bar pressure. Due to leakage, the mixture of gases starts effusion. Find the composition of the mixture in the initial stage. (Hint: According to Graham's law of effusion, the rate of effusion of a gas in the mixture r prop Partial pressure sqrt((1)/( "molecular mass"))

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> :`He:CH_4 =8:1`</body></html>
Discussion
49130.

A vessel contains He and H_(2) in the molar ratio 1 : 5. The ratio of mean transitional kinetic energies, at the same tempreatures 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> : 1`<br/>`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>`<br/>`2 : 1`<br/>`1 : 5`<br/></p>Solution :(d) `<a href="https://interviewquestions.tuteehub.com/tag/ke-527890" style="font-weight:bold;" target="_blank" title="Click to know more about KE">KE</a>=(3)/(2)<a href="https://interviewquestions.tuteehub.com/tag/pv-593601" style="font-weight:bold;" target="_blank" title="Click to know more about PV">PV</a>=(3)/(2)nRT`<br/> `((KE)He)/((KE)H_(2))=((3)/(2)RT)/((3)/(2)RT)((1)/(5))=1 : 5`</body></html>
Discussion
49131.

A vessel contains helium and methane in 4:1 mole ratio at 20 bar pressure. Find the ratio of initial rates of effusion through a small aperture.

Answer» <html><body><p>&lt;<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>&gt;</p>Solution :Ratio of moles of He and `CH_(4)` = 4 : 1 <br/> Ratio of partialpressure of He and `Ch_(4) = 4 :1` <br/> Total pressure of the mixture = <a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a> bar <br/> Partial pressure of He ` = 20 xx (4)/(5) = 16` bar <br/> partial pressure of `CH_(4) = 20 xx (1)/(5) =4 ` bar <br/> Ratio of intial <a href="https://interviewquestions.tuteehub.com/tag/rates-25465" style="font-weight:bold;" target="_blank" title="Click to know more about RATES">RATES</a> of effusion is ` (r_(1))/(r_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)) = (P_(1))/(P_(2)) sqrt((M_(2))/(M_(1))) = (16)/(4) sqrt((16)/(4)) = (4)/(1) sqrt((4)/(1))` <br/> The ratio of <a href="https://interviewquestions.tuteehub.com/tag/initial-499269" style="font-weight:bold;" target="_blank" title="Click to know more about INITIAL">INITIAL</a> rates of effusion of gases, helium and methane ` =4 sqrt(4) : 1 = 8 : 1`</body></html>
Discussion
49132.

A vessel contains equal weights of He and CH_(4)gases at 20 bar pressure. Due to the leakage, the gases in the vessel started effusing out. What is the volume ratio of He and CH_(4)gases coming out initally

Answer» <html><body><p>&lt;<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>&gt;</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_(He) = x/4 = n_1 , n_(CH_4) = x/16 = n_2 , (n_1)/(n_2) = 4/1`. <br/> `(gamma_(He))/(gamma_(CH_4)) = (P_(He))/(P_(CH_4)) <a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(16/4) = 4/1 sqrt(16/4) = 8:1`.</body></html>
Discussion
49133.

A vessel contains equal volumes of SO_2 and CH_4Through a small hole, the gases effused into vacuum. After 200sec, the total volume is reduced to half. What is the ratio of SO_2and CH_4remaining in the vessel?

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> :<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>`<br/>`2 :1`<br/>`1 :1`<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>`</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>
Discussion
49134.

A vessel contains equal number of moles of helium and methane. Through a small orifice the half of gas effused out. The ratio of the number of mole of helium and methane remaining in the vessel is

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> :<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/>`1:4`<br/>`4:1`</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/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html>
Discussion
49135.

A vessel contains equal masses of helium and oxygen at 35^@C. What is the ratio of kinetic energies of the components?

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> :`8:1`</body></html>
Discussion
49136.

A vessel contains a mixture of equal weights of oxygen and SO_2 at a pressure of 600mm of Hg. The partial pressure of oxygen in mm is

Answer» <html><body><p>200<br/>300<br/>400 <br/><a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</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 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>
Discussion
49137.

A vessel contains a mixture of equal weights of methane and sulphur dioxide al pressure of 600 mm. Find the partial pressures of each of these gases in the mixture.

