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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.
| 51351. |
24 g of magnesium in the vapour state absorb 1200 kJ of energy. If IE_1 and IE_2 of magnesium are 750 and 1450 kJ mol^(-1) respectively, the final composition of mixture is |
| Answer» <html><body><p>0.69 <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/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a>^(+)` and 0.31 mol `Mg^(2+)`<br/>0.59 mol `Mg^(+)` and 0.41 mol `Mg^(2+)`<br/>0.49 mol `Mg^(+)` and 0.51 mol `Mg^(2+)`<br/>0.29 mol `Mg^(+)` and 0.71 mol `Mg^(2+)`<br/></p>Solution :Molar <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of Mg =24 g `mol^(-1)` <br/> `:.` Amount of Mg=1 mol <br/> <a href="https://interviewquestions.tuteehub.com/tag/thus-2307358" style="font-weight:bold;" target="_blank" title="Click to know more about THUS">THUS</a>, 750 kJ of energy is used up to convert 1 mol of Mg to `Mg^(+)`<br/> Energy used to convert `Mg^(+)` to `Mg^(2+)` <br/> `=1200-750 =<a href="https://interviewquestions.tuteehub.com/tag/450-317308" style="font-weight:bold;" target="_blank" title="Click to know more about 450">450</a> kJ` <br/> `:.` Amount of magnesium converted to `Mg^(2+)` <br/> `=(450 kJ)/(1450) =0.31 mol` <br/> `:. ` Amount of magnesium converted to `Mg^(+)` <br/> =1-0.31 =0.69 mol</body></html> | |
| 51352. |
23g of Na will react with ethanol to give |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/one-585732" style="font-weight:bold;" target="_blank" title="Click to know more about ONE">ONE</a> <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> of oxygen<br/>one mole of `H_(2)`<br/>`(1)/(2)` mole of `H_(2)`<br/>None of these</p>Solution :`Na+C_(2)H_(5)<a href="https://interviewquestions.tuteehub.com/tag/oh-585115" style="font-weight:bold;" target="_blank" title="Click to know more about OH">OH</a> rarr C_(2)H_(5)ONa+1//2 H_(2)` <br/> 23gof Na gives `H_(2) = 0.5 <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a>`.</body></html> | |
| 51353. |
23g of sodium will react with methyl alcohol to give |
| Answer» <html><body><p>one mole of oxygen<br/>`22.4 dm^(3)` of hydrogen as at N.T.P<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> mole of `H_(2)`<br/>11.2 L of hydrogen gas at N.T.P</p>Solution :`{:(CH_(3)OH+<a href="https://interviewquestions.tuteehub.com/tag/na-572417" style="font-weight:bold;" target="_blank" title="Click to know more about NA">NA</a>,rarr,CH_(3)<a href="https://interviewquestions.tuteehub.com/tag/ona-2889570" style="font-weight:bold;" target="_blank" title="Click to know more about ONA">ONA</a>,+(1)/(2)H_(2)),("" 23g,,,((1)/(2)mol)),(""(1 mol),,,):}` <br/> `23g` of `Na -= 1 mol Na -= (1)/(2) mol H_(2)` <br/> `= (1)/(2) xx 22.4 = 11.2 L` of `H_(2)` at N.T.P.</body></html> | |
| 51354. |
2,3-dimethylhexane contains......tertiary......secondary and...primary carbons respectively. |
| Answer» <html><body><p>2,2,4<br/>2,4,3<br/>4,3,2<br/>3,2,4</p>Answer :A</body></html> | |
| 51355. |
2,3-Dimethyl-2-butene can be prepared by heating which of the following compounds with a strong acid ? |
| Answer» <html><body><p>`(CH_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>))_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)C=<a href="https://interviewquestions.tuteehub.com/tag/ch-913588" style="font-weight:bold;" target="_blank" title="Click to know more about CH">CH</a>-CH_(2)-CH_(3)`<br/>`(CH_(3))_(2)CH-CH_(2)-CH=CH_(2)`<br/>`(CH_(3))_(2)CH-<a href="https://interviewquestions.tuteehub.com/tag/underset-3243992" style="font-weight:bold;" target="_blank" title="Click to know more about UNDERSET">UNDERSET</a>(CH_(3))underset(|)(CH)-CH=CH_(2)`<br/>`(CH_(3))_(3)C-CH=CH_(2)`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_CHE_XI_P2_C13_E03_120_S01.png" width="80%"/></body></html> | |
| 51356. |
2,3-Dimethyl-2-butene can be prepared by heated which of the following compounds with a strong acid ? |
