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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.
| 52101. |
0.15 g of an organic compound gave 0.12 g of silver bromide by carius by carius method. Find out the percentage of bromine in the compound. |
| 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> :<a href="https://interviewquestions.tuteehub.com/tag/percentage-13406" style="font-weight:bold;" target="_blank" title="Click to know more about PERCENTAGE">PERCENTAGE</a> of <a href="https://interviewquestions.tuteehub.com/tag/bromine-904632" style="font-weight:bold;" target="_blank" title="Click to know more about BROMINE">BROMINE</a> `=80/188xx ("Mass of AgBr")/("Mass of <a href="https://interviewquestions.tuteehub.com/tag/compound-926863" style="font-weight:bold;" target="_blank" title="Click to know more about COMPOUND">COMPOUND</a>")xx100=80/188xx0.12/0.15xx100=34%`</body></html> | |
| 52102. |
0.15 g of iodoform gave 0.2682 g of AgI. Calculate the percentage of iodine. |
| Answer» <html><body><p></p>Solution :Weight of compound = 0.15 <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/> Weight of silver <a href="https://interviewquestions.tuteehub.com/tag/iodide-1051130" style="font-weight:bold;" target="_blank" title="Click to know more about IODIDE">IODIDE</a> = 0.2682 g <br/> Molecular weight of silver iodide `= 108 + 127 = <a href="https://interviewquestions.tuteehub.com/tag/235-1826120" style="font-weight:bold;" target="_blank" title="Click to know more about 235">235</a>` <br/> 235 g of silver iodide contains 127 g of iodine <br/> 0.2682 g of AgI contains `=(127xx0.2682)/(235)` <br/> `=0.144 g` iodine <br/> 0.15 g of compound contains 0.144 g of iodine <br/> <a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a> g of compound contains `=(100xx0.1449)/(0.15)=96.6g` <br/> `therefore"Percentage of iodine"=96.6`</body></html> | |
| 52103. |
0.14kg of nitrogen at 300K is expanded isothermally and reversibly until its volume becomes doubled. Calculate the work done by the gas. |
| 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> :`1.729 <a href="https://interviewquestions.tuteehub.com/tag/kj-1063034" style="font-weight:bold;" target="_blank" title="Click to know more about KJ">KJ</a>`</body></html> | |
| 52104. |
0.14 g of an element on combustion gives 0.28 g of its oxide. What is that element ? |
| Answer» <html><body><p>Nitrogen<br/>Carbon<br/>Fluorine<br/>Sulphur</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> | |
| 52105. |
0.14 g of an element A combines with 0.16 g of the elements B. Also 0.05 g of another element C combines with 0.35 g of A. It was also found that 0.2857 g of C combines with 2.2857 g of B. Show that the data illutrates the law of reciprocal proportions. |
| 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> :In the compound AB <br/> 0.14 g of A are combined with `B = 0.16 g` <br/> 1.0 g of A are combined with `B=(0.16)/(0.14)g=1.143g` <br/> In the compound <a href="https://interviewquestions.tuteehub.com/tag/ac-361271" style="font-weight:bold;" target="_blank" title="Click to know more about AC">AC</a> <br/> 0.35 of A are combined with C = 0.05 g <br/> 1.0 g of A are combined with `C=(0.05)/(0.35)g=0.143 g` <br/> <a href="https://interviewquestions.tuteehub.com/tag/ratio-13379" style="font-weight:bold;" target="_blank" title="Click to know more about RATIO">RATIO</a> by weight of B and C combined with 1.0 g of C in the two compounds `= (1.143)/(0.143)` <br/> In the compound BC <br/> Ratio by weight of B and C `= (2.2857)/(0.2857)` <br/> Comparing (i) and (<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>), <br/> `(1.143)/(0.143):(2.2857)/(0.2857)or 8:8or 1:1` <br/> This illustrates the Law of Reciprocal Proportions. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/SKM_COMP_CHM_V01_XI_C01_E01_206_S01.png" width="80%"/></body></html> | |
| 52106. |
0.132g of an organic compound gave 50ml of N_(2) at NTP. The weight percentage of nitrogen in the compound is close to |
| Answer» <html><body><p>15<br/>20<br/>48.9<br/>47.43</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/wt-1462289" style="font-weight:bold;" target="_blank" title="Click to know more about WT">WT</a>% of `N_(2) = (28)/(22400) <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/vol-723961" style="font-weight:bold;" target="_blank" title="Click to know more about VOL">VOL</a> of" N_(2)(ml) "at stp")/("wt. of compound") xx 100`</body></html> | |
| 52107. |
0.13 g of Cu, when treated with AgNO_3 solution, displaced 0.433 g of Ag. 0.13 g of AI, when treated with CuSO_4 solution displaced 0.47g of Cu. 1.17 g of AI displaces 0.13 g of hydrogen from an acid. Find the equivalent weight of Ag if equivalent weights of Cu and Al are not known. |
