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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.
| 51701. |
100 ml aqueous 0.1 molar M(CN)_(2), (80% ionized) solution is mixed with 100 ml of 0.05 molar H_(2),SO_(2) solution (80% ionized). (K_(6), of CN" = 10^(-6)) |
| Answer» <html><body><p>`6` <br/>`8` <br/>`9` <br/>`7` </p>Solution :`M(<a href="https://interviewquestions.tuteehub.com/tag/cn-907325" style="font-weight:bold;" target="_blank" title="Click to know more about CN">CN</a>)_2 hArr M^(+2)+ 2CN^(-) ` <br/>`10m " moles" ` <br/> ` {:( 10(1- alpha ) , 10 alpha =8 , 2 xx 10 alpha = 16 ),( H_2SO_4 hArr, 2H^(+) , SO_4^(2-)),( 5 " moles" ,,),(5(1- beta ), 2 xx 5 beta =8, 5 beta =4 ), (<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>^(+) + , CN^(-) hArr, HCN),( 8,16,):}` <br/> `8,8 ` <br/>i.e., <a href="https://interviewquestions.tuteehub.com/tag/buffer-905159" style="font-weight:bold;" target="_blank" title="Click to know more about BUFFER">BUFFER</a> of HCN & `CN^(-) `<br/> ` therefore <a href="https://interviewquestions.tuteehub.com/tag/ph-1145128" style="font-weight:bold;" target="_blank" title="Click to know more about PH">PH</a> =pKa +<a href="https://interviewquestions.tuteehub.com/tag/log-543719" style="font-weight:bold;" target="_blank" title="Click to know more about LOG">LOG</a> ""( S)/(A)=pKa =14 -6=8`</body></html> | |
| 51702. |
100 ml, 1.5M Fe(NO_(3))_(3). 100 ml, 2M FeCl_(3), 100 ml 2M Mg(NO_(3))_(2) are mixed and 700 ml water is added. The molar concentration of total ions in the solution in ______M |
| Answer» <html><body><p><br/></p>Solution :Total <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> of ions from `Fe(NO_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>))_(3)` <br/> `=0.1xx1.5xx4=0.6` <br/> From `FeCl_(3)=0.1xx2xx4=0.8` <br/> From `<a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a>(NO_(3))_(2)=0.1xx2xx3=0.6` <br/> `therefore` Total moles of ions = 2 <br/> `therefore M=(2)/(1)=2M`</body></html> | |
| 51703. |
100 ml of 0.1 M NaCl and 100 ml of 0.2 MNaOH are mixed. What is the change in pH of NaCI solution ? |
| Answer» <html><body><p> Some `KMnO_4`is <a href="https://interviewquestions.tuteehub.com/tag/left-1070879" style="font-weight:bold;" target="_blank" title="Click to know more about LEFT">LEFT</a> unreacted<br/>0.02 M `<a href="https://interviewquestions.tuteehub.com/tag/mn-548487" style="font-weight:bold;" target="_blank" title="Click to know more about MN">MN</a>^(2+)` ions are present in the solutions <br/>`K_2SO_3` is <a href="https://interviewquestions.tuteehub.com/tag/partially-7321885" style="font-weight:bold;" target="_blank" title="Click to know more about PARTIALLY">PARTIALLY</a> left unreacted <br/>Newly formed `SO_4^(2-)` has 0.05 M cocentration</p>Solution :`100xx0.2 =<a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a>` meq `O_3^(2-)` <br/> `100xx0.5 =50`meq `MnO_4^(-)` <br/> `:. SO_3^(2-)` in limiting <a href="https://interviewquestions.tuteehub.com/tag/reagent-1178480" style="font-weight:bold;" target="_blank" title="Click to know more about REAGENT">REAGENT</a> <br/> `MnO_4^(-)` left = 30 meq<br/> `:.[Mn^(+2)]=(20//5)/200=0.05M`<br/> `[SO_4^(-2)]=(20//2)/200 = 0.05M`</body></html> | |
| 51704. |
100 ml 0.1 M urea solution is diluted upto 200 ml than the molarity is ...............M. |
| Answer» <html><body><p>0.<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a><br/>0.<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/>0.05<br/>0.025</p>Solution :`M_(1)V_(1) = M_(2)V_(2)` <br/> `0.1xx100=M_(2)xx200 "" :.M_(2)=0.05M`</body></html> | |
| 51705. |
100 ml, 0.1 M K_(2)SO_(4) is mixed with 100 ml, 0.1M Al_(2)(SO_(4))_(3) solution. The resultant solution is |
| Answer» <html><body><p>`0.1NK^(+)` ions <br/>`0.4NK^(+)` ions <br/>`0.2MSO_(4)^(-2)` ions<br/>`0.1MAl^(+<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)` ions </p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`[K^(+)]=(100xx0.1xx2)/(200)=0.1M=0.1N` <br/> `[SO_(4)^(2-)]=(100xx0.1xx1+100xx0.1xx3)/(200)` <br/> `=0.2M=0.4N` <br/> `[<a href="https://interviewquestions.tuteehub.com/tag/al-370666" style="font-weight:bold;" target="_blank" title="Click to know more about AL">AL</a>^(+3)]=(100xx0.1xx2)/(200)=0.1M=0.3N`</body></html> | |
| 51706. |
100 ml 0.1 M H_3PO_4 solution is being titrated with 0.1 M NaOHsolution. The pH of the reaction mixture keeps increasing with addition of NaOH. The successive dissociation constant of H_3PO_4 " are "10^(-3) , 10 ^(-6) and 10^(-14)respectively.What is the pH of the solution after adding 150 ml of NaOH solution? |
| Answer» <html><body><p>`<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/>`6`<br/>`6. <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/>`5.3`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`<a href="https://interviewquestions.tuteehub.com/tag/ph-1145128" style="font-weight:bold;" target="_blank" title="Click to know more about PH">PH</a> = P^(Ka_2)= 6`</body></html> | |
| 51707. |
