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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your Class 11 knowledge and support exam preparation. Choose a topic below to get started.
| 49801. |
When a mixture of LiO_(2)CO_(3) and Na_(2)CO_(3) 10H_(2)O is heated strongly, there occurs a loss of mass due to |
| Answer» <html><body><p>it has <a href="https://interviewquestions.tuteehub.com/tag/lower-1080637" style="font-weight:bold;" target="_blank" title="Click to know more about LOWER">LOWER</a> m.pt than <a href="https://interviewquestions.tuteehub.com/tag/na2co3-572455" style="font-weight:bold;" target="_blank" title="Click to know more about NA2CO3">NA2CO3</a> and converts metal <a href="https://interviewquestions.tuteehub.com/tag/salts-1193872" style="font-weight:bold;" target="_blank" title="Click to know more about SALTS">SALTS</a> to carbonates which decompose to metal oxides<br/> it has <a href="https://interviewquestions.tuteehub.com/tag/higher-1022060" style="font-weight:bold;" target="_blank" title="Click to know more about HIGHER">HIGHER</a> m.pt than K2CO3 and converts metal salts to carbonates, which decompose to metal oxides<br/> it has lower melting <a href="https://interviewquestions.tuteehub.com/tag/point-1157106" style="font-weight:bold;" target="_blank" title="Click to know more about POINT">POINT</a> than both Na2CO3 and K,CO, and converts the metal salts to carbonates, which decompose to metal oxides<br/>it has higher melting point than both Na2CO3 and K2CO3 and converts the metal salts to carbonates which decompose to metal oxide</p>Answer :C</body></html> | |
| 49802. |
A mixture of Na_(2)CO_(3) and NaHCO_(3) having a total weight of 100 gm on heating produced 11.2L of CO_(2) under STP conditions. The percentage of Na_(2)CO_(3) in the mixture is |
| Answer» <html><body><p>`55.8 %`<br/>`44.2 %`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/84-339262" style="font-weight:bold;" target="_blank" title="Click to know more about 84">84</a> %`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/16-276476" style="font-weight:bold;" target="_blank" title="Click to know more about 16">16</a> %`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :D</body></html> | |
| 49803. |
A mixture of N_(2) and H_(2) in the ratio 1 : 3 is allowed to attain equilibrium. At equilibrium, the total pressure is 5xx10^(-5)Nm^(-2) and the mixture contains 40% by volume of NH_(3). Calculate K_(p). |
| Answer» <html><body><p></p>Solution :Total pressure of the mixture = `5xx10^(5) Nm^(-2)` <br/> `therefore` Partial pressure of `NH_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)=(40xx5xx10^(5))/100=2xx10^(5)Nm^(-2)` <br/> The <a href="https://interviewquestions.tuteehub.com/tag/sum-1234400" style="font-weight:bold;" target="_blank" title="Click to know more about SUM">SUM</a> of the partial pressure of `N_(2)andH_(2)=5xx10^(5)-2xx10^(5)=<a href="https://interviewquestions.tuteehub.com/tag/3xx10-1865443" style="font-weight:bold;" target="_blank" title="Click to know more about 3XX10">3XX10</a>^(5)` <br/> Since `N_(2)andH_(2)` are in the <a href="https://interviewquestions.tuteehub.com/tag/ratio-13379" style="font-weight:bold;" target="_blank" title="Click to know more about RATIO">RATIO</a> 1 : 3 <br/> `P_(N_(2))=1/4xx3xx10^(5)=0.75xx10^(5)Nm^(-2),P_(H_(2))=3/4xx3xx10^(5)=2.25xx10^(5)Nm^(-2)` <br/> `K_(p)=(P_(NH_(3))^(2))/(P_(N_(2))xxP_(H_(2))^(3))=((2xx10^(5))^(2))/((0.75xx10^(5))xx(2.25xx10^(5))^(3))=0.468xx10^(-10)(Nm^(-2))^(-2)`</body></html> | |
| 49804. |
A mixture of N_(2) and H_(2) in the molar ratio1:3 attains equilibrium when 50% of the mixture has reacted , If P is the total pressure of the mixture , then partial pressure of NH_(3) formed is |
| Answer» <html><body><p>`P//2`<br/>`P//3`<br/>`P//4`<br/>`P//6`</p>Solution :`{:(,N_(2),+,3 H_(2),hArr,2 NH_(3)),("Intial",1 " mole",,3 "moles",,0),("At eqm." ,1-x,,3-3x,,2 x):}` <br/> But out of4 moles of the mixture , 2 moles have reacted and <a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a> 2 moles are <a href="https://interviewquestions.tuteehub.com/tag/left-1070879" style="font-weight:bold;" target="_blank" title="Click to know more about LEFT">LEFT</a> .<br/> ` :. (1-x) + ( 3-3x) = 2 <a href="https://interviewquestions.tuteehub.com/tag/or4-583925" style="font-weight:bold;" target="_blank" title="Click to know more about OR4">OR4</a> x = 2 or 4 x = 2 or x = 0.5` <br/> ` :. " Total moles at equilibrium " ` <br/> ` = (1-x) + (3-3x )+ 2 x ` <br/> ` = 4 - 2 x = 4 - 1 = 3 ` <br/> No . of moles of `NH_(3)` at equilibrium = 2 x = 1<br/> ` :. p_(NH_(3)) = 1/3 xx P = P/3 `</body></html> | |
| 49805. |
A mixture of MgO and Mg weighing 10 g is treated with excess of dil HCl. Then 2.24 lit of H_(2) gas was liberated under STP conditions . The mass of MgO present in the sample 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>.4 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> <br/>7.6 g <br/><a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a> g <br/>2 g</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> | |
| 49806. |
A mixture of methyl alochol and acetone can be separated by |
| Answer» <html><body><p> <a href="https://interviewquestions.tuteehub.com/tag/distillation-15402" style="font-weight:bold;" target="_blank" title="Click to know more about DISTILLATION">DISTILLATION</a><br/> Fractional distillation<br/> <a href="https://interviewquestions.tuteehub.com/tag/stream-11292" style="font-weight:bold;" target="_blank" title="Click to know more about STREAM">STREAM</a> distillation<br/> Distillation under <a href="https://interviewquestions.tuteehub.com/tag/reduced-1181337" style="font-weight:bold;" target="_blank" title="Click to know more about REDUCED">REDUCED</a> pressure</p>Answer :A::B::<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>::D</body></html> | |
