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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.
| 51501. |
1.725 g of a metal carbonate is mixed with 300 mL of N/10HCI. 10 mL of N/2 sodium hydroxide were required to neutralise excess of the acid. Calculate the equivalent mass of the metal carbonate. |
| 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/69-331759" style="font-weight:bold;" target="_blank" title="Click to know more about 69">69</a></body></html> | |
| 51502. |
17.1 gms of Al_(2)(SO_(4))_(3) is present in 500 ml of aqueous solution. It.s concentration can be |
| Answer» <html><body><p>0.1 M<br/>0.6 N<br/>1 N<br/>0.5 M </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> :`M=(17.1)/(342xx0.5)=0.1M=0.6N`</body></html> | |
| 51503. |
17 gm of H_2O_2is present in 100 ml of an aqueous solution. When 10 ml of this solution is decomposed, the STP volume of oxygen obtained is |
| Answer» <html><body><p>56ml<br/>5.6L<br/>0.56L<br/>560L</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> | |
| 51504. |
1.7 g of an unknown gas occupies 2.24 L at N.T.P Calculate its molecular mass. Identify the gas. |
| 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> :2.24 L of the <a href="https://interviewquestions.tuteehub.com/tag/gas-1003521" style="font-weight:bold;" target="_blank" title="Click to know more about GAS">GAS</a> of N.T.P has mass = 1.7 <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/> 22.4 L of the gas at N.T.P has mass `= ((1.7g))/((2.24L))xx(22.4L)=17g`<br/> The molecular mass of gas `= 17 "g mol"^(-1)` <br/> The gas is <a href="https://interviewquestions.tuteehub.com/tag/ammonia-859284" style="font-weight:bold;" target="_blank" title="Click to know more about AMMONIA">AMMONIA</a> `(NH_(3))`</body></html> | |
| 51505. |
1.7 g of AgNO_(3) dissolved in 100 g of water wire mixed with 0.585 g of NaCl dissolved in 100 g of water when 1.435 g of AgCl and 0.85 g of NaNO_(3) were formed. Justify that the data illustrates law of conservation of mass. |
| Answer» <html><body><p></p>Solution :Since water has not taken <a href="https://interviewquestions.tuteehub.com/tag/part-596478" style="font-weight:bold;" target="_blank" title="Click to know more about PART">PART</a> in the reaction, its mass (<a href="https://interviewquestions.tuteehub.com/tag/200-288914" style="font-weight:bold;" target="_blank" title="Click to know more about 200">200</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>) will count towards both <a href="https://interviewquestions.tuteehub.com/tag/sides-1207029" style="font-weight:bold;" target="_blank" title="Click to know more about SIDES">SIDES</a> <br/> Mass 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> `= 1.7 + 0.585 + 200 = 202.285 g` <br/> Mass of products `= 1.435 + 0.85 + 200 = 202.285 g` <br/> `:.` Law of conservation of mass is verified.</body></html> | |
| 51506. |
16gm of oxygen and 3gm of hydrogen are present in a vessel at 0^@C and 760mm of Hg pressure. Volume of the vessel is |
| Answer» <html><body><p>`22.4 <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>`<br/>`44.8 L`<br/>`11.2 L`<br/>`5.6 L`</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> | |
| 51507. |
1.64 g of a mixture of calcium carbonate and magnesium carbonate were dissolved in 50 mL of 0.8 N hydrochloric acid. The excess of the acid required 16 mL N//4 sodium hydroxide solution for neutralisation. Find out the percentage composition of the mixture of two carbonates. |
| 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> | |
| 51508. |
16.26 milligram of sample of an element X contains 1.66 xx 10^(20) atoms. What is the atomic mass of the element X ? |
| Answer» <html><body><p><br/></p>Solution :`1.66 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10^(20)` atoms <a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a> have <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> = 16.26 <a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a> <br/> `6.022 xx 10^(23)` atoms of X have mass `= ((6.022xx10^(23)"atoms"))/((1.66xx20^(20)"atoms"))xx(16.26mg)` <br/> `=59.0xx10^(3)mg=59.0g`.</body></html> | |
| 51509. |
1.60g of a metal were dissolved in HNO_3to prepare its nitrate. The nitrate on strong heating gives2g oxide. The equivalent weight of metal is |
| Answer» <html><body><p>16<br/>32<br/>48<br/>12</p>Solution :E.w.t of <a href="https://interviewquestions.tuteehub.com/tag/metal-1094457" style="font-weight:bold;" target="_blank" title="Click to know more about METAL">METAL</a> = `("<a href="https://interviewquestions.tuteehub.com/tag/wt-1462289" style="font-weight:bold;" target="_blank" title="Click to know more about WT">WT</a> of metal ")/("wt of "O_2) <a href="https://interviewquestions.tuteehub.com/tag/xx8-3292693" style="font-weight:bold;" target="_blank" title="Click to know more about XX8">XX8</a> =(1.6)/(0.4) xx8=3`</body></html> | |
| 51510. |