| Answer» <html><body><p></p>Solution :`underset(0.13g)(Cu)+ AgNO_3 to underset(0.433g)(Ag) ""...(1)` <br/> `underset(0.13g)(AI)+ CuSO_4 to underset(0.47g)(Cu)""...(2)` <br/> `underset(1.17g)(AI) + "Acid" to underset(0.13g)(H_2)""...(3)` <br/> For Eqn. (3) : <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a>. of <a href="https://interviewquestions.tuteehub.com/tag/al-370666" style="font-weight:bold;" target="_blank" title="Click to know more about AL">AL</a> = eq. of hydrogen <br/> `therefore (1.17)/("eq. <a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a>. of AI")= (0.13)/(1),` eq. <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 Al is 9. <br/> For Eqn. (2) : eq. of Al = eq. of Cu <br/> `therefore (0.13)/(9) = (0.47)/("eq. wt. of Cu"),` eq. wt. of Cu= 32.5. <br/> For Eqn. (1) : eq. of Cu = eq. of Ag <br/> `therefore (0.13)/( 32.5) = (0.433)/("eq. wt. of Ag")` <br/> eq. wt. of Ag= 108.25.</body></html> | |
| 52108. |
0.12g organic compound gave 0.22g Mg_(2)P_(2)O_(7). What is the percentage of phosphorus in compound? (P= 31) (Molar mass of Mg_(2)P_(2)O_(7)= 222g) |
| Answer» <html><body><p></p>Solution :`2P rarr Mg_(2)P_(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>)` <br/> % `P =("<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 2P")/("<a href="https://interviewquestions.tuteehub.com/tag/molar-562965" style="font-weight:bold;" target="_blank" title="Click to know more about MOLAR">MOLAR</a> mass of" Mg_(2)P_(2)O_(7)) xx ("mass of "Mg_(2)P_(2)O_(7))/("molar mass of compound") xx 100` <br/> `=(<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) xx (0.22)/(0.12) xx 100= 51.20%`</body></html> | |
| 52109. |
0.12 g of an organic compound containing phosphorus gave 0.22 g of Mg_(2)P_(2)O_(7) by the usual analysis. Calculate the percentage of phosphorus in the compound. |
| Answer» <html><body><p></p>Solution :Here, the mas of the compound taken = 0.13 <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/> Mass of `Mg_(2)P_(2)O_(7) -= 2g` atoms of P or `(2 xx 24 + 2 xx 31 + 16 xx 7) = 222g` of `Mg_(2)P_(2)O_(7) -= 62 g` of P<br/> i.e., 222g of `Mg_(2)P_(2)O_(7)` <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/phosphorus-1153299" style="font-weight:bold;" target="_blank" title="Click to know more about PHOSPHORUS">PHOSPHORUS</a> = 62 g <br/> `:.` 0.22 g of `Mg_(2)P_(2)O_(7)` will contain phosphorus `= (62)/(222) xx 0.22 g` <br/> But this is the amount of phosphorus present in 0.12 g of the organic compound. <br/> `:.` Percentage of phosphorus `= (62)/(222) xx (0.22)/(0.12) xx 100 = 51.20`</body></html> | |
| 52110. |
0.12 g of an organic compound gave on combustion 0.18 g of water and 0.11 g of CO_(2). Calculate the percentage of C and H in the organic compound. |
| 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> :`C=25%, H=16.66%`</body></html> | |
| 52111. |
0.12 g of an organic compound containing phosphorus gave 0.22 g of Mg_2 P_2 O_7 by usual analysis. Calculate the percentage of phosphorus in the compound. |
| 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> :0.512</body></html> | |
| 52112. |
0.11 g of colourless oxide is nitrogen occupies56 cc at 273^(@)C and 2 atm. Name the gas. |
| 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/nitrous-2196967" style="font-weight:bold;" target="_blank" title="Click to know more about NITROUS">NITROUS</a> <a href="https://interviewquestions.tuteehub.com/tag/oxide-1144484" style="font-weight:bold;" target="_blank" title="Click to know more about OXIDE">OXIDE</a></body></html> | |
| 52113. |
0.108g of finely divided copper was treated with an excess of ferric sulphate solution until copper was comnpletely dissolved. The solutionafter the addition of excess dilute sulphuric acidrequired 33.7mL of 0.1N KMnO_(4) for complete oxidation. Find theequaction whichrepresentsthe reaction between metallic copper and ferric sulphate solution. At wt. of Cu = 63.6 and Fe = 56. |
| Answer» <html><body><p><br/></p>Answer :`Cu + Fe_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)(SO_(4))_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>) <a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> CuSO_(4) + 2FeSO_(4)`</body></html> | |
| 52114. |
0.10 moole of AgCI_((s)) is added to 1 litre of H_(2)O. Next crystal of NaBr are added until 75% of the AgCI is concerted to AgBr_((s)), the less soluble silver halide. What is Br^(-) at this point? K_(SP) ofAgCI is 1.78xx10^(-10) and K_(SP) of AgBr is 5.25xx10^(-13). |
| 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> :`2.2xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)M;`</body></html> | |
| 52115. |