| 49807. |
A mixture of methane and ethylene in the volume ratio x : y has a total volume of 30 ml . On complete combustion it gave 40 ml of CO_(2) . If the ratio had been y :x , instead of x : y , what volume of CO_(2) could have been obtained ? |
| Answer» <html><body><p>50 <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/>100 ml <br/><a href="https://interviewquestions.tuteehub.com/tag/25-296426" style="font-weight:bold;" target="_blank" title="Click to know more about 25">25</a> ml <br/><a href="https://interviewquestions.tuteehub.com/tag/75-334971" style="font-weight:bold;" target="_blank" title="Click to know more about 75">75</a> ml </p>Answer :A</body></html> | |
| 49808. |
A mixture of KOH and Na_(2)CO_(3) solution required 15 mL of N//20 HCl using phenolphthalein as indicator. The same amount of alkali mixture when titrated using methyl orange as indicator required 25 mL of same acid. Calculate amount of KOH and Na_(2)CO_(3) present in solution. |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :N//A</body></html> | |
| 49809. |
A mixture of K_(2)C_(2)O_(4) and KHC_(2)O_(4) required equal volumes of 0.1 M K_(2)Cr_(2)O_(7) for oxidation and 0.1 M NOH for neutralisation is separate titratiosn. The molar ratio of K_(2)CrO_(4) and KHC_(2)O_(4) in the mixture is |
| Answer» <html><body><p>`1:1`<br/>`2:1`<br/>`1:2`<br/>`3:1`</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/let-11597" style="font-weight:bold;" target="_blank" title="Click to know more about LET">LET</a> no. of moles of `K_(2)C_(2)O_(4)=a&` <br/> no. of moles of `KHC_(2)O_(4)=b` <br/> Eqts of `K_(2)C_(2)O_(4)+` Eqts of `KHC_(2)O_(4)` <br/> = Eqts of `K_(2)Cr_(2)O_(7)` <br/> `(axx2)+(bxx2)=0.1xx6xxv....(1)` <br/> Eqts of `KHC_(2)O_(4)` = Eqts of NaOH <br/> `(bxx1)=0.1xx1xxv....(1)` <br/> On <a href="https://interviewquestions.tuteehub.com/tag/solving-1217196" style="font-weight:bold;" target="_blank" title="Click to know more about SOLVING">SOLVING</a> (1), (2) <br/> a : b = 2 : 1</body></html> | |
| 49810. |
A mixture of hydrocarbon C_(2)H_(2).C_(2)H_(4) & CH_(4) in mole ration of 2:1:2 is burnt completely in the pressence of air containing 80%N_(2) % 20% O_(2) by volume. The mass of air required for the complete combustion of the one gm of mixture is |
| Answer» <html><body><p>`(1728)/(112)`<br/>`(1528)/(73)`<br/>`(1920)/(120)`<br/>`(112)/(1728)`</p>Solution :`{:(,C_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)H_(2),:,C_(2)H_(4),:,CH_(4)),(,2,:,1,:,2),(,2x,:,x,:,2x):}` <br/> `52x+28x+32x=1` <br/> `112x=1` <br/> `x=(1)/(112)` <br/> `{:(C_(2)H_(2)+(<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)/(2)O_(2)rarr2CO_(2)+H_(2)O(l)),((2)/(112)""(5)/(112)),(CH_(4)+3O_(2)rarr2CO_(2)+2H_(2)O(l)),((1)/(112)""(4)/(112)),(CH_(4)+2O_(2)rarrCO_(2)+2H_(2)O(l)),((2)/(112)""(4)/(112)):}` <br/> Total moles of`O_(2) = (5)/(112)+(3)/(112)+(4)/(112) = (<a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a>)/(112)` <br/> moles of `O_(2)= (12)/(112)xx32 = (384)/(112)gm` <br/> moles of `N_(2) =(12)/(112)xx4 = (48)/(112)`<br/> moles of `N_(2) = (48)/(112)xx28=(1344)/(112)gm` <br/> mass of air `=(48+1344)/(112) = (1728)/(112) gm`</body></html> | |
| 49811. |
A mixture of hydrogen and helium is prepared such that the number of collisions on the wall per unit time by molecules of each gas is same. Which gas has higher concentration? |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/helium-484142" style="font-weight:bold;" target="_blank" title="Click to know more about HELIUM">HELIUM</a> <br/>hydrogen <br/>both have same concentration <br/>can't be determined </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> :More diameter `<a href="https://interviewquestions.tuteehub.com/tag/implies-1037962" style="font-weight:bold;" target="_blank" title="Click to know more about IMPLIES">IMPLIES</a>` less conc.</body></html> | |
| 49812. |
A mixture of hydrazine, hydrogen and Cu(II) catalyst is used as rocked fuel. Why? |
| Answer» <html><body><p></p>Solution :In the presence of `Cu(<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>)` catalyst, hydrogen peroxide hydraine `(NH_(2)-NH_(2))` to <a href="https://interviewquestions.tuteehub.com/tag/nitrogen-1118291" style="font-weight:bold;" target="_blank" title="Click to know more about NITROGEN">NITROGEN</a>. The reaction is <a href="https://interviewquestions.tuteehub.com/tag/highly-2102664" style="font-weight:bold;" target="_blank" title="Click to know more about HIGHLY">HIGHLY</a> exothermic and in accompanied by large increasein volume. Therefore, the mixture provides an upward and thrust and is used to a rocket <a href="https://interviewquestions.tuteehub.com/tag/fuel-1001053" style="font-weight:bold;" target="_blank" title="Click to know more about FUEL">FUEL</a>. <br/> `NH_(2)-NH_(2)(l)+2H_(2)O_(2)(l) <a href="https://interviewquestions.tuteehub.com/tag/underset-3243992" style="font-weight:bold;" target="_blank" title="Click to know more about UNDERSET">UNDERSET</a>("Heat")overset("Cu(II)")to N_(2)(g)+4H_(2)O(g)+"heat"`</body></html> | |
| 49813. |
A mixture of hydrazine and ……. With a copper (II) catalyst is used as rocket propellant. |
| 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> :`H_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)O_(2)`</body></html> | |
| 49814. |
A mixture of hydrazine and H_(2)O_(2) with Cu(II) catalyst is used as a rocket prepellant . Why ? |