100 ml 0.1 M H_3PO_4 solution is being titrated with 0.1 M NaOHsolution. The pH of the reaction mixture keeps increasing with addition of NaOH. The successive dissociation constant of H_3PO_4 " are "10^(-3) , 10 ^(-6) and 10^(-14)respectively. How much volume of the given NaOH must be added such that a buffer H_2PO_4^(-) // HPO_4^(-2)of maximum capacity is formed. |
| Answer» <html><body><p>100 ml <br/>150 ml <br/>200 ml <br/>250 ml </p>Solution :` {:(H_3PO_4 +, <a href="https://interviewquestions.tuteehub.com/tag/naoh-572531" style="font-weight:bold;" target="_blank" title="Click to know more about NAOH">NAOH</a> to, NaH_2PO_4, +H_2O ),(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> m " moles " , 10 m " moles " ,0,),(-,-, ,10 m " moles " ):}` <br/> ` {:(NaH_2PO_4 +, NaOH to , Na_2HPO_4+ H_2O ),(10 m " moles " , 5 m " moles " ,"-" ),( 5 m " moles " , , ):}` <br/>To get <a href="https://interviewquestions.tuteehub.com/tag/buffer-905159" style="font-weight:bold;" target="_blank" title="Click to know more about BUFFER">BUFFER</a> with maximum buffer capacity <a href="https://interviewquestions.tuteehub.com/tag/half-1014510" style="font-weight:bold;" target="_blank" title="Click to know more about HALF">HALF</a> of the `NaH_2PO_4` should be consumed. <br/> In 1st <a href="https://interviewquestions.tuteehub.com/tag/step-25533" style="font-weight:bold;" target="_blank" title="Click to know more about STEP">STEP</a>, To get 10 m moles NaOH, volume should be 100 ml. <br/>In 2nd, step , To get 5 m moles NaOH , volume should be 50 ml <br/> Total vol = 150 ml</body></html> | |
| 51708. |
100 ml 0.1 M H_3PO_4 solution is being titrated with 0.1 M NaOHsolution. The pH of the reaction mixture keeps increasing with addition of NaOH. The successive dissociation constant of H_3PO_4 " are "10^(-3) , 10 ^(-6) and 10^(-14)respectively. What is the pH of the reaction mixture after adding 50 ml of NaOH ? |
| Answer» <html><body><p>`3`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>`<br/>`<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/>`5.3`</p>Solution :After <a href="https://interviewquestions.tuteehub.com/tag/adding-2399902" style="font-weight:bold;" target="_blank" title="Click to know more about ADDING">ADDING</a> <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 NaOH, buffer ` (H_3PO_4 + NaH_2 PO_4) ` with <a href="https://interviewquestions.tuteehub.com/tag/maximum-556915" style="font-weight:bold;" target="_blank" title="Click to know more about MAXIMUM">MAXIMUM</a> capacity is obtained. <br/> ` therefore pH =P^(Ka_2) =3`</body></html> | |
| 51709. |
100 mL of 0.01M KMnO_4 oxidises 100mL of acidified H_2O_2 . The volume of Oxygen in (MnO_4^(-)) changes to Mn^(2+)in acidic medium and to MnO_4 in alkaline medium). |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/26-298265" style="font-weight:bold;" target="_blank" title="Click to know more about 26">26</a> <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a><br/>260 mL<br/>`2.6 mL`<br/>26 Lt</p>Solution :`(N_(1)V_(1))_(KMnO_(4))=(N_(2)V_(2))_(H_(2)O_(2))` <br/> `N=0.05 rArr 11.2` <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 `H_(2)O_(2)=(11.2xx0.05)/(2)` <br/> Vol of `O_(2)=(11.2xx0.05)/(2)xx100 mL`</body></html> | |
| 51710. |
100 gm oleum sample (labelled as 107.8%) is mixed with 7.8 gm water and requires, 1.1 L of x molar aq. Solution of NaOH for complete neutratization. The value of x is: |
| 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/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a></body></html> | |
| 51711. |
112% labelled oleum is diluted with sufficient water. The solution on mixing with 5.3xx102gm Na_(2)CO_(3) liberates CO_(2). The volume of Co_(2) given out at 1 atm at 273 K will be |
| Answer» <html><body><p>1.12 lit<br/>1.23 lit<br/>2.2 lit <br/>25.5 lit</p>Solution :Final weight of `H_(2)SO_(4)=<a href="https://interviewquestions.tuteehub.com/tag/112-268456" style="font-weight:bold;" target="_blank" title="Click to know more about 112">112</a>` gms <br/> `Na_(2)O_(3)+H_(2)SO_(4)rarrNa_(2)SO_(4)+H_(2)O+CO_(2)` <br/> 530 <a href="https://interviewquestions.tuteehub.com/tag/gm-1008640" style="font-weight:bold;" target="_blank" title="Click to know more about GM">GM</a> 112 gm <br/> = 5 <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> 1.14 moles <br/> <a href="https://interviewquestions.tuteehub.com/tag/implies-1037962" style="font-weight:bold;" target="_blank" title="Click to know more about IMPLIES">IMPLIES</a> <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> of `CO_(2)=1.14xx22.4=25.5` lit.</body></html> | |
| 51712. |
10.0 g sample of Cu_(2)O is disoolved dissolved in dil. H_(2)SO_(4) where it undergoes disproportionation quantitatively. The solution is filtered off and 8.3 g pure KI cyrstals are added to clear filtrate in order to precipitate CuI with the liberation of I_(2). The solution is again filtered and boiled till all the I_(2) is removes. Now excess of an oxidising agent is added to the rfiltrate which liberates I_(2) again. The liberated I_(2) now percentrage by mass of Cu_(2)O in the sample. |
