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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.
| 51101. |
40 ml of a hydrocarbon undergoes combustion in 260 ml oxygen and gives 160 mlof CO_(2). If all volumes are measured under similar conditions of temperature and pressue, the formula of the hydrocarbon is |
| Answer» <html><body><p>`C_(3)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_(8)`<br/>`C_(6)H_(<a href="https://interviewquestions.tuteehub.com/tag/14-272882" style="font-weight:bold;" target="_blank" title="Click to know more about 14">14</a>)`<br/>`C_(4)H_(10)`</p>Solution :`C_(4)H_(10)+(13)/(<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)rarr4CO_(2)+5H_(2)O` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/40-314386" style="font-weight:bold;" target="_blank" title="Click to know more about 40">40</a> ml-260 ml-160 ml`</body></html> | |
| 51102. |
40 ml of 0.1 Mammonia is mixed with 20 ml of 0.1 M HCl. What is the pH of the mixture ? ( pK_(b) of ammoniasolution is 4.74) |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>.74<br/><a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>.26<br/>9.26<br/>`5.00`</p>Solution :40 ml of 0.1 M `NH_(3)` solution `= 40xx0.1` millimole <br/> = 4 millimoles<br/> `NH_(4)OH+HCl rarr NH_(4)<a href="https://interviewquestions.tuteehub.com/tag/cl-408888" style="font-weight:bold;" target="_blank" title="Click to know more about CL">CL</a> + H_(2)O` <br/> 2 millimole of HCl will <a href="https://interviewquestions.tuteehub.com/tag/neutralize-7700166" style="font-weight:bold;" target="_blank" title="Click to know more about NEUTRALIZE">NEUTRALIZE</a> 2 millimoles of `NH_(4)OH` to form 2 millimoles of `NH_(4)Cl`. <br/> `NH_(4)OH` left = 60 ml <br/> `:. [NH_(4)OH]=(2)/(60)M, [NH_(4)Cl]=(2)/(60)M` <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> .([NH_(4)Cl])/([NH_(4)OH])` <br/> `=4.74 + log .(2//60)/(2//60)=4.74` <br/> `:. pH=14-4.4.74=9.26`</body></html> | |
| 51103. |
40 ml gaseous mixture of CO, CH_(4) and Ne was exploded with 10 ml of oxygen. On cooling, the gases occupied 36.5 ml. After treatment with KOH the volume reduced by 9 ml and again on treatment with alkaline pyrogallol, the volume further reduced, percentage of CH_(4) in the original mixture is |
| Answer» <html><body><p>a. `22.4`<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>. `77.5`<br/>c. `7.5`<br/>d. `15`</p>Solution :`underset(x)(CO)+(1)/(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)underset((x)/(2))(O_(2))rarrunderset(x)(CO_(2))` <br/> `underset(y)(CH_(4))+underset(2y)(2O_(2))rarrunderset(y)(CO_(2))+2H_(2)O` <br/> On passing through KOH, volume decrease due to `CO_(2)` absorption <br/> `impliesx+y=9….(1)` <br/> Final volume after reaction = <br/> `CO_(2)+Ne+"left over "O_(2)` <br/> `=(x+y)+(40-x-y)+(10-x//2-2y)` <br/> `36.5=50-(x)/(2)-2y` <br/> `x+4y=27....(2)` <br/> solving, `y=6,x=3` <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> %" of "CH_(4)=(6)/(40)xx100=15%`</body></html> | |
| 51104. |
40 mL of 0.05 M Na_(2)CO_(3). NaHCO_(3). 2H_(2)O (sesquicarbonate ) is titrated against 0.05 M HCl. X and of HCl is used when phenolphthalein is the indicator in two separate titrations , hence (y-x) is : |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/80-337972" style="font-weight:bold;" target="_blank" title="Click to know more about 80">80</a> mL<br/>30 mL<br/>120 mL<br/>none of these</p>Solution :N//A</body></html> | |
| 51105. |
40 gm of a carbonate of an alkali metal or alkaline earth metal containg some insert impurities was made to react with excess HCl solution. The liberated CO_(2) occupied 12.315 litre at 1 atm and 300K. The corrrect option is: |
| Answer» <html><body><p>Mass of <a href="https://interviewquestions.tuteehub.com/tag/impurity-1039200" style="font-weight:bold;" target="_blank" title="Click to know more about IMPURITY">IMPURITY</a> of 1gm and <a href="https://interviewquestions.tuteehub.com/tag/metal-1094457" style="font-weight:bold;" target="_blank" title="Click to know more about METAL">METAL</a> is Be<br/>Mass of impurity is 3gm and metal is Li<br/>Mass of impurity is <a href="https://interviewquestions.tuteehub.com/tag/6gm-1912279" style="font-weight:bold;" target="_blank" title="Click to know more about 6GM">6GM</a> and metal is Li<br/>Mass of impurity is 2gm and metal is Mg</p>Answer :B</body></html> | |