| Answer» <html><body><p></p>Solution :The reaction between hydrazine and `H_(2)O_(2)` is highl <a href="https://interviewquestions.tuteehub.com/tag/exothermic-2066005" style="font-weight:bold;" target="_blank" title="Click to know more about EXOTHERMIC">EXOTHERMIC</a> and is accoumpanied by a large <a href="https://interviewquestions.tuteehub.com/tag/increase-1040383" style="font-weight:bold;" target="_blank" title="Click to know more about INCREASE">INCREASE</a> in the volume of the <a href="https://interviewquestions.tuteehub.com/tag/products-18828" style="font-weight:bold;" target="_blank" title="Click to know more about PRODUCTS">PRODUCTS</a> and hence this <a href="https://interviewquestions.tuteehub.com/tag/mixture-1098735" style="font-weight:bold;" target="_blank" title="Click to know more about MIXTURE">MIXTURE</a> is used as a rocket propellant. <br/>`NH_(2)NH_(2)(l) + 2H_(2)O_(2)(l) overset( Cu (II))to N_(2)(g)uarr + 4H_(2)O(g) uarr`</body></html> | |
| 49815. |
A mixture of hydrazine and H_(2)O_(2) is |
| Answer» <html><body><p>antiseptic <br/><a href="https://interviewquestions.tuteehub.com/tag/rocket-1190549" style="font-weight:bold;" target="_blank" title="Click to know more about ROCKET">ROCKET</a> <a href="https://interviewquestions.tuteehub.com/tag/fuel-1001053" style="font-weight:bold;" target="_blank" title="Click to know more about FUEL">FUEL</a><br/>germicide<br/>insecticide </p>Solution :A <a href="https://interviewquestions.tuteehub.com/tag/mixture-1098735" style="font-weight:bold;" target="_blank" title="Click to know more about MIXTURE">MIXTURE</a> of `N_2H_4` and `H_2O_2` is <a href="https://interviewquestions.tuteehub.com/tag/used-2318798" style="font-weight:bold;" target="_blank" title="Click to know more about USED">USED</a> as rocket fuel</body></html> | |
| 49816. |
A mixture of H_(2)O_(2) and hydrzine with copper (II) is used as a rocket propellant. Why ? |
| Answer» <html><body><p></p>Solution :The reaction between <a href="https://interviewquestions.tuteehub.com/tag/hydrazine-493075" style="font-weight:bold;" target="_blank" title="Click to know more about HYDRAZINE">HYDRAZINE</a> `(NH_(2)NH_(2))andH_(2)O_(2)` in the presence of Cu(II) is highly exothermic and is <a href="https://interviewquestions.tuteehub.com/tag/accompanied-7662390" style="font-weight:bold;" target="_blank" title="Click to know more about ACCOMPANIED">ACCOMPANIED</a> by a <a href="https://interviewquestions.tuteehub.com/tag/large-1066424" style="font-weight:bold;" target="_blank" title="Click to know more about LARGE">LARGE</a> <a href="https://interviewquestions.tuteehub.com/tag/increase-1040383" style="font-weight:bold;" target="_blank" title="Click to know more about INCREASE">INCREASE</a> in energy as well as in volume of the <a href="https://interviewquestions.tuteehub.com/tag/products-18828" style="font-weight:bold;" target="_blank" title="Click to know more about PRODUCTS">PRODUCTS</a>. Therefore, this mixture is used as a rocket propellant. <br/> `2H_(2)O_(2)(I)+H_(2)NH_(2)(I)overset(Cu(II))toN_(2)(g)uarr+4H_(2)O(g)uarr`</body></html> | |
| 49817. |
A mixture of H_2,N_2 and NH_3 with molar concentrations 5.0xx10^(-3) mol L^(-1) , 4.0xx10^(-3) mol L^(-1) and 2.0xx10^(-3) mol L^(-1)respectively was prepared and heated to 500 K. The value of K_c for the reaction: 3H_2(g)+N_2(g) hArr 2NH_3(g) at this temperature is 60. Predict whether ammonia tends to form or to decompose at this stage of concentration. |
| 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> :Reaction quotient, `Q_c` for the reaction, <br/> `N_2(g) +3H_2(g) hArr 2NH_3(g)` <br/> `Q_c=([NH_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>)/([N_2][H_2]^3)` <br/> `[NH_3]=2.0xx10^(-3) <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> L^(-1) , [H_2]=5.0xx10^(-3) mol L^(-1)`<br/> `[N_2]=4.0xx10^(-3) mol^(-1)`<br/> `Q_c=(2.0xx10^(-3))^2/((4.0xx10^(-3))(5.0xx10^(-3))^(3))=8.0xx10^3 L^2 mol^(-2)`<br/> `K_c=60` <br/>Since `Q_c > K_c `, reaction will <a href="https://interviewquestions.tuteehub.com/tag/go-468886" style="font-weight:bold;" target="_blank" title="Click to know more about GO">GO</a> in the <a href="https://interviewquestions.tuteehub.com/tag/left-1070879" style="font-weight:bold;" target="_blank" title="Click to know more about LEFT">LEFT</a> direction and ammonia will decompose.</body></html> | |
| 49818. |
A mixture of gases having different molecular weights is seperated by which method ? |
| Answer» <html><body><p>Atomlysis<br/>Metathesis<br/>Ostwald and <a href="https://interviewquestions.tuteehub.com/tag/walker-1448634" style="font-weight:bold;" target="_blank" title="Click to know more about WALKER">WALKER</a> method<br/>Reverse osmosis<br/></p>Solution :(a) Diffusion method is <a href="https://interviewquestions.tuteehub.com/tag/used-2318798" style="font-weight:bold;" target="_blank" title="Click to know more about USED">USED</a> to <a href="https://interviewquestions.tuteehub.com/tag/separate-1201161" style="font-weight:bold;" target="_blank" title="Click to know more about SEPARATE">SEPARATE</a> a mixture of gases having different <a href="https://interviewquestions.tuteehub.com/tag/molecular-562994" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULAR">MOLECULAR</a> weight as rate of diffusion <a href="https://interviewquestions.tuteehub.com/tag/varies-1443299" style="font-weight:bold;" target="_blank" title="Click to know more about VARIES">VARIES</a> inversely with molecular mass, i.e. <br/> `rprop(1)/(sqrtM)` <br/> This method is called atmolysis.</body></html> | |
| 49819. |
A mixture of gases having different molecular weights is separated by which method ? |