| Answer» <html><body><p></p>Solution :`Cu_(2)OtoCu^(2+)+Cu^(0)` <br/> The solution after dissolution of `Cu_(2)O` in dil `H_(2)SO_(4)` <a href="https://interviewquestions.tuteehub.com/tag/contains-11473" style="font-weight:bold;" target="_blank" title="Click to know more about CONTAINS">CONTAINS</a> `Cu^(2+)` <a href="https://interviewquestions.tuteehub.com/tag/ions-1051295" style="font-weight:bold;" target="_blank" title="Click to know more about IONS">IONS</a> and `Cu`. `Cu^(2+)` ions react with KI to give `CuI_(2)` which is converted to CuI (or `CU_(2)I_(2))` and `I_(2)`. <br/> `Cu^(2+)+<a href="https://interviewquestions.tuteehub.com/tag/2i-300397" style="font-weight:bold;" target="_blank" title="Click to know more about 2I">2I</a>^(ɵ)toCuI_(2)to(Cu_(2)I_(2) or CuI)+(1)/(2)I_(2)` <br/> Millimoles of KI taken`=(8.3)/(166)xx10^(3)=50(mE of KI=166)` <br/> Now, KI left unused reacts with oxidising agent to liberate `I_(2)` again. <br/> `2I^(ɵ)overset("oxidising agent")toI_(2)overset(2S_(2)O_(3)^(2-))toS_(4)O_(6)^(2-)+I^(ɵ)` <br/> Millimoles of KI left`=` millimoles of `S_(2)O_(3)^(2-)` used <br/> `(n=(2)/(2)=1)=(n=(2)/(2)=1)` <br/> `=10xx1.0xx1` (n factor)`=10` <br/> Therefore, millimoes of KI used for `Cu^(2+)=50-10=40` <br/> Therefore, <a href="https://interviewquestions.tuteehub.com/tag/milli-560824" style="font-weight:bold;" target="_blank" title="Click to know more about MILLI">MILLI</a> moles of `Cu_(2)O=20` <br/> `{:(Coverset(+1)(u_(2))to2overset(+2)(Cu)),(mol ratio of overset(+1)(Cu_(2)):overset(+2)(Cu)=1:2):}` <br/> `(Weight)/(Mw)xx10^(3)=20` (Mw of `Cu_(2)O=143)` <br/> `(weight)/(143)xx10^(3)=20` <br/> `thereforeW_(Cu_(2)O)=2.86` <br/> `% of Cu_(2)O=(2.86xx100)/(10.0)=28.6%`</body></html> | |
| 51713. |
1.00 g of oxygen combine with 0.126 g of hydrogen to form H_2O .1.00 g of nitrogen combine with 0.216g of hydrogen to form NH_3. Predict the weight of oxygen required to combine with 1.00 g of nitrogen to form an oxide. |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> : 1.67 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> or its <a href="https://interviewquestions.tuteehub.com/tag/simple-1208262" style="font-weight:bold;" target="_blank" title="Click to know more about SIMPLE">SIMPLE</a> <a href="https://interviewquestions.tuteehub.com/tag/multiple-1105557" style="font-weight:bold;" target="_blank" title="Click to know more about MULTIPLE">MULTIPLE</a></body></html> | |
| 51714. |
1.00 g of EuCl_(2), is treated with excess of aqueous AgNO_(3) and all the chlorine is recovered as 1.29 g of AgCl. Calculate the atomic weight of Eu. (Cl = 35.5, Ag = 108) |
| Answer» <html><body><p></p>Solution :`underset(1g)(<a href="https://interviewquestions.tuteehub.com/tag/eu-446864" style="font-weight:bold;" target="_blank" title="Click to know more about EU">EU</a>)CI_(2)+AgNO_(3) rarr underset(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.29 g)(<a href="https://interviewquestions.tuteehub.com/tag/agci-1969988" style="font-weight:bold;" target="_blank" title="Click to know more about AGCI">AGCI</a>)` <br/> Since <a href="https://interviewquestions.tuteehub.com/tag/ci-408488" style="font-weight:bold;" target="_blank" title="Click to know more about CI">CI</a> atoms are conserved applying POAC for CI atoms moles of CI in `EuCI_(2)` = moles of CI in AgCI <br/> `2xx ` moles of `EuCI_(2)=<a href="https://interviewquestions.tuteehub.com/tag/1xx-1804644" style="font-weight:bold;" target="_blank" title="Click to know more about 1XX">1XX</a>` moles of AgCI <br/> `2xx(1)/((x+35.5xx2))=1 xx(1.29)/((108+35.5))`<br/> ( where x = atomic weight of Eu ) <br/> `:. x= 152.48`</body></html> | |
| 51715. |
100 g of an ideal gas (m01 wt 40) is pressent in a cylinder at 27^(@)C and 2 atm pressure During transportation cylinder The valve attached to cylinder cannot keep the pressure greater than 2 atm and therefore 10g of gas leaked out through cylinder Calculate (i) the volme of cylinder before and after dent (ii) the pressure inside the cylinder . |
| 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> :`30.79 <a href="https://interviewquestions.tuteehub.com/tag/litre-1075864" style="font-weight:bold;" target="_blank" title="Click to know more about LITRE">LITRE</a>, 27.71 litre 2.22 <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a>`</body></html> | |
| 51716. |
100 g of a solution of hydrochloric acid (sp.gr. 1.18) contains 36.5 g of the acid. The normality of the solution is: |
| Answer» <html><body><p>`1.18`<br/>`11,8`<br/>118<br/>10</p>Solution :Volume `=("<a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a>")/("<a href="https://interviewquestions.tuteehub.com/tag/density-17451" style="font-weight:bold;" target="_blank" title="Click to know more about DENSITY">DENSITY</a>") = 100/1.18 = 84.7 mL` <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> <a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a> = (<a href="https://interviewquestions.tuteehub.com/tag/nev-3742739" style="font-weight:bold;" target="_blank" title="Click to know more about NEV">NEV</a>)/1000` <br. n="(w" v="" xx=""></br.></body></html> | |
| 51717. |
1.0 g of non - electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.4 K. Find the molar mass of the solute. [Given : Freezing point depression constant of benzene = 5.12 K. kg mol]. |