| Answer» <html><body><p>`(CH_3)_3C-CH=CH_2`<br/>`CH_3)_2C=CH-CH_2CH_3`<br/>`(CH_3)_2CH-CH_2-CH=CH_2`<br/>`(CH_3)_2 CH-undersetunderset(CH_3)|CH-CH=CH_2`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PR_CHE_02_XI_C13_E20_071_S01.png" width="80%"/></body></html> | |
| 51357. |
2,3-Butanediol has the 2R, 3R configuration Identify the correct statement among the following. |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/2r-300472" style="font-weight:bold;" target="_blank" title="Click to know more about 2R">2R</a>, <a href="https://interviewquestions.tuteehub.com/tag/3s-311333" style="font-weight:bold;" target="_blank" title="Click to know more about 3S">3S</a> - is its <a href="https://interviewquestions.tuteehub.com/tag/enantiomers-16554" style="font-weight:bold;" target="_blank" title="Click to know more about ENANTIOMERS">ENANTIOMERS</a><br/><a href="https://interviewquestions.tuteehub.com/tag/2s-301125" style="font-weight:bold;" target="_blank" title="Click to know more about 2S">2S</a>, <a href="https://interviewquestions.tuteehub.com/tag/3r-311113" style="font-weight:bold;" target="_blank" title="Click to know more about 3R">3R</a> - is its enantiomers<br/>2S, 3S - is its enantiomers<br/>2S, 3R - Is its diastereoisomer</p>Solution :2S. 3S - is its enantiomers , 2S, 3R - is its diastereoisomer</body></html> | |
| 51358. |
.^(227)Ac has a half-line of 22 year with respect to radiioactive decay. The decay follows two parallel paths, one leading to .^(227)Th and the other leading to .^(223)Fr. The percentage yiedls of these two daughter nuclides are 2% and 98.0% respectovely. What is the rate constant in "year"^(-1), for each of the separate paths? |
| Answer» <html><body><p></p>Solution :We know, <br/> `lambda_(av) = (0.693)/(<a href="https://interviewquestions.tuteehub.com/tag/22-294057" style="font-weight:bold;" target="_blank" title="Click to know more about 22">22</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>.15xx10^(-2) "year"^(-1)` <br/> For the decay involving <a href="https://interviewquestions.tuteehub.com/tag/two-714195" style="font-weight:bold;" target="_blank" title="Click to know more about TWO">TWO</a> <a href="https://interviewquestions.tuteehub.com/tag/parallel-1146369" style="font-weight:bold;" target="_blank" title="Click to know more about PARALLEL">PARALLEL</a> paths,<br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/DPB_PHY_CHM_IX_C04_E01_050_S01.png" width="80%"/> <br/> We have `lambda_(<a href="https://interviewquestions.tuteehub.com/tag/ac-361271" style="font-weight:bold;" target="_blank" title="Click to know more about AC">AC</a>) = lambda_("Th path") + lambda_("Fr path")` <br/> `:. Lambda_(Ac) xx "Fraction of Th" = lambda_("Th path")` ....(1) <br/> `lambda_(AC) xx "Fraction of Fr" = lambda_("Fr path")` .....(2) <br/> or `lambda_(AC) xx (1 - "Fraction of Th") = lambda_("Fr path")` .....(3) <br/>Thus by eqs. (1) and (3) we get <br/> `lambda_(AC) =lambda_("Th path") + lambda_("Fr path")` <br/> Thus by eqs. (1) and (3) we get <br/> `lambda_(AC) = lambda_("Th path")+ lambda_("Fr path")` <br/>Thus, Fractionlal yield of `Th = (lambda_("Th path"))/(lambda_("Ac path"))` <br/> or `lambda_("Th path")=3.15xx10^(-2)xx(2)/(100)` <br/> `=6.30xx10^(-4)yr^(-1)` <br/> Also Fractional yeild of Fr`=(lambda_("Fr path"))/(lambda_(Ac path))` <br/> `:. lambda_(Fr) = 3.15xx10^(-2) xx (98)/(100) =3.087xx10^(-2) yr^(-1)`</body></html> | |
| 51359. |
22.71 Lit means ………. meter ? |
| Answer» <html><body><p></p>Solution :`22.71 L = 22.71 xx10^(-3)m^(3)`<br/>1 <a href="https://interviewquestions.tuteehub.com/tag/lit-537267" style="font-weight:bold;" target="_blank" title="Click to know more about LIT">LIT</a> `= 1 (<a href="https://interviewquestions.tuteehub.com/tag/dm-432223" style="font-weight:bold;" target="_blank" title="Click to know more about DM">DM</a>)^(3)`<br/>`1 dm = 0.1 m = 1xx10^(-1)m`<br/>`1 (dm)^(3)=(0.1 m)^(3)=1xx10^(-3)m^(3)`<br/>`1L=1(dm)^(3)=1xx10^(-3)m^(3)`<br/>`therefore 22.71 L =(22.71 L xx1xx10^(-3)m^(3))/(12)=22.71xx10^(-3)m^(3)` gas<br/> Molar <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> `= 22.71 L = 22.71xx10^(-3)m^(3)`</body></html> | |