Answer» <html><body><p>&lt;<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>&gt;</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_(CH_4) = 480mm ,P_(SO_2) = <a href="https://interviewquestions.tuteehub.com/tag/120-270396" style="font-weight:bold;" target="_blank" title="Click to know more about 120">120</a> <a href="https://interviewquestions.tuteehub.com/tag/mm-1098795" style="font-weight:bold;" target="_blank" title="Click to know more about MM">MM</a>`</body></html>
Discussion
49138.

A vessel contains a mixture of equal masses of helium and oxygen at a pressure of 600 torr. Calculate the partial pressures of components in the mixture.

Answer» <html><body><p></p>Solution :Let the mass of each component gas is x. <br/> Total number of moles in the mixture = <br/> `(<a href="https://interviewquestions.tuteehub.com/tag/xg-747279" style="font-weight:bold;" target="_blank" title="Click to know more about XG">XG</a> " of " He)/(<a href="https://interviewquestions.tuteehub.com/tag/4g-318729" style="font-weight:bold;" target="_blank" title="Click to know more about 4G">4G</a> " of " mol^(-1)) + (xg " of " O_2)/(<a href="https://interviewquestions.tuteehub.com/tag/32-307188" style="font-weight:bold;" target="_blank" title="Click to know more about 32">32</a> g " of " mol^(-1)) = (9x)/(32 g mol^(-1))` <br/> Partial pressure = pressure of the mixture x mole <a href="https://interviewquestions.tuteehub.com/tag/fraction-458259" style="font-weight:bold;" target="_blank" title="Click to know more about FRACTION">FRACTION</a> <br/> Partial pressure of He `= 600 xx (x//4)/(9x//32) = 600 xx 8/9` <br/> `= 5333.33"<a href="https://interviewquestions.tuteehub.com/tag/torr-711044" style="font-weight:bold;" target="_blank" title="Click to know more about TORR">TORR</a>"` <br/>Partial pressure of `O_2 = 600 xx (x//32)/(9x//32)` <br/> `= 600 xx 1/9 = 66.67` torr <br/> (or) Partial pressure of `O_2 = 600 - 533.37 ` <br/> = 66.67 torr</body></html>
Discussion
49139.

A vessel contains a mixture of different types of gases. Which of the following satements is correct?

Answer» <html><body><p>On the average, the heavier molecules have higher molecules have higher speed <br/>The average transaltional energy of <a href="https://interviewquestions.tuteehub.com/tag/different-951434" style="font-weight:bold;" target="_blank" title="Click to know more about DIFFERENT">DIFFERENT</a> <a href="https://interviewquestions.tuteehub.com/tag/types-15918" style="font-weight:bold;" target="_blank" title="Click to know more about TYPES">TYPES</a> of molecules is the same<br/>The average speed of different molecules is same<br/>On an average, the heavier moleules have higher <a href="https://interviewquestions.tuteehub.com/tag/transitional-7331850" style="font-weight:bold;" target="_blank" title="Click to know more about TRANSITIONAL">TRANSITIONAL</a> energy<br/></p>Solution :(b) The average <a href="https://interviewquestions.tuteehub.com/tag/translational-3229278" style="font-weight:bold;" target="_blank" title="Click to know more about TRANSLATIONAL">TRANSLATIONAL</a> energy of different types of molecules is the same because it depends only on temperature.</body></html>
Discussion
49140.

A vessel contains 1.6 g of dioxygen at STP (273.15 K, 1 atm pressure). The gas is now transferred to another vessel at constant temperature, where pressure becomes half of the original pressure. Calculate (A) volume of the new vessel. (B) number of molecules of dioxygen.

Answer» <html><body><p> </p>Solution :(A) `p_(1)=1` atm, `p_(2)=1/2 = 0.5` atm, `T_(1)=273.15, V_(2)=(?)` <br/> 32g of dioxygen occupies `=22.4 L` volume at STP <br/> `:.1.6g` dioxygen will occupy `=(22.4Lxx1.6g)/(32g)` <br/> `=1.12L` <br/> `V_(1)=1.12L` <br/> From Boyle.s law <br/> `p_(1)V_(1)=p_(2)V_(2)` <br/> `V_(2)=(p_(1)V_(1))/(p_(2))= (1 "atm"xx1.12L)/(0.5 "atm")=2.24L` <br/> Number of moles of dioxygen `=("<a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of dioxygen")/("Molar mass of dioxygen")` <br/> `n_(O_(2))=(1.6)/(<a href="https://interviewquestions.tuteehub.com/tag/32-307188" style="font-weight:bold;" target="_blank" title="Click to know more about 32">32</a>)=0.05` mol <br/> 1 mol of dioxygen contains `= 6.022xx10^(23)` molecules <br/> `:.` 0.5 mol of dioxygen <a href="https://interviewquestions.tuteehub.com/tag/contain-409810" style="font-weight:bold;" target="_blank" title="Click to know more about CONTAIN">CONTAIN</a> <br/> `= 0.05xx6.022xx10^(23)` molecule <br/> `=3.011xx10^(22)` molecules</body></html>
Discussion
49141.