0.1 N solution of Na_(2)CO_(3) is being titrated with 0.1 N HCl, the best indicator to be used is : |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/potassium-603438" style="font-weight:bold;" target="_blank" title="Click to know more about POTASSIUM">POTASSIUM</a> <a href="https://interviewquestions.tuteehub.com/tag/ferricyanide-2082569" style="font-weight:bold;" target="_blank" title="Click to know more about FERRICYANIDE">FERRICYANIDE</a>`<br/>phenolphthalein<br/>methyl orange<br/>litmus</p>Solution :N//A</body></html> | |
| 52116. |
0.1 N solution of a dibasic acid can be prepared by dissolving 0.45 g of the acid in water and diluting to 100 mL. The molecular mass of the acid is : |
| Answer» <html><body><p>45<br/>90<br/>135<br/>180</p>Solution :N//A</body></html> | |
| 52117. |
0.1 moles of Hydrocaron on complete combustion produced 17.6 gms of CO_(2).How many Carbon atoms are present in each molecule of the hydrocarbon. |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`C_(4)H_(8)+6O_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)rarr4CO_(2)+4H_(2)O` <br/> 1 mole _________ `4xx44=176g` <br/> 0.1 mole _________ 17.6 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a></body></html> | |
| 52118. |
0*1 mole of PCl_(5) is vaporised in a litrevessel at 260^(@)C.Calculate the concentration of Cl_(2) at equilibrium, if the equilibrium constant for the dissociation of PCl_(5) is 0.0414. |
| Answer» <html><body><p></p>Solution : `{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),(" Intial conc. ",0.1 "mole",,0,,0),(" Conc. at <a href="https://interviewquestions.tuteehub.com/tag/eqm-446398" style="font-weight:bold;" target="_blank" title="Click to know more about EQM">EQM</a>. (moles/litres)",(0.1 -x),,x,,x):}` <br/> Applying the law of chemical equilibrium, we get`K_(c) = ([PCl_(3)][Cl_(2)])/([PCl_(5)]` <br/> Here `K_(c) = 0.0414 " (Given) "` <br/> `:. 0.0414 = (x <a href="https://interviewquestions.tuteehub.com/tag/xxx-1463707" style="font-weight:bold;" target="_blank" title="Click to know more about XXX">XXX</a>)/((0.1-x)) or x^(2)/(0.1 - x) = 0.0414 or x^(2) + 0.0414 x - 0.00414 = 0` <br/> `(x= (-0.0414 pm sqrt((0.0414)^(2) - <a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a> xx 1xx (-0.00414)))/2 ""[ "Using the formula " x = (-b pm sqrt(b^(2) - 4ac))/(<a href="https://interviewquestions.tuteehub.com/tag/2a-300249" style="font-weight:bold;" target="_blank" title="Click to know more about 2A">2A</a>)]` <br/> ( The negative value of x is meaningless and hence is rejected ) <br/> Thus, the concentration of `Cl_(2) " at equilibrium will be " 0.0468 mol L^(-1)`</body></html> | |
| 52119. |
0*1 mole of N_(2)O_(4) (g) was sealed in a tube under atmospheric conditions at 25^(@)C. Calculate the number of moles of NO_(2) (g) present if the equilibrium N_(2)O_(4) (g) hArr 2 NO_(2)is reached after some time (K_(p) = 0 *14). |
| Answer» <html><body><p></p>Solution :` {: (,N_(<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_(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>),hArr,2NO_(2)(g),) ,("Intial amounts",0.1 "mole",,0,), ("At equilibrium " ,(0*1 -x),,2x," <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> " 0*1 +x "moles"):}`<br/> As P=1atm, ` p_(N_(2)O_(4)) = (0*1 -x)/(0*1 +x), p_(NO_(2))= 2/(0*1 + x) atm `<br/>` K_(p) - (p_(N_(O_(4)))^(2))/p_(N_(2)O_(4)) = (2x // (0*1 + x))^(2)/((0*1-x) //( 0*1 +x))= (4x^(2))/((0*1- x) ( 0*1 +x) )= (4x^(2))/(0*01 - x^(2) )` <br/> ` :. (4x^(2))/(0*01 - x^(2) = 0*14 )or 4*14x^(2) = 0*0014or x = 0* 018 ` <br/> ` :. " No. of moles of " NO_(2) " at equilibrium " = 2x = 2 xx 0* 018= 0* 036 "mole" `</body></html> | |
| 52120. |
0.1 mole of KMnO_(4) present in 500 ml solution got converted into K_(2)MnO_(4). The concentration of KMnO_(4) is |
| Answer» <html><body><p>0.2 M<br/>0.2 <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> N <br/>0.04 N</p>Solution :n-factor = `1,M=(0.1)/(0.5)=0.2M=0.2N`</body></html> | |
| 52121. |
0.1 mole of CH_(3)NH_(2)(K_(b)=5xx10^(-4)) is mixed with 0.08 mole of HCl and the solution diluted to one litre. The H^(+) ion concentration in the solution will be |
| Answer» <html><body><p>`1.6xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/11-267621" style="font-weight:bold;" target="_blank" title="Click to know more about 11">11</a>)`<br/>`8xx10^(-11)`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/5xx10-1901302" style="font-weight:bold;" target="_blank" title="Click to know more about 5XX10">5XX10</a>^(-5)`<br/>`8xx10^(-2)`</p>Solution :`{:(,CH_(3)NH_(2),+,HCl,rarr,CH_(3)N overset(+)H_(3)Cl^(-),,,),("Initial conc.",0.1,,0.08,,0 "mol" L^(-1),,,),("Conc. in solution after mixing",0.1-0.08 ,,0,,0.08 "mol " L^(-1),,,):}` <br/> `=0.02` (HCl being limiting <a href="https://interviewquestions.tuteehub.com/tag/reactant-1178091" style="font-weight:bold;" target="_blank" title="Click to know more about REACTANT">REACTANT</a>) <br/> Thus, the final solution is abasic buffer of `CH_(3)NH_(2) and CH_(3)NH_(3)^(+)Cl` <br/> `:. pOH = pK_(b) + <a href="https://interviewquestions.tuteehub.com/tag/log-543719" style="font-weight:bold;" target="_blank" title="Click to know more about LOG">LOG</a>. (["Salt"])/(["Base"])`<br/> i.e., `-log [OH^(-)]=-log K_(b) + log. (["Salt"])/(["Base"])` <br/> or `[OH^(-)]=K_(b) (["Salt"])/(["Base"]) = 5xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)xx(0.02)/(0.08) ` <br/> `=(5)/(4) xx 10^(-4)M = 1.25 xx 10^(-4)M` <br/> `:. [H^(+)]=(K_(w))/([OH^(-)])=(10^(-14))/(1.25xx10^(-4))=8xx10^(-11)`</body></html> | |