| 51106. |
40 gm NaOH, 106 gm Na_(2)CO_(3) and 84 gm NaHCO_(3) is dissolved in water and the solution is made 1 lit, 20 ml of this stock solution is titrated with 1 N HCl, hence which of the followign statements are correct? |
| Answer» <html><body><p>The burette reading of HCl will be 40 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a>, if phenolphthalein is used as indiator from the beginning <br/>The burette reading of HCl will be 60 ml, if phenolphthalein is used as indicator form the beginning. <br/>The burette readin of HCl will be 40ml, if methyl orange is used as indicator after the first end point <br/>The burette reading of HCl will be 80 ml, if methyl <a href="https://interviewquestions.tuteehub.com/tag/organe-2898437" style="font-weight:bold;" target="_blank" title="Click to know more about ORGANE">ORGANE</a> is used as indicator from the very beginning. </p>Solution :`n_(NaOH)=n_(Na_(2)CO_(3))=n_(NaHCO_(3))=1` <br/> a) In presence of Hph, <br/> Eqts of `NaOH+(1)/(2)` <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a> `N_(2)CO_(3)` = Eq HCl <br/> `(200xx1xx1)+(1)/(2)(20xx1xx2)=1xxV` <br/> `V_(HCl)=40ml` <br/> c) In presence of methyl orange at first end point <br/> `(1)/(2)` eq `NA_(2)CO_(3)+` eq `NAHCO_(3)=` eq HCl <br/> `((1)/(2)xx20xx1xx2)+(20xx1xx1)=V_(HCl)<a href="https://interviewquestions.tuteehub.com/tag/xx1-1463705" style="font-weight:bold;" target="_blank" title="Click to know more about XX1">XX1</a>` <br/> `V_(HCl)=40ml` <br/> d) If methyl orange used <br/> Eq `NAOH+` Eq `NAl_(2)CO_(3)+` Eq `NaHCO_(3)` <br/> = Eq HCl <br/> `(20xx1xx1)+(20xx1xx2)+(20xx1xx1)` <br/> `=V_(HCl)xx1impliesV_(HCl)=80`</body></html> | |
| 51107. |
4.0 g of NaOH are dissolved per litre. Find (i) molarity of the solution (ii) OH^(-)ion concentration (iii) pH value of the solution (At. Masses : Na = 23, O=16, H=1). |
| Answer» <html><body><p></p>Solution :(i) Calculation of molarity : Mass of NaOH <a href="https://interviewquestions.tuteehub.com/tag/dissolved-956358" style="font-weight:bold;" target="_blank" title="Click to know more about DISSOLVED">DISSOLVED</a> = 4.0 <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/litre-1075864" style="font-weight:bold;" target="_blank" title="Click to know more about LITRE">LITRE</a> <br/> Mol . Mass of NaOH = 40 `:.` Molarity of the solution `= ("Strength in g // litre")/("Mol . Mass") = (4.0)/(40) = 0.1 M` <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>) Calculation of the `OH^(-)` ion conc. <br/> NaOH completely ionizes as : `NaOH rarr Na^(+) + OH^(-)" " :. [OH^(-)]=[NaOH]=0.1 M = 10^(-1)M` <br/> (iii) Calculationof pH :<br/> We know that `[H_(3)O^(+)][OH^(-)]=K_(w)=1.0xx10^(-14)` <br/> `:. [H_(3)O^(+)]=(K_(w))/([OH^(-)])=(1.0xx10^(-14))/(10^(-1))=10^(-13)M" " :. pH = - <a href="https://interviewquestions.tuteehub.com/tag/log-543719" style="font-weight:bold;" target="_blank" title="Click to know more about LOG">LOG</a> [H_(3)O^(+)]= - log 10 ^(-13) = 13`.</body></html> | |
| 51108. |
4.0 g of argon (at mass = 40) in a bulb at a temperature of TK had a pressure P atm. When the bulb was placed in hotter bath at a temperature 50^(@) more than the first one, 0.8 g of gas had to be removed to get the original pressure. T is equal is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/510-324427" style="font-weight:bold;" target="_blank" title="Click to know more about 510">510</a> K<br/>200 K<br/>100 K<br/>73 K</p>Solution :According to ideal gas eqatio <br/> `<a href="https://interviewquestions.tuteehub.com/tag/pv-593601" style="font-weight:bold;" target="_blank" title="Click to know more about PV">PV</a> = w (RT)/(M) " or " <a href="https://interviewquestions.tuteehub.com/tag/wt-1462289" style="font-weight:bold;" target="_blank" title="Click to know more about WT">WT</a> = (PV)/(R) .M` <br/> or `w_(1)T_(1) = w_(2) T_(2)` when P and V are constant <br/> Hence `4 xx T = 3.2 (T + <a href="https://interviewquestions.tuteehub.com/tag/50-322056" style="font-weight:bold;" target="_blank" title="Click to know more about 50">50</a>)` <br/> or `T = 200 K`</body></html> | |