1.60 g of a metal were dissolved in HNO_3 to prepare its nitrate. The nitrate was strongly heated when 2.0 g of the metal oxide was obtained. Calculate the equivalent weight of the metal. |
| Answer» <html><body><p></p>Solution :Mass of the <a href="https://interviewquestions.tuteehub.com/tag/metal-1094457" style="font-weight:bold;" target="_blank" title="Click to know more about METAL">METAL</a> taken = 1.60 g <br/> Mass of <a href="https://interviewquestions.tuteehub.com/tag/oxide-1144484" style="font-weight:bold;" target="_blank" title="Click to know more about OXIDE">OXIDE</a> formed = 2.0 g <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>` Mass of oxygen that combined with 1.60 g of the metal `=2.0- 1.60 = 0.40 g` <br/> `therefore`Equivalent weight of the metal `=("Mass of the metal")/("Mass of oxygen") xx 8 = 1.60/0.40xx 8 = 32`</body></html> | |
| 51511. |
1.6 g of pyrolusite was treated with 60 mL of normal oxalic acid and some H_(2)SO_(4). The oxalic acid left undecomposed was made up to 250 mL, 25 mL of this solution required 32 mL of 0.1 N potassium permangante (KMnO_(4)). Calcualte the percentage of pure MnO_(2) in pyrolusite. |
| 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> | |
| 51512. |
16 g. of oxygen occupies a volume of 22.4L at 1 atm and |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>^@C`<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`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/273-1832932" style="font-weight:bold;" target="_blank" title="Click to know more about 273">273</a>^@C`<br/>`273 K`</p>Answer :C</body></html> | |
| 51513. |
16 g of oxygen has same number of molecules as in |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/16-276476" style="font-weight:bold;" target="_blank" title="Click to know more about 16">16</a> g of CO<br/>28 g of `N_(2)`<br/>14 g of `N_(2)`<br/>1.0 g of `H_(2)`</p>Solution :The calculation of molecule can be done as follows : <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 molecules `= ("Mass")/("Molar mass") <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> "Avogadro"` <br/> No. of molecules , in 16 g oxygen `=(16)/(32) xx N_(A)=(N_(A))/(3)` <br/> In 16 g of `CO=(16)/(32)xxN_(A)=(N_(A))/(1.75)` <br/> In 28 g of `N_(2) = (28)/(28) xx N_(A)=N_(A)` <br/> In 14 g of `N_(2) = (14)/(28)xx N_(A) = (N_(A))/(2)` <br/> In 1 g of `H_(2)=(1)/(2) xx N_(A) = (N_(A))/(2)` <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>, 16 g of `O_(2)= 14 g` of `N_(2) = 1.0 g ` of `H_(2)`</body></html> | |
| 51514. |
1.6 g of an organic compound gave 2.6 g of magnesium pyrophosphate. The percentage of phosphorus in the compound is |
| Answer» <html><body><p>0.4538<br/>0.5438<br/>0.3776<br/>0.1902</p>Solution :% of `P=(<a href="https://interviewquestions.tuteehub.com/tag/62-330265" style="font-weight:bold;" target="_blank" title="Click to know more about 62">62</a>)/(222)<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/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>)/(w)<a href="https://interviewquestions.tuteehub.com/tag/xx100-3292680" style="font-weight:bold;" target="_blank" title="Click to know more about XX100">XX100</a>=(62)/(222)xx(2.6)/(1.6)xx100=45.38%`</body></html> | |
| 51515. |
16 g of an ideal gas SO_(X) occupies 5.6 L at S.T.P. What is its molecular mass ? What is the value of X ? |
| 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> :5.6 <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a> of the <a href="https://interviewquestions.tuteehub.com/tag/gas-1003521" style="font-weight:bold;" target="_blank" title="Click to know more about GAS">GAS</a> at S.T.P weigh = 16 g <br/> 22.4 L of the gas at S.T.P. weigh `= ((1.6g))/((5.6L))xx(22.4L)=<a href="https://interviewquestions.tuteehub.com/tag/64g-1909876" style="font-weight:bold;" target="_blank" title="Click to know more about 64G">64G</a>` <br/> By definition, molar mass of gas `(SO_(X))=64g` <br/> `:. 32+x(16)=64or16x=64-32=32orx=2` <br/> Value of x = 2 and the gas `= SO_(2)`</body></html> | |
| 51516. |
1.6-Diethy1 cyclohexene 2. 4-Hydrocy-5-isopropy1 hept-6-yn-2-one 3. 6-Ethy1-4-isopropy1-2-methy1 decane 4. 2,-Dimethy1 cyclo pent-3-en-1-one 5. Benzonitrile 6. Bicyclo [4.1.0] hept-3-en-2-one 7.Sprio [3.5] non-5-en-2-ol 8. 1-Ethy1-2-methy1 cyclo hexane 9. 4-Bromo-2-ethy-1-methy1 cyclohexane 10. 3-Isopropy1 cyclopetane carbaldehyde 11. 4-cyclo propy1-3,6-diethy1 undecane 12. 6-(Cyclobut-2-enl1) hex-2-ene 13. 1,4-Dicyclopenty1 butane 14. Ethylidene bromide 15. Ethylene dichloride 16. trans.-1,3 -Dibromo cyclobutance 17. 2-exo-3-endo Dichloro bicylco [2.2.1] heptane 18. Benzocyclopentene 19. 2,5-Dimethy1 oxalone or 2,5-dimethy1 oxa cyclopentane 20. Methy1 viny1 carbino1 21. 3-Cyclopenteny1 ethy1 ether 22. sec-Buty1 isopropy1 ether 23. N,N-Diethy1 butan-1-amine 24. o-Toluidine 25. Isoamy1 school 26. sec-Buty1 alcohol 27. isobuty1 alcohol 28. Maleic acid 29. Caproic acid Z-Crotonalddehyde |