| 52122. |
0.1 mol acetic acid and 0.1 mol sodium acetate in 500 mL solution pH is 4.74. Find ionization constant. |
| 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> :`1.82xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)`</body></html> | |
| 52123. |
0.1 M-KMnO_(4) is used for the following titration. How much volume of the solution in ml will be required to react with 0.158 gm of Na_(2)S_(2)O_(3)? S_(2)O_(3)^(2-)+MnO_(4)^(-)+H_(2)OtoMnO_(2)(s)+SO_(4)^(2-)+OH^(-) |
| Answer» <html><body><p>80 <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/>26.67 ml<br/>13.33 ml<br/>16 ml</p>Solution :eq. `KMnO_(4)` = eq. `Na_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)S_(2)O_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)` <br/> `(0.1xxV_(mL))/(1000)xx3=(0.158)/(158)xx8impliesV_(mL)=26.67mL`</body></html> | |
| 52124. |
0.1 M HQ acid has pH=3 then findits ionization constant. |
| Answer» <html><body><p>`3xx10^(-1)`<br/>`1xx10^(-3)`<br/>`1xx10^(-5)`<br/>`1xx10^(-7)`</p>Solution :`{:(,HQoversettolarr, H^+ +, Q^-),("Initial " , 0.1 M, O,O),("Ionized <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</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>,x,x),(,darr,darr,darr),("Mole at <a href="https://interviewquestions.tuteehub.com/tag/equilibrium-974342" style="font-weight:bold;" target="_blank" title="Click to know more about EQUILIBRIUM">EQUILIBRIUM</a>",0.1-x,x,x):}` <br/> pH=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> [H^+]=10^(-3)` <br/> `therefore x=10^(-3)` <br/> `therefore K_a=((x)xx(x))/(0.1-x)` <br/> `=(10^(-3))^2/(0.1-10^(-3)) approx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)/0.1 = 1xx10^(-5)`</body></html> | |
| 52125. |
0.1 MHCl and 0.1 M H_(2)SO_(4), each of volume 2 ml are mixed and the volume is made up to 6 ml by adding 2 ml of 0.01 N NaCl solution. The pH of the resulting mixture is |
| Answer» <html><body><p>`1.17`<br/>`1.<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>`<br/>`0.3`<br/>log 2 - log 3</p>Solution :2 <a href="https://interviewquestions.tuteehub.com/tag/mlof-2837377" style="font-weight:bold;" target="_blank" title="Click to know more about MLOF">MLOF</a> 0.1 M HCl =0.2 <a href="https://interviewquestions.tuteehub.com/tag/millimole-2833492" style="font-weight:bold;" target="_blank" title="Click to know more about MILLIMOLE">MILLIMOLE</a> of HCl <br/> = 0.2 millimole of `<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>^(+)` ion<br/>2 ml of 0.1 M `H_(2)SO_(4)= 0 .2` millimole of `H_(2)SO_(4)` <br/> = 0.4 millimole of `H^(+)` ion<br/> Total `H^(+)` ions = 0.6 millimole<br/> Total volume = 6 ml <br/> `:. [H^(+)]=(0.6)/(6)M=0.1 M =<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(-1)M:. pH = 1`.</body></html> | |
| 52126. |
0.1 M HA is titrated against 0.1 M NaOH. Find the pH at the end point. Dissociation constant for the acid HA is 5xx10^(-6) and the degree of hydrolysis, h lt 1. |
| Answer» <html><body><p></p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/underset-3243992" style="font-weight:bold;" target="_blank" title="Click to know more about UNDERSET">UNDERSET</a>("weak")HA+NaOH rarr <a href="https://interviewquestions.tuteehub.com/tag/naa-572547" style="font-weight:bold;" target="_blank" title="Click to know more about NAA">NAA</a>+H_(2)O` <br/> At the <a href="https://interviewquestions.tuteehub.com/tag/end-971042" style="font-weight:bold;" target="_blank" title="Click to know more about END">END</a> <a href="https://interviewquestions.tuteehub.com/tag/point-1157106" style="font-weight:bold;" target="_blank" title="Click to know more about POINT">POINT</a>, theirequivalent <a href="https://interviewquestions.tuteehub.com/tag/amounts-374805" style="font-weight:bold;" target="_blank" title="Click to know more about AMOUNTS">AMOUNTS</a> react together <br/> `:.` In thefinal solution. `[NaA]=(0.1)/(2)=0.05M` <br/> AsNaA is a salt of weak acid and strong base, it hydrolyses as <br/> `a^(-) + H_(2)O hArr HA+ OH^(-)` <br/> For such a salt, <br/> `pH = 7 + (1)/(2) [pK_(a) + log c]=7+(1)/(2) [-log(5xx10^(-6))+log 0.05]` <br/>`=7+(1)/(2) [ 6-0.6990+0.6990-2]=9`</body></html> | |
| 52127. |