| 51109. |
40 g f a gas occupies 20 dm^(3) at 300 K and 100 kPa pressure. If the pressure is changed to 50 lPa without changing to temperature, what would be its volume ? |
| Answer» <html><body><p></p>Solution :`P_(1)v_(1)=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>)v_(2)` at constant temperature. <br/> `P_(1)=<a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a> kPa, v_(1)=<a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a> dm^(<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_(2) =50 kPa, V_(2) ?` <br/> `V_(2)=(P_(1)V_(1))/(P_(2)),v_(2)=(100xx20)/(50)=40 dm^(3)` <br/> `:. V_(2)=40 dm^(3)`</body></html> | |
| 51110. |
4-Oxobutanoic acid is reduced with Na-borohydride and the product is treated with aqueous acid. The final product is : |
| Answer» <html><body><p><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/RES_CHM_ROHR_E03_095_O01.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/RES_CHM_ROHR_E03_095_O02.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/RES_CHM_ROHR_E03_095_O03.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/RES_CHM_ROHR_E03_095_O04.png" width="30%"/></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> | |
| 51111. |
4-Pentenoic acid when treated with I_(2) and NaHCO_(3) gives: |
| Answer» <html><body><p>4,5-diiodopentanoic acid<br/>5-iodomethyl-dihydrofuran-2-one<br/>5-<a href="https://interviewquestions.tuteehub.com/tag/iodo-2747950" style="font-weight:bold;" target="_blank" title="Click to know more about IODO">IODO</a>-tetrahydropyran-2-one<br/>4-pentenolyiodide</p>Solution :Iodo lactonization.</body></html> | |
| 51112. |
4 mole of NH_(4) OH and 1 mole of H_(2) SO_(4) are mixed and dilute to 1 lit solution . The pK_(b) of NH_(4) OH is 4.8 . The pH of solution is |
| Answer» <html><body><p>`4.8`<br/>`9.2`<br/>`9.5`<br/>`8.9`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 51113. |
4-Methylbenzenesulphoic acid reacts with sodium acetate to give |
| Answer» <html><body><p><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/DIN_OBJ_CHM_V02_C7_2_E01_350_O01.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/DIN_OBJ_CHM_V02_C7_2_E01_350_O02.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/DIN_OBJ_CHM_V02_C7_2_E01_350_O03.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/DIN_OBJ_CHM_V02_C7_2_E01_350_O04.png" width="30%"/></p>Solution :4-Methylbenzenesulphonic <a href="https://interviewquestions.tuteehub.com/tag/acid-847491" style="font-weight:bold;" target="_blank" title="Click to know more about ACID">ACID</a> is <a href="https://interviewquestions.tuteehub.com/tag/stronger-1229999" style="font-weight:bold;" target="_blank" title="Click to know more about STRONGER">STRONGER</a> than acetic acid <a href="https://interviewquestions.tuteehub.com/tag/thus-2307358" style="font-weight:bold;" target="_blank" title="Click to know more about THUS">THUS</a> it will <a href="https://interviewquestions.tuteehub.com/tag/release-1183763" style="font-weight:bold;" target="_blank" title="Click to know more about RELEASE">RELEASE</a> acetic acid from sodiumm acetate</body></html> | |
| 51114. |
4-methyl-pent-2-yne has how many sigma and pi bond? |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/15-274069" style="font-weight:bold;" target="_blank" title="Click to know more about 15">15</a> sigma, 2pi`<br/>`12 sigma, 2pi`<br/>`13 sigma, 2pi`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/14-272882" style="font-weight:bold;" target="_blank" title="Click to know more about 14">14</a> sigma, 2pi`</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> | |
| 51115. |
4-methyl benzene sulphonic acid reacts with sodium acetate to give |