| Answer» <html><body><p></p>Solution :`DeltaT_(f)=0.40K` <br/> `K_(f)="5.12 kg <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a>"^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)` <br/> `W_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)=1g` <br/> `W_(1)="50.5 gm "=(50.5)/(1000)kg` <br/> `M_(2)=(K_(f))/(DeltaT_(f))xx(W_(2))/(W_(1))` <br/> `therefore""M_(2)=(5.12)/(0.40)xx(1)/(50.5)xx1000` <br/> `="2.56 g mol"^(-1)` <br/> Thus, the molecule mass of the solute `=" 256 g mol"^(-1)`.</body></html> | |
| 51718. |
1.00 g of a non-electrolyte dissolved in 50.5g of benzene lowered its freezing point by 0.40K. The freezing point depression constant of benzene is "5.12K.kg mol"^(-1). Find the molecular mass of the solute. |
| Answer» <html><body><p></p>Solution :`DeltaT_(f)=0.40K` <br/> `K_(f)="5.12 kg mol"^(-1)` <br/> `W_(2)=<a href="https://interviewquestions.tuteehub.com/tag/1g-282964" style="font-weight:bold;" target="_blank" title="Click to know more about 1G">1G</a>` <br/> `W_(1)="50.5 <a href="https://interviewquestions.tuteehub.com/tag/gm-1008640" style="font-weight:bold;" target="_blank" title="Click to know more about GM">GM</a> "=(50.5)/(1000)kg` <br/> `M_(2)=(K_(f))/(DeltaT_(f))xx(W_(2))/(W_(1))` <br/> `therefore""M_(2)=(5.12)/(0.40)xx(1)/(50.5)xx1000` <br/> `="2.56 g mol"^(-1)` <br/> Thus, the molecule mass of the solute `=" 256 g mol"^(-1)`.</body></html> | |
| 51719. |
1.00 g of a chloride of an element contains 0.835 g of chlorine. If the vapour density of the chloride is 85, find the atomic weight of the element and its valency. |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/let-11597" style="font-weight:bold;" target="_blank" title="Click to know more about LET">LET</a> the formula of the chloride be `MCI_(<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>)` where x is the valency of the <a href="https://interviewquestions.tuteehub.com/tag/element-969236" style="font-weight:bold;" target="_blank" title="Click to know more about ELEMENT">ELEMENT</a> M. <br/> `x <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> ` moles of M =1 `xx` moles of CI <br/> or `x xx (1-0.835)/("at . Wt . Of M)= (0.835)/(35.5)` <br/> Now we know <br/> molecular weight `= 2xx ` vapour density <br/> `=2xx85 =170`<br/> For `MCI_(x)` <br/> mol . wt. of `MCI_(x)` = at . wt. of M +x `xx` at . wt. of CI <br/> or 170 =at .wt . of M + 35.5 x. <br/> `:. ` at . wt. of M = 170 - 35.5 x <br/>or `x = (170-"at .wt.of M")/(35.5)` <br/> From (1) and (2), we get, at. wt. of M = 28.05. Substituting at. wt. of M in the eqn. (1) we get, x = <a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>.</body></html> | |
| 51720. |
100 cm^3 of a centimolar solution of an acid contain 0.098 g of the acid. Find the molecular mass of the acid. |
| 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/98-342802" style="font-weight:bold;" target="_blank" title="Click to know more about 98">98</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> <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a>^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)`</body></html> | |
| 51721. |
100cc of gas is collected over water at 15^@C and 750torr. If the gas occupies 91.9 mL in dry state at STP, find the aqueous tension at STP. |
| 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> :Let aqueous tension at `15^(@)C`is .f. <br/> `{:("<a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> conditions",,"STP conditions"),(P_(1) = (750-f)"mm",,P_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) = 760"mm"),(V_(2) = 100 "mL",,V_(2) = 91.9"mL"),(T_(1) = 288"K",,T_(2) = <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/> <a href="https://interviewquestions.tuteehub.com/tag/applying-1982651" style="font-weight:bold;" target="_blank" title="Click to know more about APPLYING">APPLYING</a> equation of state at two conditions, <br/> `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2)) = ((750-f) xx100)/(288) = (760 xx 91.9)/(273)` <br/> ` = 750-f = 736.8` <br/> Aqueous tension .f. at `15^(@)C = 750-736.8 = 13.2` mm</body></html> | |
| 51722. |
10% W/V NaOH is same as |
| Answer» <html><body><p>2.5 M<br/>2.5 N<br/>1.5 m<br/>10% W/W NaOH</p>Solution :`M=(<a href="https://interviewquestions.tuteehub.com/tag/10v-267349" style="font-weight:bold;" target="_blank" title="Click to know more about 10V">10V</a>%)/("M.wt")=(10xx10)/(<a href="https://interviewquestions.tuteehub.com/tag/40-314386" style="font-weight:bold;" target="_blank" title="Click to know more about 40">40</a>)=2.5M=2.5N`</body></html> | |
| 51723. |
When P_4O_10 is hydrolysed with water ,it forms :- |
| Answer» | |
| 51724. |
10% PCl_5 decompose at definite temperature and 4 atm pressure. At same temperature if 20% PCl_5 decompose than find pressure. (Temperature not change.) |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`K_p`= 0.04 <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a>, 0.96 atm</body></html> | |
| 51725. |
10 moles of SO_(2) and 4 moles of O_(2) are mixed in a closed vessel of volume 2 litres. The mixture is heated in the presence of Pt catalyst. Following reaction takes place: 2SO_(2)(g)+O_(2)(g)to2SO_(3)(g) Assuming the reaction proceeds to completion. Select the correct statement. |