| 51360. |
2.26 g of impure ammonium chloride were boiled with 100 mL of NaOH solution till no more ammonia was given off. The excess of NaOH solution left over required 30 mL 2N H_(2)SO_(4) for neutralisation. Calculate the percentage purity of the salt. (H=1, N=14, O=16, Na=23, S=32, Cl=35.5) |
| 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> | |
| 51361. |
22.44 kJ energy is required to convert 8 gm of gaseous atom of metal M to M_((g))^+ if I.E._1 of metal M = 374 kJ/mole. Select correct for above metal M |
| Answer» <html><body><p>0.6 <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> gaseous ion (`M^+`) are formed<br/>Same <a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a> can convert all `M_((<a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>))^+` to `M_((g))^(<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/>Atomic <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of metal = 133.33 <br/>`3.613xx10^22` atoms of M are converted to `M_((g))^(+)` </p>Answer :C::D</body></html> | |
| 51362. |
22.4 litres of water vapour at NTP, when condensed to water, occupies an approximate volume of: |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/18-278913" style="font-weight:bold;" target="_blank" title="Click to know more about 18">18</a> litres<br/>1 litre<br/>1 mL<br/>18 mL</p>Answer :D</body></html> | |
| 51363. |
Fr has half life period of |
| Answer» <html><body><p> <a href="https://interviewquestions.tuteehub.com/tag/21-293276" style="font-weight:bold;" target="_blank" title="Click to know more about 21">21</a> secs<br/>21minutes <br/> 21 hours<br/>21 days</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> | |
| 51364. |
2.20 g of an ammonium salt were boiled with 75 mL of NaOH till the emission of ammonia gas ceased. The excess of unused NaOH solution required 70 mL of N//2 sulphuric acid for neutralisation. Calculate the percentage of ammonia in the salt. |
| 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> | |
| 51365. |
2.2 moles of phosphorus penta chloride were taken in a closed vessel and dissociated into phosphorus trichloride and chlorine . At equilibrium , the total number of moles of the reactants and the products was 2.53 . The degree of dissociation is , |
| Answer» <html><body><p>0.33<br/>0.165<br/>0.15<br/>0.3</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> | |
| 51366. |
Both 22 grams of CO_(2) and 8 grams of methane contain the following carbon atoms |
| Answer» <html><body><p>`3 xx <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(23)`<br/>`12 xx 10^(23)`<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> xx 10^(23)`<br/>`1.5 xx 10^(23)`</p>Answer :C</body></html> | |
| 51367. |
22 g of CO_(2) contains ……….. molecules of CO_(2). |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<a href="https://interviewquestions.tuteehub.com/tag/44-316683" style="font-weight:bold;" target="_blank" title="Click to know more about 44">44</a> g of `CO_(2)` contains `6.023 xx <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(<a href="https://interviewquestions.tuteehub.com/tag/23-294845" style="font-weight:bold;" target="_blank" title="Click to know more about 23">23</a>)` molecules. <br/>`:.`22 g of `CO_(2)` will contain = `(6.023xx10^(23))/44xx22 = 3.0115 xx 10^(23)`</body></html> | |
| 51368. |
22 g of a gas occupies 11.2 litres of volume at STP. The gas is |
| Answer» <html><body><p>CH4<br/>CO2<br/>NO<br/>CO</p>Solution :22 g of 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> <a href="https://interviewquestions.tuteehub.com/tag/occupies-1127820" style="font-weight:bold;" target="_blank" title="Click to know more about OCCUPIES">OCCUPIES</a> 11.<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/litres-1075876" style="font-weight:bold;" target="_blank" title="Click to know more about LITRES">LITRES</a>.<br/>11.2 litres is occupied by 22 g of a gas.`:.` Molar volume 22.4 <a href="https://interviewquestions.tuteehub.com/tag/litre-1075864" style="font-weight:bold;" target="_blank" title="Click to know more about LITRE">LITRE</a> will be occupied by`22/11.2xx22.4 = 44` g<br/>`:.` The gas is C`O_(2)` .</body></html> | |