| Answer» <html><body><p><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/DIN_OBJ_CHM_V02_C7_2_E01_485_O01.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/DIN_OBJ_CHM_V02_C7_2_E01_485_O02.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/DIN_OBJ_CHM_V02_C7_2_E01_485_O03.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/DIN_OBJ_CHM_V02_C7_2_E01_485_O04.png" width="30%"/></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 <a href="https://interviewquestions.tuteehub.com/tag/methyl-1095247" style="font-weight:bold;" target="_blank" title="Click to know more about METHYL">METHYL</a> <a href="https://interviewquestions.tuteehub.com/tag/benzene-895203" style="font-weight:bold;" target="_blank" title="Click to know more about BENZENE">BENZENE</a> sulphonic acid is <a href="https://interviewquestions.tuteehub.com/tag/stronger-1229999" style="font-weight:bold;" target="_blank" title="Click to know more about STRONGER">STRONGER</a> than acetic acid hence it will release acetic acid from sodium acetate <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/DIN_OBJ_CHM_V02_C7_2_E01_485_S01.png" width="80%"/></body></html> | |
| 51116. |
4-Methyl-1,3-pentadiene react with HBr, The possible products obtained are |
| Answer» <html><body><p>4-Bromo-2-methyl-2-pentene <br/>4-Bromo-4-methyl-1-pentene <br/>4-Bromo-4-methyl-2-pentene <br/>All the above</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/cumulative-428447" style="font-weight:bold;" target="_blank" title="Click to know more about CUMULATIVE">CUMULATIVE</a> <a href="https://interviewquestions.tuteehub.com/tag/dienes-439163" style="font-weight:bold;" target="_blank" title="Click to know more about DIENES">DIENES</a></body></html> | |
| 51117. |
4 litres of water are added to 2 litres 6 molar hydrochloric acid solution. What is the molarity of the resulting solution ? |
| Answer» <html><body><p><br/></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> volume of the solution after <a href="https://interviewquestions.tuteehub.com/tag/mixing-17204" style="font-weight:bold;" target="_blank" title="Click to know more about MIXING">MIXING</a> `=(4+2)=6L` <br/> By <a href="https://interviewquestions.tuteehub.com/tag/applying-1982651" style="font-weight:bold;" target="_blank" title="Click to know more about APPLYING">APPLYING</a> molarity equation : <br/> `overset(("dilute"))(M_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)V_(1))-=overset(("conc."))(M_(2)V_(2))` <br/> `M_(1)xx6=6xx2orM_(1)=(6xx2)/(6)=2M`.</body></html> | |
| 51118. |
4 L of 0.02 M aqueous solution of NaCl was diluted by adding one litre of water.The molality of the resultant solution is : |
| Answer» <html><body><p>`0.004`<br/>`0.008`<br/>`0.012`<br/>`0.016`</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> | |
| 51119. |
4-hydroxyl phenol reacts with acidified potassium dichromate to give ……………. |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/quinol-1175118" style="font-weight:bold;" target="_blank" title="Click to know more about QUINOL">QUINOL</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/cyclohexanone-942565" style="font-weight:bold;" target="_blank" title="Click to know more about CYCLOHEXANONE">CYCLOHEXANONE</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/phenol-1152951" style="font-weight:bold;" target="_blank" title="Click to know more about PHENOL">PHENOL</a> <br/>P-Benzoquinone</p>Solution :P-Benzoquinone</body></html> | |
| 51120. |
4-hydroxy phenol reacts with acidified potassium dichromate to gives .............. |
| 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> :p-Benzoquinone</body></html> | |
| 51121. |
4- heptanoneoverset(KMnO_(4)//H^(+)//Delta)rightarrowA+B. Identify A and B |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/ethanoic-2073639" style="font-weight:bold;" target="_blank" title="Click to know more about ETHANOIC">ETHANOIC</a> <a href="https://interviewquestions.tuteehub.com/tag/acid-847491" style="font-weight:bold;" target="_blank" title="Click to know more about ACID">ACID</a> pentanoic acid<br/>Ethanal andbutanone<br/>Butanoic acid and <a href="https://interviewquestions.tuteehub.com/tag/propanoic-2957733" style="font-weight:bold;" target="_blank" title="Click to know more about PROPANOIC">PROPANOIC</a> acid<br/>Acetic acid andpentanoic acid</p>Solution :<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NAR_CHM_XII_V04_C03_E01_066_S01.png" width="80%"/></body></html> | |