0.1 M CH_(3)CO OH (pH = 3) is titrated with 0.05 M NaOH solution. Calculate the pH when (i) 1/4th of the acid has been neutralized.(ii) 3/4th of the acid has been neutralized. |
| Answer» <html><body><p></p>Solution :Calculation of dissociation constant of the acid <br/> `CH_(3)CO OH hArr CH_(3)CO O^(-) + H^(+)` <br/> As `pH = 3, :. [H^(+)] = 10^(-3)M, [CH_(3)CO O^(-)]=[H^(+)]=10^(-3)M` <br/> `K_(a) = ([CH_(3)CO O^(-)][H^(+)])/([CH_(3)CO OH])=(10^(-3)xx10^(-3))/(0.1) = 10^(-5)` <br/> (i) When 1/4th of the acid has been neutralized <br/> `{:(,CH_(3)CO OH,+,NaOH,rarr,CH_(3)CO ONa ,+,H_(2)O),("Initial <a href="https://interviewquestions.tuteehub.com/tag/conc-927968" style="font-weight:bold;" target="_blank" title="Click to know more about CONC">CONC</a>.",0.1 M,,,,,,),("After 1/4th <a href="https://interviewquestions.tuteehub.com/tag/neutralization-15018" style="font-weight:bold;" target="_blank" title="Click to know more about NEUTRALIZATION">NEUTRALIZATION</a>",0.1xx(3)/(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>),,,,0.1xx(1)/(4),,),(,=0.075 M ,,,,=0.025 M ,,):}` <br/> `:. pH = pK_(a) + <a href="https://interviewquestions.tuteehub.com/tag/log-543719" style="font-weight:bold;" target="_blank" title="Click to know more about LOG">LOG</a>. (["Salt"])/(["Acid"])=-log 10^(-5) + log. (0.025)/(0.075) = 5 - 0.4771 = 4.5229` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>) When 3/4th of the acid has been neutralized <br/> `{:(,CH_(3)CO OH,+,NaOH,rarr,CH_(3)CO ONa , +,H_(2)O),("Initial conc.",0.1 M,,,,,,),("After 3/4th",0.1xx1/4M,,,,0.1xx3/4M,,),("neutralization",=0.025 M,,,,=0.075 M,,):}` <br/> `:. pH = - log 10^(-5) + log. (0.075)/(0.025) = 5 + 0.4771=5.4771`</body></html> | |
| 52128. |
0.1 litre of 0.01 M KMnO_(4) is used by 100 mL of H_(2)O_(2) in acidic medium. Volume of same KMnO_(4) required in alkaline medium to oxidise 0.1 litre of some H_(2)O_(2) will be : |
| Answer» <html><body><p>`(100)/(3) <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/>`(<a href="https://interviewquestions.tuteehub.com/tag/500-323288" style="font-weight:bold;" target="_blank" title="Click to know more about 500">500</a>)/(3) mL`<br/>`(<a href="https://interviewquestions.tuteehub.com/tag/300-305868" style="font-weight:bold;" target="_blank" title="Click to know more about 300">300</a>)/(5) mL`<br/>`(<a href="https://interviewquestions.tuteehub.com/tag/400-315233" style="font-weight:bold;" target="_blank" title="Click to know more about 400">400</a>)/(3) mL`</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> | |
| 52129. |
0.1 gram mole of urea is dissolved in 100g. of water. The molality of the solution is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> m<br/>0.01 M<br/>0.01 M<br/>1.0 M</p>Answer :A</body></html> | |
| 52130. |
0.1 g of metal combines with 46.6 mL of oxygen at STP. The equivalent weight of metal is |
| Answer» <html><body><p>12<br/>24<br/>18<br/>36<br/></p>Solution :a) 1 mole of `O_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)` = 4 equivalent of <a href="https://interviewquestions.tuteehub.com/tag/oxygen-1144542" style="font-weight:bold;" target="_blank" title="Click to know more about OXYGEN">OXYGEN</a> <br/> 22400 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> of `O_(2)` = 4 equivalent of oxygen <br/> 46.6 mL of `O_(2)` = (4/22400) xx 46.6 = 0.00832 eq. <br/> Equivalent of metal = Equivalent of oxygen <br/> (<a href="https://interviewquestions.tuteehub.com/tag/weight-1451304" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHT">WEIGHT</a>)/(Equivalent) = 0.00832 <br/> 0.1/E = 0.00832 `rARr` E= (0.1)/(0.00832) = 12.0</body></html> | |
| 52131. |
0.1 g of an organic compound gave 0.222g of Mg_(2)P_(2)O_(7). What is the percentage of phosphorus in the compound ? |
| Answer» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>>0.32<br/>0.42<br/>0.52<br/>0.62</p>Solution :Amountof `P=(0.<a href="https://interviewquestions.tuteehub.com/tag/222-294559" style="font-weight:bold;" target="_blank" title="Click to know more about 222">222</a> <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 62)/(222) = 0.062 <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/> `therefore " % of" P= (0.062)/(0.1) xx 100 =62.`</body></html> | |
| 52132. |
0.1 g KIO_(3) and excess KI when treated with HCl, the iodine is liberated. The liberated iodine required 45 mL solution thiosulphate for titration . The molarity of sodium thisoulphate will be equivalent to : |
| Answer» <html><body><p>0.252 M <br/>0.126 M<br/>0.0313 M<br/>0.0623 M</p>Solution :The reaction involved are : <br/> `IO_(3)^(-)+<a href="https://interviewquestions.tuteehub.com/tag/5i-326324" style="font-weight:bold;" target="_blank" title="Click to know more about 5I">5I</a>^(-)+6H^(+) to 3I_(2)+H_(2)O` <br/> `2Na_(2)S_(2)O_(3)+I_(2) to 2 NaI+Na_(2)S_(4)O_(<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/> Number of moles of `I_(2)=3xx " Number of moles of " KIO_(3)` <br/> `=(3xx0.1)/(214)` <br/> Number of moles of `Na_(2)S_(2)O_(3)=2xx " Number of moles of " I_(2)` <br/> `(Mxx45)/(1000)=(2xx3xx0.1)/(241)` <br/> M=0.623</body></html> | |