A vessel contains 100 litres of liquid X. Heat is suppplied to the liquid in such a fashion that, Heat given = change in ethalpy. The volume of the liquid is incresed by 2 litres. If the external pressure is one atm, and 202.6joules of heat were supplied then, [U= total internal energy] :

Answer» <html><body><p>`DeltaU=0,DeltaH=0`<br/>`DeltaU=+202.6J,DeltaH=+202.6J`<br/>`DeltaU=-202.6J,DeltaH=-202.6J`<br/>`DeltaU=0,DeltaH=+202.6J`</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>
Discussion
49142.

A vessel contains 100 litres of a liquid x. Heat is suppllied to the liquid in such a fashion that, Heat given = change in enthalpy. The volume of the liquid increases by 2 lites. If the external pressure is one atm and 202.6 Joules of heat were supplied then,

Answer» <html><body><p>`Delta U = <a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>, Delta <a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a> = 0`<br/>`Delta U = +202.6J, Delta H = +202.6 <a href="https://interviewquestions.tuteehub.com/tag/j-520843" style="font-weight:bold;" target="_blank" title="Click to know more about J">J</a>`<br/>`Delta U = -202.6J, Delta H = - 202.6J`<br/>`Delta U = 0, Delta H = + 202.6J`</p>Solution :from the data `Delta H= <a href="https://interviewquestions.tuteehub.com/tag/q-609558" style="font-weight:bold;" target="_blank" title="Click to know more about Q">Q</a>= 202.6J` <br/> from first law of T. D `Delta U = Q + W` <br/> `Delta U = 202.6 + (-P Delta V)` <br/> `=202.6 + (-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>) xx 101.3J) = 0`</body></html>
Discussion
49143.

A vessel contains 1 mole PCI_5 (g) at 4 atm and 0.5 mole PCI_3 formed at equilibrium. Now, equilibrium pressure of mixture is (assume ideal behavior)

Answer» <html><body><p>16 <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/>6 atm<br/>2 atm<br/><a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>.5 atm</p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_B_C04_E01_058_S01.png" width="80%"/> <br/> given <a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a> = 0.5 <br/> total moles of equilibrium = 1+ x =1.5 <br/> moles `alpha` pressure <br/> `1 to 4`atm <br/> `1.5 to ? implies` pressure = 6atm</body></html>
Discussion
49144.

A vessel contains 0.25 moles of Helium and0.15 moles of neon at 298K and 2.4 atmosphere pressure. Calculate the partial pressure of each gas in the mixture.

Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</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> .P.=2.4atm <br/> No. of moles of He=0.<a href="https://interviewquestions.tuteehub.com/tag/25-296426" style="font-weight:bold;" target="_blank" title="Click to know more about 25">25</a> <br/> No. of moles of Ne=0.15 <br/> Total moles of n=0.25+0.15=0.4 <br/> Partial pressure = Total pressure <a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a> Mole-fraction <br/> `P_(He) = (n_(He))/(n) xx P = (0,.25)/(0.4) xx 2.4 = 1.5 atm` <br/> `P_(N e) = (n_(N e))/(n) xx P = (0.15)/(0.4) xx 2.4 = 0.9 atm`.</body></html>
Discussion
49145.

A vesselat 1000 K containsCO_(2) with a pressure of 0.5 atm . Some of the CO_(2) is converted intoCO on the addition of graphite . If the total pressure at equilibrium is 0*8 atm , the value of K is

Answer» <html><body><p>3 atm <br/>`<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>*3 `atm<br/>`0*18 atm`<br/>`1*8 ` atm</p>Solution :`{:(,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>)(<a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>),+,C(s),hArr,2CO(g)),(" <a href="https://interviewquestions.tuteehub.com/tag/intial-2746226" style="font-weight:bold;" target="_blank" title="Click to know more about INTIAL">INTIAL</a> ",0.5 atm,,,,),("At eqm.",0.5 -p,,,,2 p):}` <br/> `:. (0.5 -p) +2 p = 0.8" atm "or p=0.3 " atm "`<br/> `K=(p^(2) CO)/p_(CO_(2)) = (0.6)^(2)/((0.5 -0.3)) = (0.6 xx0.6)/0.2 = 1.8 " atm", `</body></html>
Discussion
49146.