| 51122. |
4^(@) hydrogen is not possible, give reasons….. |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/carbon-16249" style="font-weight:bold;" target="_blank" title="Click to know more about CARBON">CARBON</a> has <a href="https://interviewquestions.tuteehub.com/tag/four-464592" style="font-weight:bold;" target="_blank" title="Click to know more about FOUR">FOUR</a> covalent bond.<br/>Centre carbon is attach with other four carbon and so there is no valence for attachement of <a href="https://interviewquestions.tuteehub.com/tag/hydrogen-22331" style="font-weight:bold;" target="_blank" title="Click to know more about HYDROGEN">HYDROGEN</a><br/>`1^(@), 2^(@)` and `3^(@)` carbon are stable but `4^(@)` carbon is unstable.<br/>Free <a href="https://interviewquestions.tuteehub.com/tag/radical-616075" style="font-weight:bold;" target="_blank" title="Click to know more about RADICAL">RADICAL</a> is formed with `4^(@)` carbon.</p>Solution :Centre carbon is attach with other four carbon and so there is no valence for attachement of hydrogen</body></html> | |
| 51123. |
4 grams of each pure huydrochloric acid and pure caustic soda are together dissolved in water. What weight of sodium chloride is obtained? |
| Answer» <html><body><p></p>Solution :The <a href="https://interviewquestions.tuteehub.com/tag/stoichiometric-1227930" style="font-weight:bold;" target="_blank" title="Click to know more about STOICHIOMETRIC">STOICHIOMETRIC</a> equation is given as <br/> `HCl+NaOH to NaCl+H_(2)O` <br/> 1 mole of HCl =1 mole of NaOH <br/> 36.5 grams of HCl=40 grams of NaOH <br/> NaOH is the limiting reagent. Therefore, the <a href="https://interviewquestions.tuteehub.com/tag/weight-1451304" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHT">WEIGHT</a> of NaCl obtained is <a href="https://interviewquestions.tuteehub.com/tag/limited-1074043" style="font-weight:bold;" target="_blank" title="Click to know more about LIMITED">LIMITED</a> by NaOH <br/> 1 mole of NaOH=1 mole of NaCl <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> grams of NaOH=58.5 grams of NaCl <br/> 4 grams of NaOH=? <br/> The weight of sodium <a href="https://interviewquestions.tuteehub.com/tag/chloride-915854" style="font-weight:bold;" target="_blank" title="Click to know more about CHLORIDE">CHLORIDE</a> obtained `=58.5 xx (4)/(40)=5.85"grams"`</body></html> | |
| 51124. |
4 grams of copper chloride on analysis were found to contain 1.890 g of copper (Cu) and 2.110 g of chlorine (CI). What is the empirical formula of copper chloride? |
| 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> :`CuCl_(<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> | |
| 51125. |
4 grams of each pure hydrochloric acid and pure caustic soda are together dissolved in water. What weight of sodium chloride is obtained? |
| 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> :5.85g</body></html> | |
| 51126. |
4 grams of an ideal gas occupies 5.6035 litres of volume at 546 K and 2 atm, pressure. What is its molecular weight ? |
| Answer» <html><body><p>4<br/>16<br/>32<br/>64</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> | |
| 51127. |
A container of 1 L capacity contains a mixture of 4 g of O_(2) and 2 g H_(2) at 0 .^(@)C . What will be the total pressure of the mixture ?(a) 50 . 42 atm(b) 25 . 21 atm(C) 15 . 2 atm(d) 12 . 5 atm |
| 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><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Calculation of <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> pressure : <br/> <a href="https://interviewquestions.tuteehub.com/tag/according-366619" style="font-weight:bold;" target="_blank" title="Click to know more about ACCORDING">ACCORDING</a> to Dalton.s law of partial pressure, the total pressure of the gaseous mixture is given by <br/>`P_(mixture) = P_(O_2)+P_(H_2) = 2.80 + 22.4 = 25.2` atm</body></html> | |
| 51128. |
4 gm NaOH dissolve in 250 ml water and solution is prepared. Find molarity of the solution. |