| 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> :<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/KSV_ORG_P1_C01_S01_010_S01.png" width="80%"/> <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/KSV_ORG_P1_C01_S01_010_S02.png" width="80%"/><br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/KSV_ORG_P1_C01_S01_010_S03.png" width="80%"/><br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/KSV_ORG_P1_C01_S01_010_S04.png" width="80%"/><br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/KSV_ORG_P1_C01_S01_010_S05.png" width="80%"/></body></html> | |
| 51517. |
15cm^(3) of hydrocarbonrequires 45cm^(3) of oxygen for complete combustion and 30cm^(3) of CO_(2) is formed. The formula of hydrocarbon is |
| Answer» <html><body><p>`C_(3)H_(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)`<br/>`C_(2)H_(2)`<br/>`C_(4)H_(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)`<br/>`C_(2)H_(4)`</p>Solution :`C_(x)H_(y)+(x+(y)/(4))O_(2)rarr xCO_(2)+(y)/(2)H_(2)O` <br/> `{:("15 ml","45 ml","30 ml"),("1 ml","3 ml","2 ml"):}` <br/> `therefore x=2` <br/> `x+(y)/(4)=3 rArr (y)/(4)=1` <br/> `"hydrocarbon is "C_(2)H_(4)`</body></html> | |
| 51518. |
1.575 g of oxalic acid (COOH)_(2).xH_(2)O are dissolved in water and the volume made up to 250 mL. On titration 16.68 mL of this solution requires 25 mL of N/15 NaOH solution for complete neutralisation. Calculate the value of x. |
| Answer» <html><body><p></p>Solution :Molecular mass of oxalic <a href="https://interviewquestions.tuteehub.com/tag/acid-847491" style="font-weight:bold;" target="_blank" title="Click to know more about ACID">ACID</a>, `(COOH)_2 xH_2O = <a href="https://interviewquestions.tuteehub.com/tag/24-295400" style="font-weight:bold;" target="_blank" title="Click to know more about 24">24</a> + 62 + 2 + 18x =90 + 18x` <br/> Since oxalic acid is a dibasic acid, Eq. mass of oxalic acid `=("Molecular mass")/2` <br/> `=(90 + 8x)/2 = 45 + 9x` <br/> Normality of oxalic acid solution can be calculated from the following relation : <br/> `w=(NEV)/1000` <br/> <a href="https://interviewquestions.tuteehub.com/tag/substituting-1231652" style="font-weight:bold;" target="_blank" title="Click to know more about SUBSTITUTING">SUBSTITUTING</a> the values, we get <br/> `1.575 =(N xx (45 + 9x) xx 250)/1000`<br/> or `N =(1.575 xx 1000)/((45 + 9x) xx 250) = (6.3)/(45 + 9x)` <br/> According to the normality relation, <br/> `underset("oxalic acid")(N_(1)V_(1)) = underset("NaOH")(N_(2)V_(2))` <br/> `6.3/(45 + 9x) xx 16.68 = 1/15 xx 25` <br/> `45 + 9x = (6.3 xx 16.68 xx 15)/25` <br/> `=63.05 = 63` <br/> `9x = 63 - 45 = 18` <br/> or `x=18/9 =2`</body></html> | |
| 51519. |
1.53 g of a compound containing only sulphur, oxygen and chlorine after easy hydrolysis with water yielded acid products which consumed 91 " mL of " (N)/(2) sodium hydroxide for complete neutralisation in a parallel experiment, 0.4 g of the compound after hydrolysis with water, was treated with excess of BaCl_(2) solution and 0.7 g of BaSO_(4) was precipitated. What is the formula of the compound? |
| Answer» <html><body><p></p>Solution :After hydrolysis, acid products are obtained. This suggests that the substanece is an acid <a href="https://interviewquestions.tuteehub.com/tag/chloride-915854" style="font-weight:bold;" target="_blank" title="Click to know more about CHLORIDE">CHLORIDE</a>. Reaction with `BaCl_(2)` to yield `BaSO_(4)` implies that `H_(2)SO_(4)` is one of the product. The substance is (by surmise) `SO_(2)Cl_(2)` This can be <a href="https://interviewquestions.tuteehub.com/tag/verified-2323104" style="font-weight:bold;" target="_blank" title="Click to know more about VERIFIED">VERIFIED</a> by the <a href="https://interviewquestions.tuteehub.com/tag/following-463335" style="font-weight:bold;" target="_blank" title="Click to know more about FOLLOWING">FOLLOWING</a> way: <br/> `SO_(2)Cl_(2)(135g)=BaSO_(4)(233.4g)` <br/> `therefore0.4g-=[(233.4)/(135)xx0.4]g=0.692g` <br/> This agrees very nearly with the given <a href="https://interviewquestions.tuteehub.com/tag/data-25577" style="font-weight:bold;" target="_blank" title="Click to know more about DATA">DATA</a> `=0.7g` further, 135 g of `SO_(2)Cl_(2)-=H_(2)SO_(4)+2HCl` (by hydrolysis 4 <a href="https://interviewquestions.tuteehub.com/tag/equivalents-974752" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENTS">EQUIVALENTS</a>). <br/> `1.53g-=(4)/(135)xx1.53` equivalent `=0.0453` equivalent <br/> `91 " mL of " 0.5 N NaOH-=(91)/(1000)xx0.5=0.0455` equivalent <br/> This also agrees with the given data. <br/> Therefore, the formula is `SO_(2)Cl_(2)`.</body></html> | |
| 51520. |
1.520 g of the hydroxide of a metal on ignition gave 0.995 g of oxide. The equivalent weight of metal is |