| 52133. |
0.1g of carbon dioxide occupies a volume of 320cc at certain conditions. Under similar conditions 0.2g of 3 dioxide of element 'X' occupies 440cc. Calculate the atomic weight of 'X'. |
| Answer» <html><body><p></p>Solution :Applying Avogadro.slaw,` (n_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>))/(n_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)) = (V_(1))/(V_(2))`<br/> But number of moles ` =n = w//M` <br/> `(V_(1))/(V_(2)) = (w_(1)M_(2))/(w_(2)M_(1))` <br/> Molecular weight of dioxide of X ` = M_(2) = (V_(1))/(V_(2)) <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> (w_(2)m_(1))/(w_(1)) = (320)/(440) xx (0.2)/(0.1) xx <a href="https://interviewquestions.tuteehub.com/tag/44-316683" style="font-weight:bold;" target="_blank" title="Click to know more about 44">44</a> = 64` <br/> Molecular weight of `XO_(2) = 64` <br/> Therefore, <a href="https://interviewquestions.tuteehub.com/tag/atomic-2477" style="font-weight:bold;" target="_blank" title="Click to know more about ATOMIC">ATOMIC</a> weight of X = 32</body></html> | |
| 52134. |
0.093 g of Na_2H_2 EDTA.2H_2O is dissolved in in 250 " mL of " aqueous solution. A sample of hard water containing Ca^(2+) and Mg^(2+) ions is titrated with the above EDTA solution using a buffer of NH_4OH+NH_4Cl using erochrome black-T as indicator. 10 " mL of " hard water at equivalence point. Another sample of hard water is titrated with 10 " mL of " above EDTA solution using KOH solution (pH=12). Using murexide indicator, it requires 40 " mL of " hard water at equivalence point. (a). Calculate the amount of Ca^(2+) and Mg^(2+) present in 1 L of hard water. (b). Calculate the hardness due to Ca^(2+),Mg^(2+) ions and the total hardness of water in ppm of CaCO_3 (Given: Mw(EDTA sal t)=372gmol^(-1) Mw(CaCO_3)=100gmol^(-1)) |
| Answer» <html><body><p></p>Solution :Case I (Using eriochrome black-T indicator) <br/> (a). M of EDTA solution`=(0.093xx1000)/(372xx250)=0.001M` <br/> Volume of EDTA used `=<a href="https://interviewquestions.tuteehub.com/tag/10ml-267166" style="font-weight:bold;" target="_blank" title="Click to know more about 10ML">10ML</a>` <br/> Volume of water sample `=40mL` <br/> `M_1V_1(EDTA)=M_2V_2(Ca^(2+) and Mg^(2+)` in hard water) <br/> `0.001xx10=M_2xx10` <br/> `M_2=0.001` <br/> Molarities of `(Ca^(2+)+Mg^(2+))` ions `=0.001M` <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>.0 m` moles/L <br/> Case II: (Using murexide indicator) <br/> `M_1V_1(EDTA)=M_2V_2` (Hard water) <br/> `0.001xx10=M_2xx40` <br/> `M_2=0.25xx10^(-3)=0.25mmolL^(-1)` <br/> Total m mol `L^(-1) of Ca^(2+) and Mg^(2+)=1.0` <br/> <a href="https://interviewquestions.tuteehub.com/tag/mmol-2837464" style="font-weight:bold;" target="_blank" title="Click to know more about MMOL">MMOL</a> `L^(-1) of Mg^(2+)=1.0-0.25=0.75mmolL^(-1)` <br/> mmoles `L^(-1) of Ca^(2+)=0.25 mmol L^(-1)` <br/> Amount of `Ca^(2+)L^(-1)implies0.25xx40xx10^(-3)=0.01gL^(-1)` <br/> Amount of `Mg^(2+)L^(-1)implies0.75xx24xx10^(-3)=0.018gL^(-1)` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>). `[<a href="https://interviewquestions.tuteehub.com/tag/mw-550075" style="font-weight:bold;" target="_blank" title="Click to know more about MW">MW</a>(CaCO_3)=100gmol^(-1))` <br/> Total m" mol of "`Ca^(2+)` and `Mg^(2+)` ions `L^(-1)` <br/> `=1.0=0.001 (mol)/(L)` <br/> `=0.001M` <br/> 0.1 " mol of "`Ca^(2+) and Mg^(2+)-=0.1mol CaCO_3L^(-1)` <br/> `-=(0.001xx100xx10^(6))/(10^(3))` <br/> `-=100ppm` <br/> `{:[("Total hardness due to Ca^(2+) and Mg^(2+) ions"),("of the sample in gram of CaCO_3in 10^(6) " mL of " H_2O"),(=("Total MxxMw(CaCO_3)xx10^(6))/(10^(3))]:}` <br/> Hardness due to `Ca^(2+)` ions of the sample in gram of `CaCO_3` <br/> in `10^(6)` " mL of " `H_2O=(0.25xx10^(-3)xx100xx10^(5))/(10^(3))=25ppm` <br/> Hardness due to `Mg^(2+)` ions of the sample in gram of `CaCO_3` in `10^(6)` " mL of " `H_2O=100-25=75ppm`</body></html> | |
| 52135. |
0.096kg of oxygen is compressed iso thermally and reversibly from 10^5 N.M^(-2) to 2 xx 10^(5) N.M^(2) at 300 K. Calculate the work done during the impression process. |
| 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> :12.078 <a href="https://interviewquestions.tuteehub.com/tag/kj-1063034" style="font-weight:bold;" target="_blank" title="Click to know more about KJ">KJ</a></body></html> | |
| 52136. |
0.092 g of an organic compound containing phosphorus gave 0.111 g of Mg_(2)P_(2)O_(7) by usual analysis. Calculate the percentage of phosphorus in the organic compound. |