| Answer» <html><body><p> <br/> <br/> <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> :0.4 M</body></html> | |
| 51129. |
4 g of O_2 and 2g of H_2 are confined in a vessel of capacity 1 litre at 0^@C. Calculate the partial pressure of each gas, and |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/calculation-907729" style="font-weight:bold;" target="_blank" title="Click to know more about CALCULATION">CALCULATION</a> of partial pressures : <br/> Volume of vessel = 1 L, `T = <a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>^@<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> = 273 <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>` <br/> `:. "" PV = nRT " or " p = (nRT)/<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>`<br/>`:.` Partial pressure of oxygen <br/>`Po_2=(nRT)/V =( 0.125 xx 0.0821 xx 273)/1 = 2.80` atm<br/>Similarly, partial pressure of hydrogen <br/> `P_(H_2)= (nRT)/V= (1xx0.0821 xx 273)/1= 22.4` atm</body></html> | |
| 51130. |
4 g of O_2 and 2g of H_2 are confined in a vessel of capacity 1 litre at 0^@C. Calculate the number of moles of each 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/calculation-907729" style="font-weight:bold;" target="_blank" title="Click to know more about CALCULATION">CALCULATION</a> of number of moles : <br/>Number of moles = `("Mass in grams")/("<a href="https://interviewquestions.tuteehub.com/tag/gram-1010695" style="font-weight:bold;" target="_blank" title="Click to know more about GRAM">GRAM</a> molecular mass")` <br/> `:. " Number of moles of " O_2 = 4/32 =0.125` <br/>and, Number of moles of `H_2 =2/2 =<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>`</body></html> | |
| 51131. |
4 g of NaOH are dissolved in 200 cm^3 of water. Find the molarity of the solution. |
| Answer» <html><body><p></p>Solution :Mass of NaOH dissolved = 4g <br/> Molecular mass of NaOH = 23 + <a href="https://interviewquestions.tuteehub.com/tag/16-276476" style="font-weight:bold;" target="_blank" title="Click to know more about 16">16</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> = 40 <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>` Moles of NaOH dissolved `=4/40 = 0.1` <br/> Volume of water taken `=<a href="https://interviewquestions.tuteehub.com/tag/200-288914" style="font-weight:bold;" target="_blank" title="Click to know more about 200">200</a> cm^(3) = 200 xx 10^(-3) <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/> The molarity of the solution `=("No. of moles of NaOH")/("Volume in litres")` <br/> `=0.1/(200 xx 10^(-3))` <br/> `=0.5 mol L^(-1)` <br/> Hence, the given solution is 0.5 M</body></html> | |
| 51132. |
4 g. of methane at 380 torr and 273^@C occupies a volume of |
| Answer» <html><body><p>5.6 <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a> <br/>11.2 L <br/>16.8 L <br/>22.4 L </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :D</body></html> | |
| 51133. |
4 g of a mixture of Na_(2)SO_(4) and anhydrous Na_(2)CO_(3) were dissolved in pure and volume made up to 250 mL. 20 mL of this solution required 25 mL of N//5 H_(2)SO_(4) for complete neutralisation. Calculate the percentage composition of the mixture. |
| 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> | |
| 51134. |
4 g of a mixture of NaCl and Na_(2)CO_(3) were dissolved in water and volume made up to 250 mL. 15 mL of this solution required 50 mL of N//10 HCl for complete neutralisation. Calculate the percentage composition of the original mixture. |
| 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> | |
| 51135. |
4g of mixture of Na_(2)CO_(3) and NaHCO_(3) on heating liberates 448 ml of CO_(2) at STP. The percentage of Na_(2)CO_(3) in the mixture is |
| Answer» <html><body><p>84<br/>16<br/>54<br/>80</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> | |
| 51136. |
4 g / L solutionof NaOH is given. The molarity of the solution is ........... M. (Molecular mass of NaOH =40 g / mol) |
| Answer» <html><body><p>160<br/>10<br/>0.1<br/>4</p>Answer :A::<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 51137. |