| Answer» <html><body><p>`1.520`<br/>0.995<br/>`19.00`<br/>`9.00`</p>Solution :`("wt. of <a href="https://interviewquestions.tuteehub.com/tag/metal-1094457" style="font-weight:bold;" target="_blank" title="Click to know more about METAL">METAL</a> <a href="https://interviewquestions.tuteehub.com/tag/hydroxide-493242" style="font-weight:bold;" target="_blank" title="Click to know more about HYDROXIDE">HYDROXIDE</a>")/("wt. of metal <a href="https://interviewquestions.tuteehub.com/tag/oxide-1144484" style="font-weight:bold;" target="_blank" title="Click to know more about OXIDE">OXIDE</a>") = (<a href="https://interviewquestions.tuteehub.com/tag/em-969731" style="font-weight:bold;" target="_blank" title="Click to know more about EM">EM</a>+ EOH^(-))/(EM + <a href="https://interviewquestions.tuteehub.com/tag/eo-452332" style="font-weight:bold;" target="_blank" title="Click to know more about EO">EO</a>^(-))` <br/> `:.(1.520)/(0.995) = (x+17)/(x+8)` <br/> `:.1.520x+1.520xx8=0.995x+0.995xx17` <br/> `:.1.520x+12.160=0.995x+16.915` <br/> `:.0.525x=4.755` <br/> `:.x=(4.755)/(0.525)= 9.057`</body></html> | |
| 51521. |
1.520 g of hydroxide of a metal on ignition gave 0.995 g of oxide. The equivalent mass of metal is : |
| Answer» <html><body><p>1.52<br/>0.995<br/>190<br/>9</p>Answer :D</body></html> | |
| 51522. |
15.0g of an unknown molecular material is dissolved in 450g of water. The resulting solution freezes at -0.34^(@)C. What is the molar mass of the material? K_(f) for water = 1.86K kg mol^(-1). |
| 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> :`W_(("<a href="https://interviewquestions.tuteehub.com/tag/solvent-1217160" style="font-weight:bold;" target="_blank" title="Click to know more about SOLVENT">SOLVENT</a>"))=15.0g` <br/>` W_(("solvent")) =450g` <br/> `Delta T_(f) = T_(1) ^(0) - T_(f) =0-(-0.34)=0.34^(@)<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>` <br/> `Delta T_(f)= k _(f) m` <br/> `0.34 =1.86 xx (15)/(M) xx (1000)/(450)` <br/> `M = (1.86 xx 15 xx 1000)/(0.34 xx 450) = 182. <a href="https://interviewquestions.tuteehub.com/tag/35-308504" style="font-weight:bold;" target="_blank" title="Click to know more about 35">35</a> g <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a>^(-1).`</body></html> | |
| 51523. |
15.0 mL of 0.12 M KMnO_(4) solution are required to oxidise 20.0 mL of FeSO_(4) solution in aicdic medium what is the concentration of FeSO_(4) solution ? |
| Answer» <html><body><p><br/></p>Solution :The balaced chemcial equation for the <a href="https://interviewquestions.tuteehub.com/tag/redox-2246707" style="font-weight:bold;" target="_blank" title="Click to know more about REDOX">REDOX</a> reaction is <br/> `2KMnO_(4)+10FeSO_(4)+8H_(2)SO_(4)rarrK_(2)SO_(4)+2MnSO_(4)+5Fe_(2)(SO_(4)^(3)+8H_(2)O)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/applying-1982651" style="font-weight:bold;" target="_blank" title="Click to know more about APPLYING">APPLYING</a> molarity eqautin to the above redox reaction <br/> `=(15xx0.12)/(2)KmnO_(4)=(20xxM_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>))/(10)(FeSO_(4)) or M_(1)=(15xx0.12xx10)/(2xx20)=0.45`</body></html> | |
| 51524. |
150 mL of N/10HCI are required to react completely with 1.0 g of a sample of lime stone (CaCO_3). Calculate the percentage purity of the sample. |
| Answer» <html><body><p>70</p>Solution :Eq. mass of `CaCO_(3) = (40 + <a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a> + (3 xx 16))/2 = 50`. <br/> Suppose 1.0 g of the sample contains only <a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a> g of `CaCO_3`.<br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>` Number of gram <a href="https://interviewquestions.tuteehub.com/tag/equivalents-974752" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENTS">EQUIVALENTS</a> of `CaCO_(3)` in 1.0 g sample `=("Mass")/("Eq. mass") = w/50` <br/> Number of gram equivalents of HCI reacted = `(N xx V)/1000` <br/> `=(1/10 xx 150)/(1000) = 15/1000` <br/> According to the law of equivalence, number of gram Eq. of `CaCO_(3) `=no. of gram Eq. of HCl <br/> `therefore w/50 = 15/1000` or `w =(50 xx 15)/1000 = 0.75 g` <br/> `therefore` Purity of the sample `=0.75/1 xx 100 = 75%`</body></html> | |
| 51525. |
150 cm^3 of a decimolar NaOH solution is diluted to 750 cm^3. Find the molarity of the diluted solution. |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :0.02 M</body></html> | |
| 51526. |
1.50 g of a metal on being heated in oxygen gives 2.15 g of its oxide. Calculate the equivalent mass of the metal. |
| 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> :18.46</body></html> | |
| 51527. |
150 mL M/10Ba(MnO_(4))_(2), in acidic medium can oxidize completely |
| Answer» <html><body><p> 150 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> 1M `Fe^(+2)`<br/> <a href="https://interviewquestions.tuteehub.com/tag/50-322056" style="font-weight:bold;" target="_blank" title="Click to know more about 50">50</a> mL IM `FeC_(2)O_(4 )`<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 1M `C_(2)O_(4)^(-2)`<br/>25 mL 1M `K_(2)Cr_(2)O_(<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>)` solution </p>Solution :No. of m.eqts of `Ba(MnO_4)_2`reacted <br/> `=150 xx1/10xx10=150`</body></html> | |