| Answer» <html><body><p>Atmolysis <br/>Metathesis <br/>Ostwald and <a href="https://interviewquestions.tuteehub.com/tag/walker-1448634" style="font-weight:bold;" target="_blank" title="Click to know more about WALKER">WALKER</a> method <br/>Reverse osmosis </p>Solution :Diffusion method is used to <a href="https://interviewquestions.tuteehub.com/tag/separate-1201161" style="font-weight:bold;" target="_blank" title="Click to know more about SEPARATE">SEPARATE</a> a <a href="https://interviewquestions.tuteehub.com/tag/mixture-1098735" style="font-weight:bold;" target="_blank" title="Click to know more about MIXTURE">MIXTURE</a> of gases having <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/molecular-562994" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULAR">MOLECULAR</a> weights as rate of diffusion varies inversely with molecular mass i.e., `r prop (1)/(sqrtM)` <br/> This method is called atmolysis .</body></html> | |
| 49820. |
A mixture of gases contains 4.76 mole of Ne, 0.74 mole of Ar and 2.5 mole of Xe. Calculate the partial pressure of gases, if the total pressure is 2 atm, at a fixed temperature. |
| Answer» <html><body><p></p>Solution :`P_(<a href="https://interviewquestions.tuteehub.com/tag/ne-1112263" style="font-weight:bold;" target="_blank" title="Click to know more about NE">NE</a>)=X_(Ne)^(-)P_("Total")` <br/> `X_(Ne)=(n_(Ne))/(n_(Ne)+n_(Ar)+n_(<a href="https://interviewquestions.tuteehub.com/tag/xe-747964" style="font-weight:bold;" target="_blank" title="Click to know more about XE">XE</a>))=(4.76)/(4.76+0.74+2.5)=0.595` <br/> `X_(Ar)=(n_(Ar))/(n_(Ne)+n_(Ar)+n_(Xe))=(0.74)/(4.76+0.74+2.5)=0.093` <br/> `X_(Xe)=(n_(Xe))/(n_(Ne)+n_(Ar)+n_(Xe))=(2.5)/(4.76+0.74+2.5)=0.312` <br/> `P_(Ne)=X_(Ne)P_("Total")=0.595xx2=1.19` <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/> `P_(Ar)=X_(Ar)P_(Total)=0.093xx2=0.186atm` <br/> `P_(Xe)=X_(Xe)P_(Total)=0.312xx2=0.624atm`.</body></html> | |
| 49821. |
A mixture of FeCI_(2) and SnCI_(2) can exist togethr but that of FeCI_(3) and SnCI_(3) cannot explain why ? |
| 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/since-644476" style="font-weight:bold;" target="_blank" title="Click to know more about SINCE">SINCE</a> both `FeCI_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)` and `SnCI_(2)` being reducing agents can <a href="https://interviewquestions.tuteehub.com/tag/stay-1226662" style="font-weight:bold;" target="_blank" title="Click to know more about STAY">STAY</a> together but an <a href="https://interviewquestions.tuteehub.com/tag/oxidising-2209005" style="font-weight:bold;" target="_blank" title="Click to know more about OXIDISING">OXIDISING</a> agent `(FeCI_(3))` and reducing agent `(SnCI_(2))` cannot stay together</body></html> | |
| 49822. |
A mixture of ethyl alcohol and boric acid burn with green edged flame. The green edged flame contains |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/triethyl-2311322" style="font-weight:bold;" target="_blank" title="Click to know more about TRIETHYL">TRIETHYL</a> borate<br/>ethyl boride<br/>Acetaldehyde<br/>diborane</p>Answer :A</body></html> | |
| 49823. |
A mixture of ethanol and water contains 54% water by mass. Calculate the mole fraction of alcohol in this solution. |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`X_("<a href="https://interviewquestions.tuteehub.com/tag/ethanol-975823" style="font-weight:bold;" target="_blank" title="Click to know more about ETHANOL">ETHANOL</a>") = (46//46)/(46//46+54//18)=0.25`</body></html> | |
| 49824. |
A mixture of ethane, ethylene and acetylene gases are passed through a Woulfe's bottle containingammoniacal silver nitrate solution. The gas coming out is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/ethane-975817" style="font-weight:bold;" target="_blank" title="Click to know more about ETHANE">ETHANE</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/acetylene-847427" style="font-weight:bold;" target="_blank" title="Click to know more about ACETYLENE">ACETYLENE</a> <br/>Both ethane and ethylene<br/>Original mixture</p>Solution :acetylene is <a href="https://interviewquestions.tuteehub.com/tag/absorbed-846166" style="font-weight:bold;" target="_blank" title="Click to know more about ABSORBED">ABSORBED</a></body></html> | |
| 49825. |
A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen . What would be the partial pressure of dihydrogen in bar ? |
| 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> :Suppose the total <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 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> is 100 g. <br/> `:. " Mass of " H_2 = 20 g, " Mass of " O_2` <br/> `= 100 - 20 = 80 g` <br/> <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 `H_2 = 20/2 = 10` <br/>Number of moles of `O_2 =80/32 =2.5` <br/> Total number of moles in the mixture = `10 + 2.5 = 12.5` <br/> `:. " Partial pressure of " H_2 = ("Moles of " H_2)/(" Total no. of moles")<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> " Total pressure "`. <br/> `=10/(12.5)= 0.8` bar</body></html> | |
| 49826. |
A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen. |
| Answer» <html><body><p></p>Solution :As the mixture of `H_(2)` and `O_(2)` <a href="https://interviewquestions.tuteehub.com/tag/contains-11473" style="font-weight:bold;" target="_blank" title="Click to know more about CONTAINS">CONTAINS</a> 20% by weight of dihydrogen, therefore, if `H_(2)`=20 g, then `O_(2)`=<a href="https://interviewquestions.tuteehub.com/tag/80-337972" style="font-weight:bold;" target="_blank" title="Click to know more about 80">80</a> g <br/> `n_(H_(2))=(20)/(2)=10" <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a>", n_(O_(2))=(80)/(<a href="https://interviewquestions.tuteehub.com/tag/32-307188" style="font-weight:bold;" target="_blank" title="Click to know more about 32">32</a>)=2.5" moles"` <br/> `p_(H_(2))=(n_(H_(2)))/(n_(H_(2))+n_(O_(2)))xxP_("total")=(10)/(10+2.5)<a href="https://interviewquestions.tuteehub.com/tag/xx1-1463705" style="font-weight:bold;" target="_blank" title="Click to know more about XX1">XX1</a>" bar"=0.8" bar"`</body></html> | |
| 49827. |