| 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/>4<br/>8<br/>14</p>Solution :`underset("10 moles")(2SO_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>(<a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>)))+underset("reagent")underset("limiting")underset("4 moles")(O_(2(g)))rarr2SO_(3(g))` <br/> `SO_(2)` moles remaining = `10-8=2`</body></html> | |
| 51726. |
10 moles of pure PCl_(5) gas is put into a closed container of volume 'V' and temerature 'T' and allowed to reach equilibrium, at an equilibrium pressure 20 atm. The pure PCl_(5) is found to be 50% dissociated at equilibrium. PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g) The partial pressure of PCl_(5), if equilibrium is established at new equlibrium pressure 35 atm by changing volume 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> <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/15-274069" style="font-weight:bold;" target="_blank" title="Click to know more about 15">15</a> atm<br/>18 atm <br/><a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a> atm</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> | |
| 51727. |
10 moles of pure PCl_(5) gas is put into a closed container of volume 'V' and temerature 'T' and allowed to reach equilibrium, at an equilibrium pressure 20 atm. The pure PCl_(5) is found to be 50% dissociated at equilibrium. PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g) If 20 moles of PCl_(5)(g) is added to original equilibrium mixture keeping total pressure constant to 20 atm at same tempereture 'T', then ratio of new equilibriujm volume to the initial volume V will be : |
| Answer» <html><body><p>1.5<br/>`2.0`<br/>1.8<br/>`3.0`</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> | |
| 51728. |
10 moles of pure PCl_(5) gas is put into a closed container of volume 'V' and temerature 'T' and allowed to reach equilibrium, at an equilibrium pressure 20 atm. The pure PCl_(5) is found to be 50% dissociated at equilibrium. PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g) K_(p) for the above the reaction is : |
| Answer» <html><body><p>`(<a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a>)/(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)`<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><br/>`(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)/(3)`<br/>`(20)/(3)`</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</body></html> | |
| 51729. |
10 moles of an ideal gas expand isothermally and reversibly from a pressure of5 atm to 1 atm at300 K . What is thelargest mass that can be lefted through a height of 1 metre by this expansion? |
| Answer» <html><body><p></p>Solution :`w_(exp) = - 2.303 nRT log.(P_(1))/(P_(2))=-2.303 ( <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> mol) xx( 8.314 <a href="https://interviewquestions.tuteehub.com/tag/jk-521675" style="font-weight:bold;" target="_blank" title="Click to know more about JK">JK</a>^(-1) mol^(-1)) ( <a href="https://interviewquestions.tuteehub.com/tag/300-305868" style="font-weight:bold;" target="_blank" title="Click to know more about 300">300</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> ) log. ( 5)/(1) = - 40.15 xx 10^(3)J` <br/>If M is the mass that can be lifted by this work ,through a height of 1metre, then work done `=` Mgh <br/> ` 40.15xx 10^(3)J= Mxx9.81 m s^(-2) xx 1m` <br/> or ` M =( 40.15 xx10^(3) kg m^(2) s^(-2))/( 9.18 ms^(-2)xx 1 m ) (J = kg m^(2) s^(-2))` <br/> `= 4092 .76 kg`</body></html> | |
| 51730. |
10 moles of each N_(2) and H_(2) are made to react in a closed chamber. At equilibrium 40% of H_(2) was left. The total moles in the chamber are : |
| Answer» <html><body><p>16<br/>12<br/>8<br/>10</p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_B_C04_E03_023_S01.png" width="80%"/> <br/> <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> in <a href="https://interviewquestions.tuteehub.com/tag/chamber-913753" style="font-weight:bold;" target="_blank" title="Click to know more about CHAMBER">CHAMBER</a> 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> = 8+4+4=16</body></html> | |
| 51731. |
1.0 mole of ethyl alcohol and 1.0 mole of acctic acid are mixed. At equilibrium, 0.666 mole of ester is formed. The value of equilibrium constant is |
| Answer» <html><body><p>`1//<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>`<br/>`1//2`<br/>4<br/><a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a></p>Solution :`C_(2)H_(<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)<a href="https://interviewquestions.tuteehub.com/tag/oh-585115" style="font-weight:bold;" target="_blank" title="Click to know more about OH">OH</a>+CH_(3)COOH harr CH_(3)COOC_(2)H_(5)+H_(2)O` <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_B_C04_E01_045_S01.png" width="80%"/> <br/> `K_(c)=([CH_(3)COOC_(2)H_(5)][H_(2)O])/([C_(2)H_(5)OH][CH_(3)COOH])` <br/> `=(0.666 xx 0.666)/(0.333 xx 0.333)=4`</body></html> | |
| 51732. |