| 51528. |
1*5 moles of PCl_(5)are heatedat constant temperature in a closed vessel of 4 litre capacity.At the equilibrium point, PCl_(5)" is "35 % " dissociated into " PCl_(3) and Cl_(2). Calculate the equilibrium constant. |
| Answer» <html><body><p><br/></p>Solution :`{:(,PCl_(<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>),hArr,PCl_(3),+,Cl_(2)),("Intial <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a>",1*5,,,,),("At. eqm.",(1*5 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 35/100 xx 1*5),,,,),(,=1*5 - 0*525=0*<a href="https://interviewquestions.tuteehub.com/tag/975-342773" style="font-weight:bold;" target="_blank" title="Click to know more about 975">975</a>,,0*525,,0*525),(" Molar cons.",0*975//4,,0*525//4,,0*525 //4):}`<br/> `K_(c) = (( 0* 525 // 4) (0*525//4))/((0*975 //4))=0*071`</body></html> | |
| 51529. |
15 mL 1 N H_(2) SO_(4), 25 mL of 4 N HNO_(3), and 20 mL of X M HCl were mixed and made up to 1000 mL. Prepared by dissolving 4.725 g of pure Ba(OH)_(2). 8H_(2) O in water made up to 0.25 litre. What is the molarity of HCl solution (i.e. find X) |
| Answer» <html><body><p></p>Solution :`15 mL` of `1 M H_(2) SO_(4) + 25 mL of 4 M HNO_(3) + 20 mL of <a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a> M <a href="https://interviewquestions.tuteehub.com/tag/hcl-479502" style="font-weight:bold;" target="_blank" title="Click to know more about HCL">HCL</a>` <br/> `:. N_(1) V_(1) + N_(2) + V_(2) + N_(3) V_(3) = N_(4) V_(4)``(V_(4) = 1000 mL)` <br/> `15 xx 2 + 25 xx 4 + 20 X = N_(4) xx 1000` <br/> `:. N_(4) = ((130 + 20 X)/(1000))` <br/> mEq of mixture of acid = mEq of `Ba(OH)_(2)` <br/> `Mw of Ba (OH)_(2). 8H_(2) O = 137.4 + 34 + 18 xx 8 = 315.4` <br/> `Ew = (315)/(2) = 157.7 g` <br/> `N of Ba(OH)_(2) .8H_(2) O = (W_(2) xx 1000)/(Ew_(2) xx V_(<a href="https://interviewquestions.tuteehub.com/tag/sol-1216281" style="font-weight:bold;" target="_blank" title="Click to know more about SOL">SOL</a>) ("in" mL))` <br/> `= (4.725 xx 1000)/(157.7 xx 250)` <br/> `0.1198 N ~~ 0.12 N` <br/> mEq of acid mix = mEq of `Ba(OH)_(2)` <br/> `20 xx N_(4) = 26 xx 0.12` <br/> `N_(4) = (26 xx 0.12)/(20) = 0.156 N` <br/> `implies (130 + 20 X)/(1000) = 0.156` <br/> `:. X = (0.156 xx 1000 - 130)/(20) = 1.3` <br/> `N` or `M HCl = 1.3`</body></html> | |
| 51530. |
1.5 gof brass containing Cu and Zn reasts with 3.0 M HNO_3 solution. The following reaction take place: Cu+HNO_3toCu^(2+)+NO_2(g)+H_2O Zn+H^(o+)+NO_(3)^(ɵ)toNH_(4)^(ɵ)+Zn^(2+)+H_(2)O (a). Calculate the percentage composition of brass. (b). How many " mL of " 3.0 HNO_3 will be required for completely reacting with 1.0 g of brass? |
| Answer» <html><body><p></p>Solution :`CutoCu^(2+)+2e^(-)` <br/> `undersetunderset(x=5)(1+x-6=0)[<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>^(+)+e^(-)+HNO_(3))toundersetunderset(x=4)(x-4=0)(NO_(2))+H_(2)O` <br/> `Cu+2H^(o+)+2HCO_(3)to2NO_(2)2H_(2)O+Cu^(2+)` <br/> `Cu+4HNO_(3)to2NO_(2)+2H_(2)O+Cu^(2+)+2NO_(3)^(ɵ)` <br/> or `Cu+4HNO_(3)to2NO_(2)+2H_(2)O+Cu^(2+)+2H_(2)O` <br/> 1 " mol of "`Cu=4 " mol of "HNO_(3)-=2 " mol of "NO_(2)` <br/> Mole of `NO_(2)=(PV)/(RT)=(1xx1xx0.4)/(0.082xx298)=0.0425` <br/> mole of `Cu=(0.0425)/(2)=(0.0425)/(2)xx63.5=1.3462g` <br/> Weight of `Zn=1.5-1.3462=0.1538g` <br/> `% Zn=10.25%` <br/> `% of Cu=89.75%` <br/> `[ZntoZn^(+2)+2e^(-)]xx4` <br/> `8e^(-)+undersetunderset(x=5)(x-6=-1)(NO_(3)^(ɵ))+10H^(o+)toundersetunderset(x=-3)(x+4=+1)(NH_(4)^(o+)+3H_(2)O)` <br/> or `4An+10HNO_(3)toNH_(4)^(o+)+4An^(2+)+3H_(2)O` <br/> `4Zn+10HNO_(3)toNH_(4)NO_(3)+4Zn(NO_(3))_(2)+3H_(2)O` <br/> `=3xxV` <br/> `V=0.2011 <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a> =20.11mL`</body></html> | |
| 51531. |
15 grams of a mixture of calcium carbonate and sodium carbonate on ignition liberated a gas which has occupied 2.24L at STP. What is the weight ratio of components in the given 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> :`2:1`</body></html> | |
| 51532. |
1.5 g sample of P_(2)O_(3) and some impurity was dissolved in water and warmed gentally till P_(2)O_(3) disproportionated quantitatively to PH_(3) and H_(3)PO_(4). The solutionswas then boiled to get rid off PH_(3(g)) and then cooled finally to room temperature and diluted to 100 mL. 10 mL of this solution was mixed with 20 mL of 0.3M NaOH. Now 10 mL of this solution required 3.6 mL of 0.05 M H_(2)SO_(4) for back titration. Determine % by weight of P_(2)O_(3) in sample. |
| 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/80-337972" style="font-weight:bold;" target="_blank" title="Click to know more about 80">80</a>%`;</body></html> | |
| 51533. |