4 g Cu is dissolved in concentrated HNO_(3). On heating this solution 5 g oxide will be obtained Then the equivalent weight of Cu is..... |
| Answer» <html><body><p>23<br/>32<br/>12<br/>20</p>Solution :4g <a href="https://interviewquestions.tuteehub.com/tag/cu-428205" style="font-weight:bold;" target="_blank" title="Click to know more about CU">CU</a> give <a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a> g copper oxide <br/> 4 g Cu <a href="https://interviewquestions.tuteehub.com/tag/combine-409630" style="font-weight:bold;" target="_blank" title="Click to know more about COMBINE">COMBINE</a> with `(5-4)=<a href="https://interviewquestions.tuteehub.com/tag/1g-282964" style="font-weight:bold;" target="_blank" title="Click to know more about 1G">1G</a>` oxygen <br/> `:.1 g O <a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> Cu "" :. 8 g rarr (?)` <br/> ` = (8xx4)/(1) = 32 g Cu` <br/> eq. wt `=32 "g eq. wt"^(-1)`</body></html> | |
| 51138. |
3s orbital has 2 radial nodes |
| 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> :T</body></html> | |
| 51139. |
3O_(2) harr 2O_(3). Oxygen is ozonised until the partial pressures of both gases become equal at equilibrium. Which statements are correct if P_(eq) = 10 atm ? |
| Answer» <html><body><p>60% oxygen is ozonised by volume<br/>mol. wt at equilibrium = 40<br/>`K_(P)` = 0.2 atm<br/>`K_(P)`= 0.5 atm</p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_B_C04_E02_050_S01.png" width="80%"/> <br/> <a href="https://interviewquestions.tuteehub.com/tag/partial-596556" style="font-weight:bold;" target="_blank" title="Click to know more about PARTIAL">PARTIAL</a> <a href="https://interviewquestions.tuteehub.com/tag/pressures-1164423" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURES">PRESSURES</a> are <a href="https://interviewquestions.tuteehub.com/tag/equal-446400" style="font-weight:bold;" target="_blank" title="Click to know more about EQUAL">EQUAL</a> `implies 1-<a href="https://interviewquestions.tuteehub.com/tag/alpha-858274" style="font-weight:bold;" target="_blank" title="Click to know more about ALPHA">ALPHA</a>=(2alpha)/(3) implies alpha=(3)/(5)=60%` <br/> `(D)/(d)=1+(n-1)alpha=1+((-1)/(3))alpha` <br/> `(32)/(d)=(4)/(5), d=40` <br/> `2P=P_(eqm) implies 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>)/(2)=5` <br/> `:. K_(P)=(P^(2))/(P^(3))=(1)/(P)=0.2 atm^(-1)`</body></html> | |
| 51140. |
3N H_(3)PO_(4) aqu. solutionis given = .......... gm/litre. |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/98-342802" style="font-weight:bold;" target="_blank" title="Click to know more about 98">98</a><br/>298<br/>33<br/>95</p>Solution :Equivalent weight `=("<a href="https://interviewquestions.tuteehub.com/tag/molecular-562994" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULAR">MOLECULAR</a> mass")/("<a href="https://interviewquestions.tuteehub.com/tag/basicity-394484" style="font-weight:bold;" target="_blank" title="Click to know more about BASICITY">BASICITY</a>")= (98)/(<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/> `N=("gm/lit")/("equ.wt.")= 3 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> (98)/(3) = 98`</body></html> | |
| 51141. |
3L Mixture of propane and butane on complete combustion at 298K and 1atm gave 10L CO_2. Calculate the composition of gas |
| Answer» <html><body><p>`1L C_(3)H_(8), <a href="https://interviewquestions.tuteehub.com/tag/2l-300409" style="font-weight:bold;" target="_blank" title="Click to know more about 2L">2L</a> C_(4)H_(10)`<br/>`2L C_(3)H_(8) 1L C_(4)H_(10)`<br/>`1.5 <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a> C_(3)H_(8), 1.5L C_(4)H_(10)`<br/>`0.75L C_(3)H_(8), 2.25L C_(4)H_(10)`</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> :B</body></html> | |
| 51142. |
3Fe_((s)) + 4H_2O_((g)) undersetlarrto 3Fe_3O_(4(s)) + 4H_2O_((g)) formula of K_P and K_C is ……………. |
| 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> :`{:(,3Fe_((s)) + 4H_2O_((g)) <a href="https://interviewquestions.tuteehub.com/tag/undersetlarrto-3244003" style="font-weight:bold;" target="_blank" title="Click to know more about UNDERSETLARRTO">UNDERSETLARRTO</a>,Fe_3O_(4(s)) + 4H_2O_((g))),("Current reaction:", 3Fe_((s)) + 4H_2O_((g)) undersetlarrto , Fe_3O_(4(s))+4H_(2(g))):}` <br/> `K_c=[H_2]^4/[H_2O]^4` <br/> `K_p=(p_(H_2))^4/(p_(H_2O))^4`</body></html> | |