| 51369. |
2.18g of an organic compound containing sulphur produces 1.02 g of BaSO_(4). The percentage of sulphur in the compound is |
| Answer» <html><body><p>0.0726<br/>0.0898<br/>0.1<br/>0.0642</p>Solution :% of `S=(<a href="https://interviewquestions.tuteehub.com/tag/32-307188" style="font-weight:bold;" target="_blank" title="Click to know more about 32">32</a>)/(<a href="https://interviewquestions.tuteehub.com/tag/233-1826012" style="font-weight:bold;" target="_blank" title="Click to know more about 233">233</a>)<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>(1.02)/(2.18)xx100=6.42%`</body></html> | |
| 51370. |
2.16 grams of Cu on reaction with HNO_(3) followed by ignition of the nitrate gave 2.7 gm of copper oxide. In another experiment 1.15 gm of copper oxide upon reduction with hydrogen gave 0.92 gm of copper. This data illustrate the law of |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/multiple-1105557" style="font-weight:bold;" target="_blank" title="Click to know more about MULTIPLE">MULTIPLE</a> Proportions<br/>Definite Proportions<br/>Reciprocal proportions<br/>Conservation of mass</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> | |
| 51371. |
21 Mol of FeSO_(2) (atomic weight of Fe is 55.84 g mol^(-1)) is oxidized to Fe_(2)(SO_(4))^(3) calculate theequivalent weight of ferrous ion |
| Answer» <html><body><p>55.84<br/>27.92<br/>18.61<br/>111.68</p>Solution :Since <a href="https://interviewquestions.tuteehub.com/tag/oxidation-588780" style="font-weight:bold;" target="_blank" title="Click to know more about OXIDATION">OXIDATION</a> of `Fe^(2+)rarrFe^(3+)` is a one electron <a href="https://interviewquestions.tuteehub.com/tag/change-913808" style="font-weight:bold;" target="_blank" title="Click to know more about CHANGE">CHANGE</a> therefore equivalent wt of `Fe^(2+)` = <a href="https://interviewquestions.tuteehub.com/tag/atomic-2477" style="font-weight:bold;" target="_blank" title="Click to know more about ATOMIC">ATOMIC</a> wt of `Fe^(2+)` =55.84</body></html> | |
| 51372. |
2.1 g of metal carbonate on thermal decomposition gave 1 g of metal oxide as residue. Determine theequivalent weight of metal. |
| Answer» <html><body><p></p>Solution :Let the <a href="https://interviewquestions.tuteehub.com/tag/equivalent-446407" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENT">EQUIVALENT</a> weight of metal=x <br/> Equivalent weight of metal oxide =x+8 <br/> Equivalent weight of metal <a href="https://interviewquestions.tuteehub.com/tag/carbonate-909380" style="font-weight:bold;" target="_blank" title="Click to know more about CARBONATE">CARBONATE</a> =x+30 <br/> When weight <a href="https://interviewquestions.tuteehub.com/tag/ratio-13379" style="font-weight:bold;" target="_blank" title="Click to know more about RATIO">RATIO</a> is related to equivalent weight ratio for two substances `(W_(i))/(W_(2))=(E_(1))/(E_(2)), (2.1)/(1)=(x+30)/(x+8)` <br/><a href="https://interviewquestions.tuteehub.com/tag/solving-1217196" style="font-weight:bold;" target="_blank" title="Click to know more about SOLVING">SOLVING</a> , x = 12<br/> Equivalent weight of metal = 12</body></html> | |
| 51373. |
20ml of hydrogen peroxide solution is added to excess of acidified KI solution. The iodine so liberated requires 20ml of 0.1M hypo solution. Calculate the molarity and percentage strength of H_2O_2solution. |
| 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> :0.05M, 0.17 (w/v)%</body></html> | |
| 51374. |
20mL of each 0.1M NH,OH and IM NH CI aqueous solutions form a buffer of pH 8.3. Calculate the equilibrium constant for NH_4 OH hArr NH_(4)^(+)+ OH^(-). |