A vessel at 1000 K contains CO_(2) with a pressure of 0.5 atm. Some of CO_(2) is converted into CO on addtion of graphite. The value of K' at equilibrium when total pressure is 0.8 atm will be

Answer» <html><body><p>0.18 atm<br/>1.8 atm<br/>2 atm<br/>1 atm</p>Solution :`underset(0.5)(CO_(2))+C_((s)) harr 2CO` <br/> at <a href="https://interviewquestions.tuteehub.com/tag/eqm-446398" style="font-weight:bold;" target="_blank" title="Click to know more about EQM">EQM</a> 0.5-P`""<a href="https://interviewquestions.tuteehub.com/tag/2p-300467" style="font-weight:bold;" target="_blank" title="Click to know more about 2P">2P</a>` <br/> P=0.3 <br/> `K=((0.6)^(2))/((0.2))=1.8`atm</body></html>
Discussion
49147.

A vessel at 1000 K contains CO_(2) at 0.4 atm . If certain amount of CO_(2) is converted into CO by the addition of graphite , the following equilibrium is established CO_(2) + C hArr 2CO , at a pressure of 0.6 atm, then equilibrium constant is

Answer» <html><body><p>0.2 <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/>0.05 atm <br/>0.8 atm <br/>0.15 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 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>
Discussion
49148.

A vessel at 1000 K contains carbon dioxide with a pressure of 0.5 atm. Some of the carbon dioxide is converted to carbon monoxide on addition of graphite. Calculate the value of K it the total pressure at equilibrium is 0*8 atm.

Answer» <html><body><p></p>Solution :`CO_(2) (<a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>) + C (s) hArr 2 CO (g)` <br/> Suppose decreasein pressure of `CO_(2)` after reaction = p atm <br/> Then increase in pressure due to CO= 2 p <br/> Pressure of `CO_(2)` at equilibrium `=(0.<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a> - p)` atm <br/> `:." Final <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> pressure " (0.5- p) + 2p = 0*5 + p = 0.<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a> " atm"`(Given) <br/>`:. p= 0*3 " atm.Hence we have "` <br/> `p_(CO_2) = 0*5 - 0*3 = 0*2 " atm "` <br/> `and ""p_(CO) = 2 xx 0*3 = 0*6 " atm" ` <br/> `:. K = (P_(cO)^(2))/P_(CO_(2)) = (0.6)^(2)/(0*2) = <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`</body></html>
Discussion
49149.

A vessel (A) contains I mole each of N_(2) & O_(2) and another vesser (B) contains 2 mole each of N_(2)& O_(2) . 3oth vessels are heated to same temperature till equilibrium is established in-both cases. Then, correct statement is

Answer» <html><body><p>`K_c` for `N_2+O_2 harr <a href="https://interviewquestions.tuteehub.com/tag/2no-300447" style="font-weight:bold;" target="_blank" title="Click to know more about 2NO">2NO</a>` in A &amp; B are in the ratio 1:2<br/>`K_p` for `N_2+O_2 harr 2NO` in A &amp; B are in ratio of 1:2<br/>`K_c` for `N_2+O_2 harr 2NO` in A &amp; B are equal<br/>`K_p` for `N_2+O_2 harr 2NO` in A &amp; B are in ratio of 2:1</p>Solution :`K_(1)` does not depend on concentrations of <a href="https://interviewquestions.tuteehub.com/tag/reactants-1178112" style="font-weight:bold;" target="_blank" title="Click to know more about REACTANTS">REACTANTS</a></body></html>
Discussion
49150.

A very small amoutn of radioactive isotope of Pb^(213) was mixed with a non-radioactive lead salt containg 0.01g of Pb (atomic mss 207). The whole lead was brought into solution and lead chromate was prcipitated by addition of a soluabolr chromate. Evaporation of 10cm^(3) fo the supernatentliquid gave a residue having a radioactivity (1)/(24000) of that of the original quanity if Pb^(213) caculate the solubility of lead chromate in mol dm^(-3).

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> :`2.0xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>)<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/dm-432223" style="font-weight:bold;" target="_blank" title="Click to know more about DM">DM</a>^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)`</body></html>
Discussion