1.0 mole of pure dinitrogen gas at SATP conditions was put into a vessel of volume 0.025 cm^(3), maintained at the temperature of 50^(@)C, what is the preussre of the gas in the vessel ? |
| Answer» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>></p>Solution : Volume of `1.0` mol of a gas at SATP `(V_(1))=<a href="https://interviewquestions.tuteehub.com/tag/24-295400" style="font-weight:bold;" target="_blank" title="Click to know more about 24">24</a>*<a href="https://interviewquestions.tuteehub.com/tag/8xx10-1931283" style="font-weight:bold;" target="_blank" title="Click to know more about 8XX10">8XX10</a>^(-3)m^(3)` <br/> intial condition (SATP) `""` Final condition <br/> `V_(1)=24*8xx10^(-3)m^(3)"" V_(2)=0*1 m^(3)` <br/> `P_(1)1 "bar""" P_(2) ?` <br/> `T_(2)<a href="https://interviewquestions.tuteehub.com/tag/298-1834947" style="font-weight:bold;" target="_blank" title="Click to know more about 298">298</a>*15 k"" T_(2)<a href="https://interviewquestions.tuteehub.com/tag/50-322056" style="font-weight:bold;" target="_blank" title="Click to know more about 50">50</a>^(@)C323*15K`<br/> Acoording to general gas equation. <br/> `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)) or P_(2) =(P_(1)V_(1)T_(2))/(T_(1)V_(2))` <br/> `:. P_(2)=(1*0xx24*8xx10^(-3)xx323*15)/(298*15xx0*25)=1*076` bar <br/> `:. P_(2) =1*076` bar</body></html> | |
| 51733. |
1.0 mol of a monoatomic ideal gas is expanded from state (1) to state (2) as shown in figure. Calculate the work done for the expansion of gas from state (1) to state (2) at 298 K. |
| Answer» <html><body><p></p>Solution :The given diagram represent that the <a href="https://interviewquestions.tuteehub.com/tag/prcess-2947794" style="font-weight:bold;" target="_blank" title="Click to know more about PRCESS">PRCESS</a> is <a href="https://interviewquestions.tuteehub.com/tag/carried-7257188" style="font-weight:bold;" target="_blank" title="Click to know more about CARRIED">CARRIED</a> out in infinite steps, <a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a> it is isothermal <a href="https://interviewquestions.tuteehub.com/tag/reversible-1188264" style="font-weight:bold;" target="_blank" title="Click to know more about REVERSIBLE">REVERSIBLE</a> expansion of the ideal gas from pressure 2.0 atm to <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 atm 298 K. <br/> `W = - 2.303 nRT log. (p_(1))/(p_(2))`<br/> `W = - 2.303 xx 1 mol xx 8.314 JK^(-1) xx 298 K log 2 "" ( :. (p_(1))/(p_(2)) = (2)/(1))` <br/> `W = - 2.303 xx 1 xx 8.314 xx 298 xx 0.3010 J` <br/> `W = - 1717.46 J`</body></html> | |
| 51734. |
10 mL of water requires 1.47 mg ofK_(2) Cr_(2)O_(7)(M.wt. =294) fo oxidation of dissolved organic matter . C.O.D is |
| Answer» <html><body><p>2.44 ppm<br/>24 ppm<br/>32 ppm<br/>1.6 ppm</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> | |
| 51735. |
10 ml of H_2O_2solution on decomposition liberated 200ml ofO_2at STP. Then the weight/volume percentage of that H_2O_2solution is |
| Answer» <html><body><p>3.03<br/>6.12<br/>9.1<br/>3.4</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> | |
| 51736. |
10 mL of H_(2)O_(2) solution is treated with KI and titration of liberated I_(2) required 10 mL of 1 N hypo . Thus H_(2)O_(2) is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/1n-282830" style="font-weight:bold;" target="_blank" title="Click to know more about 1N">1N</a> <br/>5.6 <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a><br/>`17 gL^(-1)`<br/>all are correct</p>Solution :Let the normality of `H_(2)O_(2)=N_(1)` <br/>`therefore N_(1)xx10(H_(2)O_(2))=1xx10(Na_(2)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/>`therefore ` Normality of `H_(2)O_(2)` solution =1N <br/> `1NH_(2)O_(2)=5.6 vol. H_(2)O_(2)=17 gL^(-1)` <br/>Therefore , all statement are correct.</body></html> | |
| 51737. |
10 ml of H_(2)O_(2) could release 224 ml of O_(2) at 273K and 2 atm pressure. What is the molarity of that H_(2)O_(2) |
| Answer» <html><body><p><br/></p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>=(PV)/(RT)= n(2xx224xx10^(-3))/(0.0821xx273)` <br/> `n=(2xx224xx10^(-3))/(22.4) rArr n=2xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) "<a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a>" of O_(2)` <br/> `2H_(2)O_(2) to 2H_(2)O+O_(2)` <br/> `4xx10^(-2)` moles `2xx10^(-2)` moles <br/> `M=(n)/(Vlit)=(4xx10^(-2)xx1000)/(10), M=4`</body></html> | |
| 51738. |
10 mLof H,A (weak diprotic acid) solutions is titrated against 0.1M NaOH. pH of the solution is plotted against volume of strong base added and following observation is made. If pH of the solution at 1^(st) equivalence point is pH_(1) and at 2^(nd) equivalnee point is pH_(2), Cal the value (pH_(2),-pH_(1)) at 25^(@)C. Given for H_(2),A, p^(Kal) =4.6 &p^(Ka2) =8 |