| 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/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> <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/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)`</body></html> | |
| 51375. |
20mL fo 0.2M MbnSO_(4) are completely oxidiz3d by 16 mL of KMnO_(4) of unknown normaliity each fromingMn^(4+) oxidation state. Find out the normality and molarity of KMnO_(4) solution. |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :`0.5N, 0.167M`</body></html> | |
| 51376. |
20g of sample containing Ba(OH)_(2) is dissolved in 10 ml of 0.5 MHCl solution. The excess of HCl was then titrated against 0.2 M NaOH. The volume of NaOH used in the titration was 10 ml. Calculate the percentage of Ba(OH)_(2) in the sample. (Mol. wt. of Ba(OH_(2))=171) |
| Answer» <html><body><p></p>Solution :Calculation of volume of HCl used in titration between <a href="https://interviewquestions.tuteehub.com/tag/naoh-572531" style="font-weight:bold;" target="_blank" title="Click to know more about NAOH">NAOH</a> and HCl. <br/> `V_(NaOH) =10 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a>, M_(NaOH) = 0.2 M, M_(HC) = 0.5 M, V_(HCl)`=? <br/> `M_(NaOH) xx V_(NaOH) = M_(HCl) xx V_(HCl), V_(HCl) = (10 xx 0.2)/0.5 = 4 ml` <br/> This is the volume of HCl left unused when excess of HCl is added to `Ba(<a href="https://interviewquestions.tuteehub.com/tag/oh-585115" style="font-weight:bold;" target="_blank" title="Click to know more about OH">OH</a>)_(2)` solution. <br/> Total volume of HCl added = 10 ml <br/> Volume of HCl used to react with `Ba(OH)_(2) = 10-4 = 6` ml <br/> `2HCl + Ba(OH)_(2) to BaCl_(2) + 2H_(2)O` <br/> M = No. of moles x `1000/("volume of solution") rArr 0.5 = "moles" xx 1000/6`. <br/> `(0.5 xx 6)/1000` = moles of HCl Moles of HCl used = 0.003 moles. <br/> Observing the molar ratio of HCl and `Ba(OH)_(2)`. <br/> Moles of `Ba(OH)_(2)` <a href="https://interviewquestions.tuteehub.com/tag/reacted-7708680" style="font-weight:bold;" target="_blank" title="Click to know more about REACTED">REACTED</a> `=1/2` x moles of HCl reacted `=1/2 xx 0.003 = 0.0015` moles. <br/> Weight of `Ba(OH)_(2)` reacted = no. of moles x 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>. =` 0.0015 xx 171 = 0.2565` g <br/> Percentage of `Ba(OH)_(2)` in the sample `=(wt. of Ba(OH)_(2) "reacted")/("Total weight of sample") xx 100` = 1.28 %</body></html> | |
| 51377. |
20CC of hydro carbon were exploded with excess of oxygen. After explosion and cooling a contraction of 20cc was noted on addition of KOH another contraction of 40CC was noted. The molecular formula of hydrocarbon is |
| Answer» <html><body><p>`C_(2)H_(6)`<br/>`C_(2)H_(4)`<br/>`C_(2)H_(2)` <br/>`CH_(4)`</p>Solution :Let formula of hydrocarbon = <a href="https://interviewquestions.tuteehub.com/tag/cxhy-428877" style="font-weight:bold;" target="_blank" title="Click to know more about CXHY">CXHY</a> <br/> `CxHY+(<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>+(y)/(4))O_(2)rarrxCO_(2)+(y)/(2)H_(2)O` <br/> `{:("1 mol",(x+(y)/(4))"mol","x mol"),("1 vol",(x+y//4)"vol","x vol"),(20C <a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>,20(x+(y)/(4))C C,20xC C):}` <br/> `CO_(2)` <a href="https://interviewquestions.tuteehub.com/tag/produced-592947" style="font-weight:bold;" target="_blank" title="Click to know more about PRODUCED">PRODUCED</a> = contraction on addition of KOH <br/> = 40 CC implies 20x = 40 implies x = 2 <br/> `[20C C+20(x+(y)/(4))C C]-20xC C=30C C` <br/> `y=2` <br/> Formula of hydrocarbon =`C_(2)H_(2)`</body></html> | |
| 51378. |
2.0g of a metallic carbonate on decomposition gave 1.5g of metallic oxide. The equivalent mass of metal is |