A mixture of dihydgrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen. |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/suppose-656311" style="font-weight:bold;" target="_blank" title="Click to know more about SUPPOSE">SUPPOSE</a>, Weight of mixture gas = 100 g<br/>`therefore` Weight of `H_(2)=20 g` and<br/>Weight of `O_(2)=80 g`<br/>`therefore` Mole of `n_(H_(2))=(20)/(2)=10` mol and<br/>Mole of `n_(O_(2))=(80)/(32)=2.5` mol<br/>Total mole of mixture gas `= (10+2.5)=12.5` mol<br/>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> of `H_(2), chi_(H_(2))=(n_(H_(2)))/("Total Moles")=(10)/(12.5)`<br/>Mole fraction of `O_(2), chi_(O_(2))=(n_(O_(2)))/("Total moles")=(2.5)/(12.5)`<br/><a href="https://interviewquestions.tuteehub.com/tag/partial-596556" style="font-weight:bold;" target="_blank" title="Click to know more about PARTIAL">PARTIAL</a> Pressure = Mole fraction `xx` Total pressure<br/>`therefore p_(H_(2))=n_(H_(2))xx P=(10xx"1 bar")/(12.5)=(10xx"1 bar")/(12.5)=0.8` bar<br/>`therefore p_(O_(2))=n_(O_(2))xxP=(2.5xx"1 bar")/(12.5)=0.2` bar</body></html> | |
| 49828. |
A mixture of CuSO_(4).5H_(2)O and MgSO_(4).7H_(2)O was heated until all the water was driven off If 5.0g if mixture gave 3g of anhydrous salts, what was the percentage by mass of CuSO_(4).5H_(2)O in the original mixture : |
| Answer» <html><body><p><br/></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/let-11597" style="font-weight:bold;" target="_blank" title="Click to know more about LET">LET</a> 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> contain `x gCuSO_(4). 5H_(2)O`.<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> (x)/(249)xx159+(5-x)/(246)xx120=3impliesx=3.72` <br/>implies Mass percentage of `CuSO_(4).5H_(2)O = 74.4`</body></html> | |
| 49829. |
A mixture of CO and CO_(2) is found to have a density of 1.50" g "L^(-1) at 20^(@)C and 740 mm pressure. Calculate the composition of the mixture. |
| Answer» <html><body><p></p>Solution :Let the mol% of CO in mixture =<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/> mol%of `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/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a>-x)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/average-13416" style="font-weight:bold;" target="_blank" title="Click to know more about AVERAGE">AVERAGE</a> molecular mass `=([(x xx28)+(100-x)xx24])/(100)` <br/> `:. ([(x xx28)+(100-x)xx24])/(100)=(dRT)/(P)""(" Because ",M=(dRT)/(P))` <br/> `=((x xx28)+(100-x)xx24)/(100)=1.50gL^(-1)xx0.0821 <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a> atm K^(-1) mol^(-1)xx (293K)/(100)` <br/> mol% of `CO=43.38" and mol % of "CO_(2)=(100-x)` <br/> `=100-43.38=56.62`.</body></html> | |
| 49830. |
A mixture of CO and CO_(2) is found to have a density of 1.50g litreat 30^(@)C and 730mm What is composition of mixture ? . |
| 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> :`<a href="https://interviewquestions.tuteehub.com/tag/co-920124" style="font-weight:bold;" target="_blank" title="Click to know more about CO">CO</a> =32.19%, 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>) =67.81%`</body></html> | |
| 49831. |
A mixture of CO and CO_2 is found to have a density of 1.5 g/L at 30^(@)C and 740 torr . What is the composition of the mixture . |
| Answer» <html><body><p>`CO = 0.3575 , CO_2 = 0.64225`<br/>`CO = 0.64225 , CO_2 = 0.3575`<br/>`CO = 0.500 , CO_(2) = 0.500`<br/>`CO = 0.2500 , CO_(2) = 0.7500` <br/></p>Solution :Useexpression , ` d = (<a href="https://interviewquestions.tuteehub.com/tag/pm-1156895" style="font-weight:bold;" target="_blank" title="Click to know more about PM">PM</a>)/(RT)` <br/> `implies M = (<a href="https://interviewquestions.tuteehub.com/tag/drt-433194" style="font-weight:bold;" target="_blank" title="Click to know more about DRT">DRT</a>)/(P) = (1.5 xx 0.0821 xx 303 xx 76)/(74)` <br/> `= 38.276` <br/> Let one mole of 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> <a href="https://interviewquestions.tuteehub.com/tag/contain-409810" style="font-weight:bold;" target="_blank" title="Click to know more about CONTAIN">CONTAIN</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> mole of CO and (1 -n) mole of `CO_(2)` <br/> `n xx 28 + (1 - n) xx 44 = 1 xx 38.276` <br/> [Molecular weights , `CO = 28 CO_(2) = 44]` since n= 0.35775 <br/> Number of moles `CO = 0.35775` & number of moles of `CO_(2) = 0.64225`</body></html> | |
| 49832. |
A mixture of camphor and NaCl can be separated by |
| Answer» <html><body><p>Sublimation<br/>Evaporation<br/>Filtration<br/>Decantation</p>Answer :A</body></html> | |
| 49833. |
A mixture of C_(4)H_(8) and C_(2)H_(4) was completely burnt in excess of oxygen yielding equal volumes of CO_(2) and steam. Calculate the percentage (by volume ) of the compounds in the original mixture: |
| Answer» <html><body><p>25% `C_(4)H_(8)` and 75% `C_(2)H_(4)` <br/><a href="https://interviewquestions.tuteehub.com/tag/30-304807" style="font-weight:bold;" target="_blank" title="Click to know more about 30">30</a>% `C_(4)H_(8)` and 70% `C_(2)H_(4)` <br/>75% `C_(4)H_(8)` and25% `C_(2)H_(4)` <br/>50% `C_(4)H_(8)` and 50% `C_(2)H_(4)` </p>Answer :A::B::<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>::D</body></html> | |
| 49834. |
A mixture of an acid anhydride (A) anda mono basic acid (B) on heating produces another mono basic acid (C) of equivalent weight 74 and m an anhydride (D). The acids and anhydrides remain in equilibrium. The anhydride (D) contains two identical fluoro alkyl groups. The acid (B) contains a trifluoro methyl group and has an equivalent weight of 128. Give structures of (A) to (D) with proper reasoning (Atomic weight of fluorine =19). |