| Answer» <html><body><p><br/></p>Solution :` H_2A +NaOH to <a href="https://interviewquestions.tuteehub.com/tag/naha-2855912" style="font-weight:bold;" target="_blank" title="Click to know more about NAHA">NAHA</a> +H_2O " at " I^(<a href="https://interviewquestions.tuteehub.com/tag/st-632849" style="font-weight:bold;" target="_blank" title="Click to know more about ST">ST</a>) " <a href="https://interviewquestions.tuteehub.com/tag/end-971042" style="font-weight:bold;" target="_blank" title="Click to know more about END">END</a> point,"` <br/> ` pH = (p^(K_(a_1))+p^(K_a_2))/(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)=( 8+ 4.6)/( 2) = 6.3` <br/> ` N_1V_1 =N_2V_2 `<br/> ` N_1xx 10=0 .1 xx 20 rArr N_1 =0.2` <br/> at ` 2^(nd) `end point , `Na_2 A` salt will be formed <br/>`{:(NaHA+ ,NaOHto ,Na_2 A + H_2O), (2m " moles " ,2m " moles",0),(-,-, 2 m " moles "):}` <br/> ` [Na_2 A]=(2)/(50)=0.04 M`</body></html> | |
| 51739. |
10 mL of H_(2) combine with 5 mL of O_(2) to form H_(2)O when 200 mL of H_(2) at S.T.P. is passed through heated CuO, latter loses 0.144 g of its weight. Does the above data correspond to the law of constant composition ? |
| Answer» <html><body><p><br/></p>Solution :In the first experiment : <br/> Ratio by volume of `H_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) and O_(2)` <a href="https://interviewquestions.tuteehub.com/tag/combining-922870" style="font-weight:bold;" target="_blank" title="Click to know more about COMBINING">COMBINING</a> to form `H_(2)O=2 :1` <br/> In the second experiment : <br/> `CuO+H_(2)overset(heat)(rarr)Cu+H_(2)O` <br/> The loss weight of CuO is due to oxygen which has been removed. It is 0.44 g <br/> Now, 32 g of `O_(2)` at S.T.P occupy = 22400 L <br/> 0.144 g of `O_(2)` at S.T.P occupy `= (("22400 mL"))/(("32 g"))xx(0.144g)=100.8mL` <br/> Ratio by volume of `H_(2) and O_(2)` combining to form `H_(2)O = <a href="https://interviewquestions.tuteehub.com/tag/200-288914" style="font-weight:bold;" target="_blank" title="Click to know more about 200">200</a> : 100.8 = 2:1` <br/> Since the <a href="https://interviewquestions.tuteehub.com/tag/two-714195" style="font-weight:bold;" target="_blank" title="Click to know more about TWO">TWO</a> ratios are the same, the data <a href="https://interviewquestions.tuteehub.com/tag/corresponds-935737" style="font-weight:bold;" target="_blank" title="Click to know more about CORRESPONDS">CORRESPONDS</a> to Law of Constant Composition.</body></html> | |
| 51740. |
10 " mL of " a solution of H_(2)O_(2) of 10 violume strength decolourises 100 " mL of " KMnO_(4) solution acidified with dil H_(2)SO_(4). The amount of KMnO_(4) in the given solution is K=39, Mn=55) |
| Answer» <html><body><p>0.282 g<br/>0.564 g<br/>1.128g<br/>0.155 g</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> of `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>)` at STP`=10mLxx10V=100mL` <br/> `22400" mL of " O_(2) at STP =1 mol=4Eq` <br/> `100 " mL of " O_(2) STP=(4)/(22400)xx100=(1)/(56)<a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a>` <br/> " Eq of "`KMnO_(4)=" Eq of "O_(2)` <br/> `=(1)/(56)" Eq of "KMnO_(4)` <br/> `=(1)/(56)xx31.5g of KMnO_(4)` <br/> `=0.564g` <br/> `(Ew of KMnO_(4)` in acidic medium `=(<a href="https://interviewquestions.tuteehub.com/tag/mw-550075" style="font-weight:bold;" target="_blank" title="Click to know more about MW">MW</a>)/(5)=31.5`)</body></html> | |
| 51741. |
10 ml of a mixture of carbon monoxide, marsh gas and hydrogen exploded with excess of oxygen gave a contraction of 6.5cc. There was further contraction of 7cc when the residual gas was treated with caustic potash. The volume of marsh gas present in original mixture as |
| Answer» <html><body><p>5cc<br/>2cc<br/>3cc<br/>4cc</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> of CO = xCC <br/> Volume of <a href="https://interviewquestions.tuteehub.com/tag/marsh-555568" style="font-weight:bold;" target="_blank" title="Click to know more about MARSH">MARSH</a> <a href="https://interviewquestions.tuteehub.com/tag/gas-1003521" style="font-weight:bold;" target="_blank" title="Click to know more about GAS">GAS</a> = yCC <br/> Volume of nitrogen = `10-(x+y)C C` <br/> `{:("2 CO",+,"<a href="https://interviewquestions.tuteehub.com/tag/o2-582782" style="font-weight:bold;" target="_blank" title="Click to know more about O2">O2</a>",rarr,2CO_(2)),("x CC",,(x)/(2)"CC",,"xCC"):}` <br/> `{:(CH_(4)(g)+2O_(2)(g)rarrCO_(2)(g)+2H_(2)O(l)),(""darr),("YCC2yCCyCCvolume negligible"):}` <br/> Contraction in volume due to bruning of CO <br/> `=x+(x)/(2)-x-(x)/(2)C C` <br/> Contraction in volume due to burning of `CH_(4)` <br/> `=y+2y-y=2yC C` <br/> `(x)/(2)+2y=6.5C Corx+4y=13C C` <br/> volume of `CO_(2)` produced = `x+y` <br/> contraction in volume with caustic <a href="https://interviewquestions.tuteehub.com/tag/potash-1161117" style="font-weight:bold;" target="_blank" title="Click to know more about POTASH">POTASH</a> = 7CC <br/> `x+4y=13impliesx+y=7impliesy=2impliesx=5` <br/> `CO=5C C""CH_(4)=2C C""N_(2)=3C C`</body></html> | |
| 51742. |
10 mL of a given solution of H_(2)O_(2) contains 0.91 g of H_(2)O_(2) . Express its strength in volume. |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/68-331600" style="font-weight:bold;" target="_blank" title="Click to know more about 68">68</a> g of `H_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)O_(2)` produce `O_(2)=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> at NTP <br/>`therefore 0.91 g ` of `H_(2)O_(2)` will produce `O_(2)=(22400xx0.91)/(68)~= 300` mL at NTP <br/> `therefore ` Volume <a href="https://interviewquestions.tuteehub.com/tag/strength-1229153" style="font-weight:bold;" target="_blank" title="Click to know more about STRENGTH">STRENGTH</a> `~=(300)/(10)~=<a href="https://interviewquestions.tuteehub.com/tag/30-304807" style="font-weight:bold;" target="_blank" title="Click to know more about 30">30</a>`</body></html> | |