1.5 g of sample of impure potassium dichromate was dissolved in water and made up to 500 mL solution . 25 mL of this solution required iodometrically 24 mL of a sodium thiosulphate solution. 26 mL of this sodium thisulphate solution required 25 mL of N//20 solution of pure potassium dichromate. Find the percentage purity of impure sample of potassium dichromate. |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/normality-1124423" style="font-weight:bold;" target="_blank" title="Click to know more about NORMALITY">NORMALITY</a> of sodium thiosulphate solution may be determined as : <br/> `N_(1)V_(1)(Na_(2)S_(2)O_(3))=N_(2)V_(2) ("pure" K_(2)Cr_(2)O_(7))` <br/> `N_(1)xx26=25xx(1)/(<a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a>)` <br/> `N_(1)=0.048` (<a href="https://interviewquestions.tuteehub.com/tag/hypo-1034629" style="font-weight:bold;" target="_blank" title="Click to know more about HYPO">HYPO</a>) <br/> The <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a> involved may be given as : <br/> `Cr_(2)O_(7)^(2-)+6O^(-)+4H^(+) to 2Cr^(3+)+3I_(2)+7H_(2)O` <br/> `3[I_(2)+2Na_(2)S_(2)O_(3) to 2NaI+Na_(2)S_(4)O_(6)]` <br/> 1 mole `K_(2)Cr_(2)O_(7)-=6 " mole" Na_(2)S_(2)O_(3)` <br/> 25 mL of solution of `K_(2)Cr_(2)O_(7)` is treated by 24 mL of 0.048 N hypo <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> 500 mL` of solution will be titrated by 480 mL of 0.048 N hypo <br/> No. of moles of hypo`=("Mass")/(M.w. (158))=(ExxNxxV)/(1000xx158)` <br/> `=(158xx0.048xx480)/(1000xx158)` <br/> =0.02304 mole <br/> No. of moles of `K_(2)Cr_(2)O_(7)=(1)/(6)`[No. of moles of hypo] <br/> `=(1)/(6)[0.02304]=3.84xx10^(-3)` <br/> Mass of `K_(2)Cr_(2)O_(7)=3.84xx10^(-3)xx294=1.12896` <br/> % purity `=(1.12896)/(1.5)xx100=75.26%`</body></html> | |
| 51534. |
15 g of ethane at 380 torr and 273^@ C occupy a volume of |
| Answer» <html><body><p>11.2 <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a><br/>44.8 L<br/> 33.6 L<br/> 22.4 L </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> | |
| 51535. |
1.5 g ofCdCl_(2) was found to contain 0.9 g of Cd.Calculate the atomic weight of Cd. |
| Answer» <html><body><p>118<br/>112<br/>106.5<br/>53.25</p>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> | |
| 51536. |
1.5 g of an organic compound in a quantitative determination of phosphorus gave 2.5090 g of Mg_(2)P_(2)O_(7). Calculate the percentage weight of phosphorus. |
| 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> :0.4671</body></html> | |
| 51537. |
15 c.c. of gaseous hydrocarbon required 45 c.c. of oxygen for complete combustion and 30 c.c. of carbondioxide is formed. The formula of the hydrocarbon is |
| Answer» <html><body><p>`CP_(3)H_(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)`<br/>`C_(<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)`<br/>`C_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)H_(10)`<br/>`C_(2)H_(4)`</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> | |
| 51538. |
15 cc of gaseous hydrocarbon required 45 cc of oxygen for complete combusion if 30 cc of CO_(2) is formed, the formula of the gaseous compound is |
| Answer» <html><body><p>`C_(3)H_(6)`<br/>`C_(2)H_(2)`<br/>`C_(4)H_(10)`<br/>`C_(2)H_(2)`</p>Solution :`C_(<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_(y)+ (x+(y)/(4)) O_(2) rarr xCO_(2) +(y)/(2) H_(2)O`</body></html> | |
| 51539. |
(1)4HCl_((g)) + O_(2(g)) overset"X"to 2Cl_(2(g)) + 2H_2O_((g)) (2) 2H_2O_((aq))underset"Heat"overset(Y) to +2H_2O_((l)) + O_(2(g)) Mention the formulas of X and Y. |
| Answer» <html><body><p>`X=CuCl_2 , Y=NO_2`<br/>`X=Cu_2Cl_2 , Y=MnO_2`<br/>`X=CuCl_2 , Y=MnO_2`<br/>`X=V_2O_5 , Y=MnO_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/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 51540. |
14g of element X combine wilh 16 g of oxygen. On the basis of this information, which of the followings is a correct statement? |
| Answer» <html><body><p>The element <a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a> could have an atomic weight of 7 and its oxide is XO<br/>The element X could have an atomic weight of 14 and its oxide <a href="https://interviewquestions.tuteehub.com/tag/formula-464310" style="font-weight:bold;" target="_blank" title="Click to know more about FORMULA">FORMULA</a> is `X_2O`<br/>The element X could have an atomic weight of 7 and its oxide is `X_2O`<br/>The element X could have an atomic weight of 14 and its oxide is `X_2O`</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> :C</body></html> | |
| 51541. |
1470 cm^3 of a gas is collected over water at 303 K and 74.4 cm of Hg. If the gas weighs 1.98 g and vapour pressure of water at 30^@C is 3.2 cm of Hg, calculate the molecular weight of the gas. |