| 51143. |
3Cl_(2)+6NaOHrarr5NaCl+NaClO_(3)+3H_(2)O, chlorine gets |
| Answer» <html><body><p>Oxidised<br/>Reduced<br/>Both <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> & 2<br/>None of these</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> | |
| 51144. |
3CIO^(-)(aq)rarrCIO_(3)+2C^(-) is an exameof |
| Answer» <html><body><p>oxidation reaction<br/> <a href="https://interviewquestions.tuteehub.com/tag/reduction-621019" style="font-weight:bold;" target="_blank" title="Click to know more about REDUCTION">REDUCTION</a> reaction <br/>disproportionation reaction <br/><a href="https://interviewquestions.tuteehub.com/tag/decompostion-2567739" style="font-weight:bold;" target="_blank" title="Click to know more about DECOMPOSTION">DECOMPOSTION</a> reaction </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> | |
| 51145. |
3Ca_(3)(PO_(4))_(2).CaF_(2) is a part of enamel on teeth. |
| 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> :T</body></html> | |
| 51146. |
3C_(2)H_(2) hArr C_(6) H_(6) the above reaction is performed in a I lit vessel. Equilibrium is established when 0.5 mole of benzene is present at certain tempe-rature. If equilibrium constant is 4 lit^(2) "mole"^(-2) The total number ot moles of the substances present at cquilibrium, |
| Answer» <html><body><p>0.5<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><br/>1.5<br/>2</p>Solution :`3C_(2)H_(2(g)) <a href="https://interviewquestions.tuteehub.com/tag/harr-2692945" style="font-weight:bold;" target="_blank" title="Click to know more about HARR">HARR</a> C_(6)H_(6(g))` <br/> `K_(<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>)=([C_(6)H_(6)])/([C_(2)H_(2)]^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)), 4=(0.5)/([C_(2)H_(2)]^(3)) implies [C_(2)H_(2)]^(3)=(1)/(8)` <br/> `[C_(2)H_(5)]=0.5=(n)/(v) implies n=0.5` <br/> Total no. of moles of `C_(2)H_(2)` & `C_(6)H_(6)` at equilibrium = 0.5+0.5=1</body></html> | |
| 51147. |
3.92 g of ferrous ammonium suphate ar dissolved in 100 Ml water 20 ml of this solution requires 18 complete oxidation the weight of KMnO_(4) present in one litre of the solution is |
| Answer» <html><body><p>34.76 <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/>12.38 g<br/>1.23 g<br/>3.476 g</p>Solution :The redox reaction involving the oxidation of `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>+)` is<br/> `MnO_(4)^(-)+5Fe^(2+)+8H^(+)rarr5Fe^(3+)+<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+)+4H_(2)O` <br/> Mol <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 ferrous ammonium suphate <br/> `(NH_(4))SO_(4)FeSO_(4)6H_(23)O=392`<br/> applyinhg molarity equation <br/> `=(M_(2)V_(2))/(n_(2))` <br/> or `(0.1xx20)/(5)=(M_(2)xx18)/(1)`<br/> or `M_(2)=(0.1xx20)/(5xx18)=(M)/(45)` <br/>Amount of `KMnO_(4)` present in one litre<br/> =Molarity x mol wt <br/> `1/45xx158=3.51` g <br/> since 3.476 g isclose to 3.51 g therefore option (d) is correct</body></html> | |
| 51148. |
38 cm^(2) N_(2) gas at 300K temperature and 96pa pressure. In 0.25g organic compound calculate the % of nitrogen. (1 atm .- 101.3 pa) |
| 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> :0.1638</body></html> | |
| 51149. |
3.7 g of a gas at 25°C occupies the same volume as 0.184 g of hydrogen at 17°C and at the same pressure. What is 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> :41.33</body></html> | |
| 51150. |
3.65% W/V HCl solution has density 1.0365 gm/cc, then its concentration 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/>1 m<br/>1 N <br/>0.5 M</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.")=(10xx3.65)/(36.5)=<a href="https://interviewquestions.tuteehub.com/tag/1m-283006" style="font-weight:bold;" target="_blank" title="Click to know more about 1M">1M</a>=1N` <br/> `m=(1000M)/(1000d-MM_(1))=(<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>)/(1036.5-36.5)=1m`</body></html> | |