| Answer» <html><body><p>58<br/>29<br/>5.8<br/>2.9</p>Solution :`("weight of <a href="https://interviewquestions.tuteehub.com/tag/metal-1094457" style="font-weight:bold;" target="_blank" title="Click to know more about METAL">METAL</a> carbonate")/("weight of <a href="https://interviewquestions.tuteehub.com/tag/metallic-547634" style="font-weight:bold;" target="_blank" title="Click to know more about METALLIC">METALLIC</a> <a href="https://interviewquestions.tuteehub.com/tag/oxide-1144484" style="font-weight:bold;" target="_blank" title="Click to know more about OXIDE">OXIDE</a>") = ("<a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a>. mass of metal + <a href="https://interviewquestions.tuteehub.com/tag/eql-446397" style="font-weight:bold;" target="_blank" title="Click to know more about EQL">EQL</a>. Mass of" CO_(3)^(2-))/("eq mass of metal + eq mass of" O^(2-))`</body></html> | |
| 51379. |
200cc of ozone diffused in 15 min through a porous membrane. How much time does 150cc of oxygen take to difffuse, under similar conditions ? |
| Answer» <html><body><p></p>Solution :`(V_(O_3) xx t_(O_2))/( V_(O_2) xx t_(o_3)) = sqrt((M_(o_2))/(M_(O_3)))= >( <a href="https://interviewquestions.tuteehub.com/tag/200-288914" style="font-weight:bold;" target="_blank" title="Click to know more about 200">200</a> xx t_(o_2))/( <a href="https://interviewquestions.tuteehub.com/tag/150-275254" style="font-weight:bold;" target="_blank" title="Click to know more about 150">150</a> xx 15 ) = sqrt((<a href="https://interviewquestions.tuteehub.com/tag/32-307188" style="font-weight:bold;" target="_blank" title="Click to know more about 32">32</a>)/(48))` <br/> `t_(O_2) = ( 15 xx 150 )/(200) =sqrt((32)/(48))=9.135 ` <a href="https://interviewquestions.tuteehub.com/tag/min-548008" style="font-weight:bold;" target="_blank" title="Click to know more about MIN">MIN</a></body></html> | |
| 51380. |
200 mL of water is added to 500 mL of 0.2 M solution. What is the molarity of the diluted solution ? |
| Answer» <html><body><p>0.5010 M<br/>0.2897 M<br/>0.7093 M<br/>0.1428 M</p>Solution :`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)-=M_(<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)` <br/> `0.2Mxx500-=M_(2)xx700` <br/> `M_(2)=((0.2M)xx500)/(<a href="https://interviewquestions.tuteehub.com/tag/700-333923" style="font-weight:bold;" target="_blank" title="Click to know more about 700">700</a>)=0.1428M`</body></html> | |
| 51381. |
200ml of pure oxygen is subjected to electric discharge, 15% of oxygen is converted into ozone. The volume of ozonized oxygen 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/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> <br/>30 ml <br/><a href="https://interviewquestions.tuteehub.com/tag/190-280910" style="font-weight:bold;" target="_blank" title="Click to know more about 190">190</a> ml <br/><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 </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> | |
| 51382. |
200 ml of O_2 gas maintained at 700mm pressure and 250ml of N_2 gas maintained at 720mm pressure are put together in one litre flask. If the temperature is kept constant, the final pressure of the mixture in mm is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/450-317308" style="font-weight:bold;" target="_blank" title="Click to know more about 450">450</a> <br/>320 <br/><a href="https://interviewquestions.tuteehub.com/tag/632-330591" style="font-weight:bold;" target="_blank" title="Click to know more about 632">632</a> <br/>316 </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> | |
| 51383. |
200 mL of a solution of mixture of NaOH and Na_(2)CO_(3) was first titrated with phenolphthalein and N//10HCl. 17.5 mL of HCl was required for the end point. After this methyl orange was added and 2.5 mL of same HCl was required for next end point. Find out amounts of NaOOH and Na_(2)CO_(3) in mixture. |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :`NaOH=0.06g <a href="https://interviewquestions.tuteehub.com/tag/per-590802" style="font-weight:bold;" target="_blank" title="Click to know more about PER">PER</a> <a href="https://interviewquestions.tuteehub.com/tag/200-288914" style="font-weight:bold;" target="_blank" title="Click to know more about 200">200</a> mL`, <br/> `Na_(2)CO_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)=0.0265 g per 200 mL`;</body></html> | |