| Answer» <html><body><p></p>Solution :`underset((A))((RCO)_2)O+2R'-underset((<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>))(COOH)tounderset("<a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a>.wt.<a href="https://interviewquestions.tuteehub.com/tag/74-334795" style="font-weight:bold;" target="_blank" title="Click to know more about 74">74</a>")underset((C))(2RCOOH)+underset((D))(R'CO)_2O` <br/> Mol. Wt. of `R-COOH=74,R+12+32+1=74,R=74-45=29(C_2H_5)` <br/> Thus, (C) is `CH_3CH_2COOH` and on the basis of above reaction, acid anhydride (A) is `(C_2H_5CO)_2O`.The acid anhydride (D) contains two identical fluoro alkyl groups. The acid (B) contians a trifluoro methyl group and has an equivalent weight of 128. Hence, mono basic acid (B) is `CF_3CH_2COOH` and so, acid anhydride (D) is `(CF_3CH_2CO)_2O`. <br/> Reactions: <br/> `underset((A))((CH_3CH_2CO)_2O)+underset("Eq.wt.=128")underset((B))(CF_3CH_2COOH)tounderset("Eq.wt.= 74")underset((C))(CH_3CH_2COOH)+underset((D))((CF_3CH_2CO)_2O)`.</body></html> | |
| 49835. |
A mixture of aluminium and zinc weighing 1.67 g was completely dissolved in acid and 1.69 litres of hydrogen measured at 0°C and 1 atmospheric pressure were evolved. What was the original weight of aluminium in the mixture? |
| Answer» <html><body><p></p>Solution :Both aluminium and zinc react with sulphuric acid to form hydrogen. Suppose, the mixture contains x g of aluminium. <a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a>, amount of zinc in the mixture = `1.67 -x g`. The corresponding equations are: <br/> `underset(53.96 g)(2Al) + H_(2)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>) to underset(3 xx 22.4 "L at S.T.P.")(Al_(2)(SO_(4))_(3)) + 3H_(2)` <br/> `underset(65.38 g)(Zn) + H_(2)SO_(4) to underset(22.4 L "at S.T.P")(ZnSO_(4)) + H_(2)` <br/> `therefore 53.96` g of Al evolve `H_(2) = 3 xx 22.4` L at S.T.P. <br/> `therefore x <a href="https://interviewquestions.tuteehub.com/tag/gm-1008640" style="font-weight:bold;" target="_blank" title="Click to know more about GM">GM</a>` of Al will evolve `H_(2) = (3 xx 22.4)/(53.96) xx x` <br/> `=1.24 x` L at S.T.P. <br/> <a href="https://interviewquestions.tuteehub.com/tag/similarly-1208242" style="font-weight:bold;" target="_blank" title="Click to know more about SIMILARLY">SIMILARLY</a>, `therefore 1.24 x + (22.4)/(65.38) xx (1.67 -x) = 1.69` <br/> which gives `x = 1.24` <br/> Therefore, the <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 aluminium in the given sample is 1.24 g.</body></html> | |
| 49836. |
A mixture of acidified K_(2)Cr_(2)O_(7) and 10% Kl is titrated againstagainst Na_(2)S_(2)O_(3) (sodium thiosulphate) solution using starch indicator. The colour of the reaction m ixture at the end point is: |
| Answer» <html><body><p>Yellow<br/>Blue<br/>Green<br/>Colourless</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> | |
| 49837. |
A mixture of acetone and methanol can be separated by |
| Answer» <html><body><p>vacuum distillation <br/>Steam distillation <br/><a href="https://interviewquestions.tuteehub.com/tag/fractional-998803" style="font-weight:bold;" target="_blank" title="Click to know more about FRACTIONAL">FRACTIONAL</a> distillation <br/>none of these </p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/acetone-847404" style="font-weight:bold;" target="_blank" title="Click to know more about ACETONE">ACETONE</a> (b.p. `56^(@)C`) and methano (b.p. `<a href="https://interviewquestions.tuteehub.com/tag/67-331473" style="font-weight:bold;" target="_blank" title="Click to know more about 67">67</a>^(@)C`) <a href="https://interviewquestions.tuteehub.com/tag/differ-951389" style="font-weight:bold;" target="_blank" title="Click to know more about DIFFER">DIFFER</a> very less in their boiling points so they are separated by fractional distillation.</body></html> | |
| 49838. |
A mixture of 7g of nitrogen and 8g of oxygen at STP occupies a volume of |
| Answer» <html><body><p>11,200 mL<br/>22, 400 mL<br/>2240 mL<br/>5600 mL</p>Solution :Total moles `=(7)/(<a href="https://interviewquestions.tuteehub.com/tag/28-299271" style="font-weight:bold;" target="_blank" title="Click to know more about 28">28</a>)+(8)/(32)=(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)/(2)` <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> V_(STP)=22.4xx(1)/(2)=11.2` <a href="https://interviewquestions.tuteehub.com/tag/lit-537267" style="font-weight:bold;" target="_blank" title="Click to know more about LIT">LIT</a></body></html> | |
| 49839. |
A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H_(2)SO_(4). The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be............. . |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`underset(underset(46 g)("Formic acid"))(HCOOH) overset(conc.H_(2)SO_(4))rarrunderset(28g)(CO)+ H_(2)O` <br/> Now 46 g of COOH evolve CO = 28 g <br/> `:. 2.3 g` of HCOOH will evolve `CO = (28)/(46) <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 2.3` <br/> = 1.4 g <br/>`{:(""COOH),("|" overset(conc.H_(2)SO_(4))rarr CO+CO_(2)+H_(2)O),(COOH""28 g),("Oxalic acid"),(""90g):}` <br/> When the <a href="https://interviewquestions.tuteehub.com/tag/gaseous-467815" style="font-weight:bold;" target="_blank" title="Click to know more about GASEOUS">GASEOUS</a> mixture of `(CO+CO_(2)` is passed through KOH pellets, `CO_(2)` is <a href="https://interviewquestions.tuteehub.com/tag/absorbed-846166" style="font-weight:bold;" target="_blank" title="Click to know more about ABSORBED">ABSORBED</a> while CO phase out <br/> `2KOH + CO_(2) rarr K_(2)CO_(3) + H_(2)O` <br/> Now, 90g of oxalic acid evolve `CO = 28 g` <br/> `:. 4.5 g` of oxalic acid will evolve CO <br/> `= (28)/(90) xx 4.5 = 1.4 g` <br/> Total amount of CO <a href="https://interviewquestions.tuteehub.com/tag/evolved-2623419" style="font-weight:bold;" target="_blank" title="Click to know more about EVOLVED">EVOLVED</a> = 1.4 + 1.4 = 2.8 g</body></html> | |
| 49840. |