| 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> :`33.7%`</body></html> | |
| 52137. |
0.08 g/2.901 L is a saturated solution of CaF_2 at 298 K temp. calculate K_(sp) of CaF_2. |
| 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> :`1.767xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)`</body></html> | |
| 52138. |
0.0755 g of selenium vapours occupying a volume of 114.2 mL at 700^(@)C and 185 mm of Hg. The vapours are in equilibrium as: Se_(6(g))hArr3Se_(2(g)) Calculate: (i) Degree of dissociation of Se, (ii) K_(p), (iii) K_(c). Atomic weight of Se is 79. |
| Answer» <html><body><p></p>Solution :(`i`) `<a href="https://interviewquestions.tuteehub.com/tag/59-326069" style="font-weight:bold;" target="_blank" title="Click to know more about 59">59</a>%`, (`<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>`) `0.168 <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a>`, (`<a href="https://interviewquestions.tuteehub.com/tag/iii-497983" style="font-weight:bold;" target="_blank" title="Click to know more about III">III</a>`) `<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>.633xx10^(-5) mol^(2) litre^(-2)` ,</body></html> | |
| 52139. |
0.075 g of a monobasic acid required 10 mL of N//12 NaOH solution for complete neutralisation. Calculate the molecular mass of the acid |
| 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/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of the organic acid taken = 0.075 g <br/> Volume of `N//12 NaOH` required for neutralisation = 10 mL <br/> <a href="https://interviewquestions.tuteehub.com/tag/step-25533" style="font-weight:bold;" target="_blank" title="Click to know more about STEP">STEP</a> I. Calculation of equivalent mass of the acid. <br/> `10 mL` of `N//12 NaOH -= 0.075 g` of acid <br/> `:.` 1000 mL of `1N NaOH = ((0.075 g) xx (1000 mL) xx (1N))/((10 mL) xx (N//12)) = 90 g` <br/> 1000 mL of 1 N NaOH solution contain one gram equivalent of base and it reacts with one gram equivalent of the acid. <br/> `:.` Equivalent mass of the acid = 90 <br/> step <a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>. Calculation of the molecular mass of the acid. <br/> Molecular mass = Equivalent mass `xx` Basicity <br/> `= 90 xx 1 = 90` [Foe a monobasic acid, basicity = 1]</body></html> | |
| 52140. |
0.06mole of KNO_(3) is added to 100cm^(3) of water at 298K . The enthalpy of KNO_(3) aq. Solution is 35.8kJ mol^(-1). After the solute is dissolved , the temperature of the solution will be |
| Answer» <html><body><p>293K<br/>298K<br/>301K<br/>304 K</p>Solution :Dissoluation of `KNO_(3)` in <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a> is endothermic, `DeltaH _(sol) = 35.8 kJ mol^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)` <br/> `:.` Heat absorbed when 0.06 mole of `KNO_(3)` is <a href="https://interviewquestions.tuteehub.com/tag/dissolved-956358" style="font-weight:bold;" target="_blank" title="Click to know more about DISSOLVED">DISSOLVED</a> `= 35.8 xx 0.06 kJ = 2148 J` <br/> `q= m xx <a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> xx DeltaT` <br/> `:. 2148 = 100 xx 4.184 xx DeltaT` or `DeltaT = 5K` <br/> `:. `<a href="https://interviewquestions.tuteehub.com/tag/temperatureof-2293939" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATUREOF">TEMPERATUREOF</a> the solution `= 298 -5= 293K`.</body></html> | |
| 52141. |
0.05g of a commercial sample of KClO_(3) on decomposition liberated just sufficient oxygen for complete oxidation of 20 mL CO at 27^(circ)C and 750 mm pressure. Calculate % of KClO_(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> :`65.4%` ;</body></html> | |
| 52142. |
0.05 g of magnesium when treated with dilute HCI gave 51 mL of hydrogen at 27°C and 780 mm pressure. Calculate the equivalent mass of magnesium. |
| 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> :11.76</body></html> | |
| 52143. |
0.05 g of a gas at 750 mm pressure and 25°C occupy a volume of 46.5 mL. Calculate the molecular mass of the gas. |
| Answer» <html><body><p></p>Solution :In the present case, <br/> `P = 750 <a href="https://interviewquestions.tuteehub.com/tag/mm-1098795" style="font-weight:bold;" target="_blank" title="Click to know more about MM">MM</a>= 750/760 atm, V = 46.5 mL = (46.5)/<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a> <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/> `T = 25^(@) C = 298 K, w=0.05 g` and `<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a> = 0.0821 L atm K^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>) "mol"^(-1)` <br/> Using the ideal gas equation, `PV = w/M RT` <br/> and substituting the values, we have <br/> `750/760 xx (46.5)/1000 = 0.05/Mxx 0.0821 xx 298 = 26.66`</body></html> | |
| 52144. |