| 51384. |
200 ml of an aqueous solution of a protein contains 1.26 g of protein. At 300 K, the osmotic pressure of this solution is found to be2.52xx 10^(-3)bar.The molar mass of protein will be (R=0.083 Lbar mol ^(-1) K^(-1)) |
| Answer» <html><body><p>`62.22` <a href="https://interviewquestions.tuteehub.com/tag/kg-1063886" style="font-weight:bold;" target="_blank" title="Click to know more about KG">KG</a> `mol^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)`<br/>`12444 g mol ^(-1)`<br/>`300 g mol ^(-1)`<br/>none of these </p>Solution :`pi = CRT` <br/> `pi = (<a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a>)/(M xxV) xx RT` <br/> ` therefore M = (<a href="https://interviewquestions.tuteehub.com/tag/wrt-1462282" style="font-weight:bold;" target="_blank" title="Click to know more about WRT">WRT</a>)/(<a href="https://interviewquestions.tuteehub.com/tag/piv-591561" style="font-weight:bold;" target="_blank" title="Click to know more about PIV">PIV</a>) = (1.26 xx 0.083 xx 300)/(2.52 xx 10 ^(-3) xx 0.2)= 62.22 kg mol ^(-1)`</body></html> | |
| 51385. |
200 mL of 3 N HCl were mixed with 200 mL of 6 N H_(2)SO_(4) solution. The final normality of H_(2)SO_(4) in the resultant solution will be: |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/9-340408" style="font-weight:bold;" target="_blank" title="Click to know more about 9">9</a> N<br/>3 N<br/>6 N<br/>2 N</p>Solution :N//A</body></html> | |
| 51386. |
20.0 L of 0.2 M weak acid (pK_(a)=5.0) is titrated against 0.2 M strong base. What is the pH at the equivalence point ? |
| Answer» <html><body><p>`5.0`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>.0`<br/>`9.0`<br/>`11.0`</p>Solution :At the equivalence <a href="https://interviewquestions.tuteehub.com/tag/point-1157106" style="font-weight:bold;" target="_blank" title="Click to know more about POINT">POINT</a>, solution will contain salt of weak <a href="https://interviewquestions.tuteehub.com/tag/acid-847491" style="font-weight:bold;" target="_blank" title="Click to know more about ACID">ACID</a> and strong base. Its solution on hydrolysis will be basic with `pH gt 7`. Value of 9 may be <a href="https://interviewquestions.tuteehub.com/tag/chosen-2515513" style="font-weight:bold;" target="_blank" title="Click to know more about CHOSEN">CHOSEN</a> as the base is strong.</body></html> | |
| 51387. |
200 joules of heatwas supplied to a system at constant volume. It resulted in the increase in temperature of the system from 298 to 323 K. What is the change in internal energy of the system ? |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/400-315233" style="font-weight:bold;" target="_blank" title="Click to know more about 400">400</a> J<br/>200 J<br/>50 J<br/>150 J</p>Solution :At constant volume, `DeltaV=0` <br/> `w=-PDeltaV=0""DeltaU=q=200J`</body></html> | |
| 51388. |
20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and8.0 g magnesium oxide. What will be the percentage of purity of magnesium carbonate in the sample? |
| Answer» <html><body><p>60<br/>84<br/>75<br/>96</p>Solution :`MgCO_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>) overset(Delta)(to) MgO + CO_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)<a href="https://interviewquestions.tuteehub.com/tag/uarr-3241817" style="font-weight:bold;" target="_blank" title="Click to know more about UARR">UARR</a>` <br/>84 g of `MgCO_(3)`gives 40 g of MgO. (100% purity) <br/>20g of `MgCO_(3)` will give `=(40xx20)/84=9.52g` <br/>9.52g of MgO is given by 100% pure `MgCO_(3)`. <br/>8.0 g of MgO will be given by=`(100xx8)/9.52=84.03%`</body></html> | |
| 51389. |
20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample ? (At. Wt of Mg = 24) |