A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H_2SO_4 . The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be |
| Answer» <html><body><p>4.4<br/>1.4<br/>2.8<br/>3</p>Solution :2.3 gm formic <a href="https://interviewquestions.tuteehub.com/tag/acid-847491" style="font-weight:bold;" target="_blank" title="Click to know more about ACID">ACID</a> (HCOOH) <br/>` = (2.3)/(46) = 1/20 ` mol HCOOH <br/>4.5 gram oxalic acid `(H_2 C_2O_4)<br/>` = (4.5)/(90) = 1/20 ` mol `H_2C_2O_4`<br/>(i) `<a href="https://interviewquestions.tuteehub.com/tag/underset-3243992" style="font-weight:bold;" target="_blank" title="Click to know more about UNDERSET">UNDERSET</a>(1/20 "mol")(HCOOH) overset(<a href="https://interviewquestions.tuteehub.com/tag/conc-927968" style="font-weight:bold;" target="_blank" title="Click to know more about CONC">CONC</a>. H_2SO_4)(to) underset(1/20 "mol")(CO + H_2O)`<br/>(ii) `H_2C_2 O_4 overset(conc. H_2SO_4)(to) CO + CO_2 + H_2O` <br/>`CO_2`will be absorbed in KOH anc CO isadditional product.<br/>Total <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> of additional `CO_2 = 1/20 + 1/20 = 1/10`mol<br/>` = 1/10 "mol" = 1/10 xx <a href="https://interviewquestions.tuteehub.com/tag/28-299271" style="font-weight:bold;" target="_blank" title="Click to know more about 28">28</a> = 2.8 g`</body></html> | |
| 49841. |
A mixture of 2 moles of N_(2) and 8 moles of H_(2) are heated in a 2 lit vessel, till equilibrium is established. At equilibrium, 04 moles of N_(2) was present. The equilibrium concentration of H_(2) will be |
| Answer» <html><body><p>2 mole/lit<br/>4 mole/lit<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>.6 mole/lit<br/>1 mole/lit</p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_B_C04_E01_048_S01.png" width="80%"/> <br/> at equlibrium no. of moles of `N_(2)`. <br/> 2-<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>=0.4, x=1.6 <br/> <a href="https://interviewquestions.tuteehub.com/tag/molar-562965" style="font-weight:bold;" target="_blank" title="Click to know more about MOLAR">MOLAR</a> con. Of `H_(2)` at equlibrium = `(8-3x)/(2)=1.6` <br/> at eqm `underset(1-x)underset(1)(SO_(2)Cl_(2)) <a href="https://interviewquestions.tuteehub.com/tag/harr-2692945" style="font-weight:bold;" target="_blank" title="Click to know more about HARR">HARR</a> underset(x)underset()(SO_(2))+underset(1+x)underset(1)(Cl_(2))`</body></html> | |
| 49842. |
A mixture of1*57 " mol of " N_(2) , 1*92 " mol of " H_(2) and8*13 " mol of " NH_(3) " is introduced into a20 L reaction vessel at 500 K. At this temperature , the equilibrium constant ," K_(c) for the reaction , N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) " is " 1*7 xx 10^(2) Is the reaction mixture at equilibrium ? If not, what is the direction of the net reaction ? |
| 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> :The reaction is : ` N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g) ` <br/> ` Q_(c) = ([NH_(3)]^(2))/([N_(2) ][H_(2)]^(3)) = ((8*<a href="https://interviewquestions.tuteehub.com/tag/13-271882" style="font-weight:bold;" target="_blank" title="Click to know more about 13">13</a>)/<a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a>"mol"L^(-1))^(2)/(((1*57)/2"mol" L^(-1))((1*92 )/20 "mol" L^(-1))^(3))=2* <a href="https://interviewquestions.tuteehub.com/tag/38-310079" style="font-weight:bold;" target="_blank" title="Click to know more about 38">38</a> xx 10^(3)` <br/>As ` Q_(c) != K_(c),` the reaction mixture is not in equilibrium .<br/> As `Q_(c) gtK_(c)`, the et reaction will be in the backward direction .</body></html> | |
| 49843. |
A mixture of 1.57 mol of N_2 1.92 mol of H_2 and 8.13 mol of NH_3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, K_c for the reaction N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g)) is 1.7 xx 10^2. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction ? |
| Answer» <html><body><p></p>Solution :`{:("<a href="https://interviewquestions.tuteehub.com/tag/equilibrium-974342" style="font-weight:bold;" target="_blank" title="Click to know more about EQUILIBRIUM">EQUILIBRIUM</a> reaction :",N_(2(g)) +, 3H_(2(g)) <a href="https://interviewquestions.tuteehub.com/tag/harr-2692945" style="font-weight:bold;" target="_blank" title="Click to know more about HARR">HARR</a> , 2NH_(3(g))),("Mol of mixture :",1.57,1.92,8.13),("mol" L^(-1) "of equili. mixture :", 1.57/20,1.92/20,8.13/20),(,0.0785,0.096,0.4065):}` <br/> The equilibrium reaction mixture quotient, `Q_c` <br/> `therefore Q_c=[NH_3]^2/([N_2][H_2]^3)=(0.4065)^2/((0.0785)(0.096)^3)` <br/> Here, `(Q_c=2.3792) lt (K_c=1.7xx10^2)` . The <a href="https://interviewquestions.tuteehub.com/tag/value-1442530" style="font-weight:bold;" target="_blank" title="Click to know more about VALUE">VALUE</a> of `Q_c` is less . So, the reaction will proceed in the direction of the products (<a href="https://interviewquestions.tuteehub.com/tag/forward-464460" style="font-weight:bold;" target="_blank" title="Click to know more about FORWARD">FORWARD</a> reaction).</body></html> | |
| 49844. |
A mixture of 1.57 mol of N_2 1.92 mol of H_2 and 8.13 mol of NH_3 is introduced into a 20 L reaction vessel at 500 K . At this temperature, the equilibrium constant K_c for the reactin. N_(2)(g)+3H_(2) (g)j hArr 2NH_(3)(g) is 1.7 xx 10^(-2) Is this reaction at equilibriuim ? if not , what is the direction of net rection ? |
| 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> :The reaction is `N_(2)(g) + 3H_(2)(g) hArr 2NH_(3)(g)`<br/> Conecentration quotient `(Q_c)= ([NH_3]^)/([N_2][H_2]^3)` <br/> `=((8.13/20 " mol "L^(-1))^1)/((1.57/20 "mol " L^(-1))xx(1.92/20 " mol "L^(-1))^3)=2.38 xx <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^3`<br/> The equilibrium contant `(K_c)` for the reaction `=1.7 xx 10^(-2)`<br/> As `Q_(<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>) <a href="https://interviewquestions.tuteehub.com/tag/ne-1112263" style="font-weight:bold;" target="_blank" title="Click to know more about NE">NE</a> K_(C)` , this means that the reaction is not in a state of equilibrium.</body></html> | |