0.049 g of H_(2)SO_(4) is dissolved per litre of thegiven solution . Calculate the pH of the 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> :`[H_(2)SO_(4)]=(0.049)/(<a href="https://interviewquestions.tuteehub.com/tag/98-342802" style="font-weight:bold;" target="_blank" title="Click to know more about 98">98</a>) "mol" L^(-1)=5xx10^(-4)M`. <br/> As `H_(2)SO_(4)rarr2H^(+)+SO_(4)^(2-), [<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>^(+)]=<a href="https://interviewquestions.tuteehub.com/tag/2xx-1840186" style="font-weight:bold;" target="_blank" title="Click to know more about 2XX">2XX</a>(5xx10^(-4))=10^(-3)M :. pH = 3`</body></html> | |
| 52145. |
0.03 mole of Ca^(2+) ions is added to a litre of 0.01 M SO_(4)^(2-) solution. Will it cause precipitation of CaSO_(4) ?K_(sp) "for" CaSO_(4)=2.4xx10^(-5). |
| Answer» <html><body><p><br/></p>Solution :In the final solution, `[Ca^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>+)]=0.03 M and [SO_(4)^(2-)]=0.01 M`. Hence, ionic <a href="https://interviewquestions.tuteehub.com/tag/product-25523" style="font-weight:bold;" target="_blank" title="Click to know more about PRODUCT">PRODUCT</a> of `CaSO_(4) = (0.03) (0.01)=3xx10^(-4)` which is greater than `K_(<a href="https://interviewquestions.tuteehub.com/tag/sp-1219706" style="font-weight:bold;" target="_blank" title="Click to know more about SP">SP</a>)`. Hence, precipitation will <a href="https://interviewquestions.tuteehub.com/tag/occur-583012" style="font-weight:bold;" target="_blank" title="Click to know more about OCCUR">OCCUR</a>.</body></html> | |
| 52146. |
0.023 g of sodium metal is reacted with 100 cm^(3) of water. The pH of the resulting solution is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a><br/>11<br/>9<br/>12</p>Solution :`Na+H_(2)OrarrNaOH=(<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)H_(2)` <br/> 1 mole of Na produces 1 mole of <a href="https://interviewquestions.tuteehub.com/tag/naoh-572531" style="font-weight:bold;" target="_blank" title="Click to know more about NAOH">NAOH</a> <br/> `0.023 gNa = (0.023)/(23) "mole" = 10^(-3) "mole"` <br/> `:. ` NaOH <a href="https://interviewquestions.tuteehub.com/tag/produced-592947" style="font-weight:bold;" target="_blank" title="Click to know more about PRODUCED">PRODUCED</a> `= 10^(-3)` mole <br/> `:. [NaOH]=(10^(-3))/(100)xx1000=10^(-2) M` <br/> `:. pOH=2 ` or pH `= 14-2=12`.</body></html> | |
| 52147. |
0.01 mole sucrose dissolve in .......... Litre water so it becomes 0.01solution. |
| Answer» <html><body><p>1<br/>0.0001<br/>1000<br/>0.01</p>Answer :A</body></html> | |
| 52148. |
0.01 mole solution of .10 volume. H_2O_2 is required to convert 0.01 mol PbS into PbSO_4 ? |
| Answer» <html><body><p>11.2<br/>22.4<br/>33.6<br/>44.8</p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/pbs-597770" style="font-weight:bold;" target="_blank" title="Click to know more about PBS">PBS</a>+4H_2O_2 to PbSO_4 + 4H_2O` <br/> `4H_2O_2 to 4H_2O + 2O_2` <br/>To convert 1 mole PbS <br/> 1 `PbSO_4` required = 4 mole `H_2O_2` <br/> `therefore` to convert 0.01 <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> PbS to `PbSO_4`= (?) <br/> `=(0.01xx4)/1`= 0.04 mol `H_2O_2` required <br/> <a href="https://interviewquestions.tuteehub.com/tag/strength-1229153" style="font-weight:bold;" target="_blank" title="Click to know more about STRENGTH">STRENGTH</a> of volume=10 <br/> `M=V/11.2 =10/11.2 `=0.8928 mol/lit `H_2O_2` <br/> 0.8928 mol `H_2O_2 to` 1 <a href="https://interviewquestions.tuteehub.com/tag/litre-1075864" style="font-weight:bold;" target="_blank" title="Click to know more about LITRE">LITRE</a> `H_2O_2` <br/> `therefore` 0.04 mol `H_2O_2 to` (?) <br/> `=(0.04xx1)/0.8928` <br/> =0.04480 litre <br/> =44.80ml `H_2O_2`</body></html> | |
| 52149. |
0.01 mole of iodoform (CHI_(3)) reacts with Ag powder to produce a gas whose volume at NTP is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/224-1824778" style="font-weight:bold;" target="_blank" title="Click to know more about 224">224</a> ml<br/>112 ml<br/>336 ml<br/>None</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> | |
| 52150. |
0.01 mole of FeS_n(iron (II) sulphide) required 0.06 mole of AO_(4)^(3-)for complete oxidation. The species formed are FeO, SO_2 and A^(2+) . Calculate the value of n. |
| 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> :`0.01xxn-f =0.06xx3`, <br/> `<a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>-f =<a href="https://interviewquestions.tuteehub.com/tag/18-278913" style="font-weight:bold;" target="_blank" title="Click to know more about 18">18</a>" "18 =(2/n+<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)<a href="https://interviewquestions.tuteehub.com/tag/xxn-3292706" style="font-weight:bold;" target="_blank" title="Click to know more about XXN">XXN</a>, n =4`</body></html> | |