| Answer» <html><body><p><br/></p>Solution :The gas collected over water is wet. The pressure of the <a href="https://interviewquestions.tuteehub.com/tag/dry-960094" style="font-weight:bold;" target="_blank" title="Click to know more about DRY">DRY</a> gas can be obtained by subtracting the vapour pressure of water from the observed pressure of the wet gas. <br/> `:. ""P = 74.4 - 3.2 = 71.2 " cm Hg " (71.2)/<a href="https://interviewquestions.tuteehub.com/tag/76-335558" style="font-weight:bold;" target="_blank" title="Click to know more about 76">76</a>` atm <br/>`V = 1470 cm^3 = 1470/1000 = 1.47 L, T = 303K` <br/> If the <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 of the gas is M, then the number of moles in 1.98 g of gas = `(1.98)/M` <br/>According to the gas equation `<a href="https://interviewquestions.tuteehub.com/tag/pv-593601" style="font-weight:bold;" target="_blank" title="Click to know more about PV">PV</a> = nRT`<br/> <a href="https://interviewquestions.tuteehub.com/tag/substituting-1231652" style="font-weight:bold;" target="_blank" title="Click to know more about SUBSTITUTING">SUBSTITUTING</a> the values, we have <br/>`(71.2)/76 xx 1.47 = (1.98)/M xx 0.0821 xx 303` <br/> `:. ""M = 35.8` <br/>Hence, the molecular weight of the given gas is 35.8.</body></html> | |
| 51542. |
14.7 g of sulphuric acid was needed to dissolve 16.8 g of a metal. Calculate the equivalent weight of the metal and the volume of hydrogen liberated at NTP. |
| 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/56-325545" style="font-weight:bold;" target="_blank" title="Click to know more about 56">56</a>, 3.36 <a href="https://interviewquestions.tuteehub.com/tag/litres-1075876" style="font-weight:bold;" target="_blank" title="Click to know more about LITRES">LITRES</a></body></html> | |
| 51543. |
1.44 g of pure Fec_(2)O_(4) was dissolved in dilH_(2)SO_(4) and th solution diluted to 100 cm^(3) calculate the volume of 0.01 M KMnO_(4) requiredto oxidise FeC_(2)O_(4) solution completely |
| 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> :Step 1 To write the balanced equation for the redox reaction <br/> both the <a href="https://interviewquestions.tuteehub.com/tag/cation-910854" style="font-weight:bold;" target="_blank" title="Click to know more about CATION">CATION</a> and anioic components of `FeC_(2)O_(4)` (ferrous oxalate) i.e `Fe^(2+)` and `C_(2)O_(4)^(2-)` are oxidised by `KmnO_(4)` to `Fe^(3+)` and `CO_(2)` respectively the comlete balanced redox equation is <br/> `5 Fe^(+)+MnO_(4)^(-) +8 H^(+)rarr5 Fe^(3+) rarr 5Fe^(3+)+Mn^(2+)+4H_(2)O` <br/> `5C_(2)O_(4)^(2-)+2 MnO_(4)^(-)+16 H^(+)rarr10CO_(2)+2 Mn^(2+)+8 H_(2)O` <br/> `5FeC_(2)O_(4)+3MnO_(4)^(-)+24H^(+)rarr5Fe^(3+)+10 CO_(2)+3Mn^(2+)+12H_(2)O` <br/> Step 2 To determine the molarity of `FeC_(2)O_(4)` solution <br/><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 of `FeC_(2)O_(4) =56+2xx12+4xx16=144 g` <br/> `"volume" =100 cm^(3)` <br/> `therefore "Molarity" =("weight" )/("mol.wrt")xx(1000)/("volume")=(1.44)/(144xx(1000)/(100)=0.1 M`<br/> Step 3 To calculate of 0.01 M `KMnO_(4)` solution <br/> Applying molarity equatin to balaced redox equation we have <br/> `(M_(12)V_(1))/(n_(1))(FeC_(2)O_(4))=(M_(2)V_(2))/(n_(2))(LMnO_(4))` <br/> substituting the values of `M_(1)(=0.1),V_(1)(=100)n_(1)=5, M_(2)=(=0.01)` andn=3 , we have <br/> `(0.1xx100)/(5)=(0.01xxV_(<a href="https://interviewquestions.tuteehub.com/tag/21-293276" style="font-weight:bold;" target="_blank" title="Click to know more about 21">21</a>))/(3) or V_(2) =(3x0.1xx100)/(5xx0.01)=600 cm^(3)` <br/> thus volume of 0.01 M `KMnO_(4)` solution required =600 `cm^(3)`</body></html> | |
| 51544. |
19.5 g of benzene on burning in the free supply of oxygen liberated 12.6 kJ of energy at constant pressure. Calculate the enthalpy of combustion of benzene. Write the thermochemical equation. |
| Answer» <html><body><p></p>Solution :19.5 <a href="https://interviewquestions.tuteehub.com/tag/grams-476111" style="font-weight:bold;" target="_blank" title="Click to know more about GRAMS">GRAMS</a> of benzene on burning liberated 126 kl. 78 grams of benzene on burning liberated 502 <a href="https://interviewquestions.tuteehub.com/tag/kj-1063034" style="font-weight:bold;" target="_blank" title="Click to know more about KJ">KJ</a>. <br/> The <a href="https://interviewquestions.tuteehub.com/tag/enthalpy-15226" style="font-weight:bold;" target="_blank" title="Click to know more about ENTHALPY">ENTHALPY</a> of combustion of benzene is` -502 kJmol ^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>).`The thermochemical equation is given as, <br/> `C _(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>) H _(6(l)) + (15)/(2) O _(2(g)) to 6CO _(2(g)) + 3H_(2(g)) + 3H _(2) O _((l)), Delta H =-502 kJ mol ^(-1)`</body></html> | |
| 51545. |
140 mm pressure is developed at equilibrium when PCl_(5) at 100 mm is subjected to dissociation . ThenK_(p) for PCl_(3) + Cl_(2) hArr PCl_(5) is (in atm^(-1)) nearly |
| Answer» <html><body><p>`0.03`<br/>`0.01`<br/>`0.04`<br/>`0.02`</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> | |
| 51546. |