| 49845. |
A mixture of 1 mole of N_(2) and 3 moles of H_(2) is allowed to react at a constant pressure of 100 bar. At equilibrium, 0.6 mole of ammonia is formed. Calculate the equilibrium constant for the reaction N_(2)+3H_(2)hArr2NH_(3). |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/initial-499269" style="font-weight:bold;" target="_blank" title="Click to know more about INITIAL">INITIAL</a> 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)(N_(2))+underset(3)(3H_(2))hArrunderset(0)(2NH_(3))` <br/> `{:("No. of moles reacting"),("at equilibrium"):}}x""<a href="https://interviewquestions.tuteehub.com/tag/3x-310805" style="font-weight:bold;" target="_blank" title="Click to know more about 3X">3X</a>""2x=0.6thereforex=0.3` <br/> `{:("No. of moles present"),("at equilibrium"):}}{:(1-0.3,3-0.9),(=0.7,=2.1):}" "0.6` <br/> `therefore` Total no. of moles present at equilibrium = 0.7 + 2.1 + 0.6 = 3.4 <br/> Partial pressure = <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> fraction `xx` total pressure <br/> Since the total pressure = 10 bar. <br/> `P_(N_(2))=0.7/3.4xx100`bar = 20.59 bar <br/> `P_(H_(2))=2.1/3.4xx100` bar = 61.76 bar <br/> `P_(NH_(3))=0.6/3.4xx100` bar = 17.65 bar <br/> `K_(p)=(P_(NH_(3))^(2))/(P_(NH_(2))xxP_(H_(2))^(3))=((17.65)^(2))/(20.59xx(61*76)^(3))=6.420xx10^(-5)`</body></html> | |
| 49846. |
A mixture of 1-iodoethane and 1-iodopropane is treated with sodium metal and dry ether to carry out Wurtz reaction. Which of the following hydrocarbons will be formed? |
| Answer» <html><body><p>Propane+<a href="https://interviewquestions.tuteehub.com/tag/hexane-485026" style="font-weight:bold;" target="_blank" title="Click to know more about HEXANE">HEXANE</a><br/>Ethane+Propane<br/>Butane+Propane<br/>Butane+<a href="https://interviewquestions.tuteehub.com/tag/pentane-1150050" style="font-weight:bold;" target="_blank" title="Click to know more about PENTANE">PENTANE</a>+Hexane</p>Solution :`CH_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)CH_(2)I+2Na+ICH_(2)CH_(2)CH_(3) <a href="https://interviewquestions.tuteehub.com/tag/overset-2905731" style="font-weight:bold;" target="_blank" title="Click to know more about OVERSET">OVERSET</a>(ether)to underset("Butane")(CH_(3)CH_(2)CH_(2)CH_(3))+ underset("Pentane")(CH_(3)CH_(2)CH_(2)CH_(2)CH_(3))+underset("Hexane")(CH_(3)CH_(2)CH_(2)CH_(2)CH_(2)CH_(3))`</body></html> | |
| 49847. |
A mixture of 0.5 mole of CO and 0.5 mole of CO_(2) is taken in a vessel and allowed to effuse out through a pin hole into another vessel which has vacum If a total of A mole has effused out in time t show that M_(1)A+M_(2)(1-A)=36 where M_(1) and M_(2) are mean molar masses of the mixture that has effused out and the mixture still remaining in vessel respectively . |
| Answer» | |
| 49848. |
A mixture is to eb anlysed for penicllin. You add 10.0 mg fo penicllin labelled with .^(14)C that has a specific activity of 0.785mu Ci mg^(-1). From this mixture you are able to isotlate only 0.42 mg of pure penicllin. The specific activity of the isolated pen,om od 0.102mu Ci mg^(-1). How much penicllin was in theoriginal mixture? |
| 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> :`66.96 <a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a>`</body></html> | |
| 49849. |
A mixture of 0.3 mole of H_(2) and 0.3 mole of I_(2) is allowed to react in a 10 lit vessel at 500^(@)C. If K_(C) of H_(2) +I_(2) |
| Answer» <html><body><p>0.15 mole<br/>0.24 mole<br/>0.03 mole<br/>0.06 mole</p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_B_C04_E02_035_S01.png" width="80%"/> <br/> `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>)=((<a href="https://interviewquestions.tuteehub.com/tag/2x-301182" style="font-weight:bold;" target="_blank" title="Click to know more about 2X">2X</a> // 10)^(2))/(((0.3Lx)/(10))^(2))=((2x)/(0.3-x))^(2)` <br/> `64=((2x)/(0.3-x))^(2) <a href="https://interviewquestions.tuteehub.com/tag/implies-1037962" style="font-weight:bold;" target="_blank" title="Click to know more about IMPLIES">IMPLIES</a> x=0.24` <br/> `:.` unreacted `I_(2)=0.3-0.24=0.06`</body></html> | |
| 49850. |
A mixture in which the mole ratio H_(2) and O_(2)is 2: 1 is used to prepare water by the reaction 2 H_(2) (g) + O_(2) (g) to 2 H_(2)O (g) The total pressure of the container is 0*8 " atm " 20^(@) Cbefore the reaction. Determine the final pressure at 120^(@) C after reaction assuming 80% yield of water. |
| 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> :` {:(,2H_(2)(g),+,O_(2)(g),to,2H_(2)O(g)),(" Intial moles" ,2a ,,a,,0):}` <br/> As pressureare in the ratio of their moles <br/> ` :. 2 a + a = 0*8 "<a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a>" or 3 a = 0*8 or a = (0*8)/3 "atm " `<br/> <a href="https://interviewquestions.tuteehub.com/tag/theoretically-3201659" style="font-weight:bold;" target="_blank" title="Click to know more about THEORETICALLY">THEORETICALLY</a> expected yield of `H_(2)O = 2 a` <br/> Actual yield ` 2 a xx 90/100 = 1*6 a` <br/> ` :. H_(2) reacted = 1*6 a " moles ",O_(2) "reacted "= 0* 8 a " moles "`<br/> Moles after reaction : ` H_(2)= 2 a - 1*6 a = 0* 4 a, O_(2)= a - 0*8 a = 0*2 a `<br/> Total no. of moles ` = 0*4 a + 0*2 + 1*6 a = 2*2a `<br/><a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a>, <a href="https://interviewquestions.tuteehub.com/tag/final-461168" style="font-weight:bold;" target="_blank" title="Click to know more about FINAL">FINAL</a> pressure` = 2*2 xx(0*8)/3= 0* 59 "atm"`</body></html> | |