| 51743. |
10 ml of a gaseous hydrocarbon is mixed with excess of oxygen and burnt . The gases are then cooled back . The reduction in volume was 25 ml . When the gases are passed into caustic potash , there is a further reduction in volume of 30 ml . If all volumes are measured under the laboratory conditions , the hydrocarbon is |
| Answer» <html><body><p>`C_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>) H_(8)`<br/>`C_(3) H_(4)`<br/>`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_(4) H_(8)`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :C</body></html> | |
| 51744. |
10ml of an alkane on complete combustion gave 40ml of CO_(2) under the same conditions. The formula of the alkane is |
| Answer» <html><body><p>`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_(<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_(6) 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_(4) H_(<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>)`<br/>`C_(4) H_(10)`</p>Answer :D</body></html> | |
| 51745. |
50 ml of 0.1 M solution of sodium acetate and 50 mlof 0.01 M acetic acid are mixed . The pKa of acetic acid is 4.76 TheP^(H)ofthe buffer solution is |
| Answer» <html><body><p></p>Solution :<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 milliequivalents of acid `= 20 xx 0.<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> = 2` <br/>Number of milliequivalents of salt `= 40 xx0.1 = <a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>`<br/> ` <a href="https://interviewquestions.tuteehub.com/tag/ph-1145128" style="font-weight:bold;" target="_blank" title="Click to know more about PH">PH</a> =pK_a +log ""(["Salt"])/("[acid]")=4.8 + log = 4.8+ 0.3= 5.1`</body></html> | |
| 51746. |
10 ml N_(2) is reacted with 20 ml H_(2) to form NH_(3). The correct statements is /are |
| Answer» <html><body><p>13.3 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> `NH_(3)` is formed<br/>20 ml `NH_(3)` is formed<br/>3.4 ml `N_(2)` is left after the completion of the reaction<br/>16.7 ml `NH_(3)` of mixture is left after the completion of the reaction</p>Solution :`{:(,N_(2(g)),+,3H_(2(g)),rarr,2NH_(3(g))),("<a href="https://interviewquestions.tuteehub.com/tag/stp-633132" style="font-weight:bold;" target="_blank" title="Click to know more about STP">STP</a>","22400 ml",,3xx22400ml,,2xx22400ml),("At <a href="https://interviewquestions.tuteehub.com/tag/start-1224710" style="font-weight:bold;" target="_blank" title="Click to know more about START">START</a>","10 ml",,"20 ml",,),("After reaction",10-6.6,,-,,"13.3ml"),(,"= 3.4ml",,,,):}`</body></html> | |
| 51747. |
10 litres of a gas at S.T.P. weigh 19.64 g. Calculate the molecular mass of the gas. |
| 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/44-316683" style="font-weight:bold;" target="_blank" title="Click to know more about 44">44</a></body></html> | |
| 51748. |
1.0 litre of a solution contains 5.3 g of Na_(2)CO_(3) and 8 g of NaOH. 20 mL of this solution are taken and titrated against N//10 HCl using separately (a) methyl orange as an indicator and (b) phenolphthalein as an indicator. What will be the titre values in these two cases ? |
| 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> | |
| 51749. |
10 L of water has 33 mg of oxygen dissolved. What is the ppm of dissolved oxygen? |
| 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> :3.3 <a href="https://interviewquestions.tuteehub.com/tag/ppm-1162221" style="font-weight:bold;" target="_blank" title="Click to know more about PPM">PPM</a></body></html> | |
| 51750. |
10 kg of cold water at 2^(@)C absorbs 50J of heatfrom the surrounding which were at a temperature of37^(@)C . What is the entropy change of (i) the system (ii) the surroundings (iii) the universe ? Assumethe change in temperature of the water and the surroundings to be negligible. |
| Answer» <html><body><p><br/></p>Solution :`(i) DeltaS_("system") ("water")= ( +50J)/( 275K) = 0.18 <a href="https://interviewquestions.tuteehub.com/tag/jk-521675" style="font-weight:bold;" target="_blank" title="Click to know more about JK">JK</a>^(-1)``(ii) DeltaS_("<a href="https://interviewquestions.tuteehub.com/tag/surroundings-1235986" style="font-weight:bold;" target="_blank" title="Click to know more about SURROUNDINGS">SURROUNDINGS</a>") =( -50J)/(310K) = - 0.16 JK^(-1)` <br/> `(<a href="https://interviewquestions.tuteehub.com/tag/iii-497983" style="font-weight:bold;" target="_blank" title="Click to know more about III">III</a>) DeltaS_("universe") = <a href="https://interviewquestions.tuteehub.com/tag/deltas-947743" style="font-weight:bold;" target="_blank" title="Click to know more about DELTAS">DELTAS</a> _("system") + DeltaS_("surroundings") = 0.18 + ( -.16) = 0.02 JK^(-1)`</body></html> | |