1,4-pentadiene reacts with excess of HCI in thepresence of benzoyl peroxide to give compound X which upon reaction with excess of Mg in dry ether forms Y. Compound Y on treatment with ethyl acetate followed by dilute acid yields Z. Identify the structures of compound X,Y and Z. |
| 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> :1,4-Pentadiene on addition with excess of HCI in the <a href="https://interviewquestions.tuteehub.com/tag/presence-25557" style="font-weight:bold;" target="_blank" title="Click to know more about PRESENCE">PRESENCE</a> of benzoyl <a href="https://interviewquestions.tuteehub.com/tag/peroxide-1151683" style="font-weight:bold;" target="_blank" title="Click to know more about PEROXIDE">PEROXIDE</a>, forms 2,4-dichloro pentane (X) because HCI does not <a href="https://interviewquestions.tuteehub.com/tag/show-642916" style="font-weight:bold;" target="_blank" title="Click to know more about SHOW">SHOW</a> peroxide effect. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/GRB_ORG_CHM_P2_S01_C17_007_S01.png" width="80%"/></body></html> | |
| 51547. |
14^(@) F is equal to....... |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/298-1834947" style="font-weight:bold;" target="_blank" title="Click to know more about 298">298</a> <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/273-1832932" style="font-weight:bold;" target="_blank" title="Click to know more about 273">273</a> K`<br/>`263 K`<br/>`245 K`</p>Solution :`^(@)<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> = (.^(@)F - 32)/(1.8) = (14-32)/(1.8) = -<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(@) C` <br/> `K = .^(@)C + 273 = (-10) + 273 = 263`</body></html> | |
| 51548. |
1,4-Dimethylbenzene on heating with anhydrous AlCl_3 and HCl produces |
| Answer» <html><body><p>1,2-dimethylbenzene<br/>1,3-dimethylbenzene<br/>1,2,3-trimethylbenzene<br/>ethylbenzene</p>Solution :1,4-Dimethylbenzene when <a href="https://interviewquestions.tuteehub.com/tag/heated-1017243" style="font-weight:bold;" target="_blank" title="Click to know more about HEATED">HEATED</a> with anhydrous `AlCl_3` and <a href="https://interviewquestions.tuteehub.com/tag/hcl-479502" style="font-weight:bold;" target="_blank" title="Click to know more about HCL">HCL</a> undergoes <a href="https://interviewquestions.tuteehub.com/tag/isomerization-1052473" style="font-weight:bold;" target="_blank" title="Click to know more about ISOMERIZATION">ISOMERIZATION</a> to give thermodynamically more <a href="https://interviewquestions.tuteehub.com/tag/stable-1223548" style="font-weight:bold;" target="_blank" title="Click to know more about STABLE">STABLE</a> 1,3-dimethylbenzene as shown below : <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PR_CHE_02_XI_C13_E20_173_S01.png" width="80%"/></body></html> | |
| 51549. |
13.8g of N_(2)O_(4) was placed in a 1L reaction vessel at 400K and allowed to attain equilibrium N_(2)O_(4)(g)hArr2NO_(2)(g) The total pressure at equilibrium was found to be 9.15 bar . Calcualate K_(c), K_(p) and partial pressure at equilibrium. |
| 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> :`2.6 <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` 85.87 <a href="https://interviewquestions.tuteehub.com/tag/bar-892478" style="font-weight:bold;" target="_blank" title="Click to know more about BAR">BAR</a>, 8.34 bar</body></html> | |
| 51550. |
13*8 g " of "N_(2)O_(4) was placed in 1 L reaction vessel in 1 L reaction vessel at 400 K and allowed to attain equilibrium : N_(2)O_(4) (g) hArr 2 NO(g).The total pressure at equilibrium was found to be 9* 15 bar. Calculate K_(c) , K_(p) and partial pressures at equilibrium. |
| Answer» <html><body><p></p>Solution :`13*8 "g"N_(2)O_(4) = (13*8)/<a href="https://interviewquestions.tuteehub.com/tag/92-342088" style="font-weight:bold;" target="_blank" title="Click to know more about 92">92</a>" mol"=0*15" mol""` `13*8 N_(2)O_(4) = (13*8)/92" mol"=0*15" mol""(Molar mass of "N_(2)O_(4) = 92 g " mol"^(-1))` <br/> PV= nRT <br/> ` :. P <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 1 L = 0* 15 mol xx 0* 083 " bar"L mol^(-1) K^(-1) xx 400 K or P = 4*98 " bar"` <br/> ` {:(,N_(2)O_(4)(g),hArr,2NO_(2)),("<a href="https://interviewquestions.tuteehub.com/tag/initial-499269" style="font-weight:bold;" target="_blank" title="Click to know more about INITIAL">INITIAL</a> pressures",4*98 "bar",,0),(" At equilibrium",(4*98-p),,2p):}` <br/>` :. 4*98 - p+ 2 p = 9*15 " bar"or p= 4*17 " bar"` <br/> `:. (p_(N_(2)O_(4)))eq = 4* 98 - 4*17 = 0*<a href="https://interviewquestions.tuteehub.com/tag/81-338942" style="font-weight:bold;" target="_blank" title="Click to know more about 81">81</a> "bar", (p_(NO_(2)))eq = 2 xx 4*17 = 8*34 "bar"` <br/> `K_(p) = p_(NO_(2))^(2)//_(P_(N_(2)O_(4)))=(8*34)^(2)//)*81=<a href="https://interviewquestions.tuteehub.com/tag/85-339428" style="font-weight:bold;" target="_blank" title="Click to know more about 85">85</a>*87,` <br/> ` K_(p) = K_(c) (RT)^(Delta n) :. 85*87 = K_(c) (0*083 xx400)^(1) or K_(c)= 2*586 = 2*6`</body></html> | |