| Answer» <html><body><p>96<br/><a href="https://interviewquestions.tuteehub.com/tag/84-339262" style="font-weight:bold;" target="_blank" title="Click to know more about 84">84</a><br/>60<br/>75</p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/underset-3243992" style="font-weight:bold;" target="_blank" title="Click to know more about UNDERSET">UNDERSET</a>(84g)(MgCO_(3))overset(<a href="https://interviewquestions.tuteehub.com/tag/heat-21102" style="font-weight:bold;" target="_blank" title="Click to know more about HEAT">HEAT</a>)(rarr)underset(40g)(MgO)+CO_(2)` <br/> 84 g of `MgCO_(3)` yield upon heating MgO = <a href="https://interviewquestions.tuteehub.com/tag/40-314386" style="font-weight:bold;" target="_blank" title="Click to know more about 40">40</a> g <br/> 20 g of `MgCO_(3)` yield upon heating `MgO=((40g)xx(20g))/((84g))` <br/> `=9.52g` <br/> Weight of MgO <a href="https://interviewquestions.tuteehub.com/tag/actually-1968034" style="font-weight:bold;" target="_blank" title="Click to know more about ACTUALLY">ACTUALLY</a> formed = 8.0g <br/> % purity of `MgCO_(3)=((8.0g))/((9.52g))xx100=84%`.</body></html> | |
| 51390. |
20 volume. of H_(2)O_(2) is equal to |
| 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>% H_(2)O_(2)` by <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> <br/>`6% H_(2)O_(2)` by mass <br/>`1.764N`<br/>`3.571N`</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> :V.S. = `Nxx5.6impliesN=(20)/(5.6)=3.5171N`</body></html> | |
| 51391. |
20 volume H_2O_2 Solution has a strength of about |
| Answer» <html><body><p>`30%`<br/>`6%`<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>%`<br/>`10%`</p>Solution :22.4 <a href="https://interviewquestions.tuteehub.com/tag/litre-1075864" style="font-weight:bold;" target="_blank" title="Click to know more about LITRE">LITRE</a> `O_2` at N.T.P. <a href="https://interviewquestions.tuteehub.com/tag/obtained-7273275" style="font-weight:bold;" target="_blank" title="Click to know more about OBTAINED">OBTAINED</a> by 68 gm of `H_2O_2` <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/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> litre `O_2` at N.T.P. obtained by `68/22.4` gm of `H_2O_2`<br/> `therefore` 20 litre `O_2` at N.T.P. obtained by `68/22.4xx20` gm of `H_2O_2` =60.71 gm of `H_2O_2` <br/> `therefore` 1000 ml `O_2` at N.T.P. obtained by =60.71 gm of `H_2O_2` <br/> `therefore` 100 ml `O_2` at N.T.P. obtained by `=60/1000xx100`=6.71%</body></html> | |
| 51392. |
20 % of N_(2)O_(4) " molecules are disociated in a sample of a gas at " 27^(@)C and 760 " torr . Calculate the density of the equilibrium mixture ". |
| Answer» <html><body><p></p>Solution :` {:(,N_(2)O_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)(g),hArr,2NO_(2)(g),),("Intial",1 " mole",,,),("At.eqm.",1-0*2 = 0*<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>" mole",,0*4 "mole,",Total = 1*2 " moles" ):}` <br/> If V is the volume of the vapour <a href="https://interviewquestions.tuteehub.com/tag/per-590802" style="font-weight:bold;" target="_blank" title="Click to know more about PER">PER</a> mole, volume of vapour before dissociation = V<br/> ` " Hence density " (D) propto1/V ` <br/> But density after dissociation <br/>`D= (" Mol.wt.of " N_(2)O_(4))/2 = 92/2=46"" ("Theoretical density")` <br/> Volume after dissociation = `1*2` V<br/>` :. "Density (d)" propto 1/(1*2 V) `<br/>` :.D/d = 1/V xx 1*2V =1*<a href="https://interviewquestions.tuteehub.com/tag/2or-1838480" style="font-weight:bold;" target="_blank" title="Click to know more about 2OR">2OR</a> d=D/(1*2)= 46/(1*2)= 38*3` <br/> Alternatively, <a href="https://interviewquestions.tuteehub.com/tag/use-1441041" style="font-weight:bold;" target="_blank" title="Click to know more about USE">USE</a> theformula directly, <br/> ` alpha = (D-d)/d`</body></html> | |
| 51393. |
20 ml of the solution containg Na_(2)CO_(3) and NaHCO_(3) is titrated with 0.1MHCI using Phenolphthalein indicator the end point was 10ml. 20ml of the same solution is titrated with 0.1M HCI, the end point was 25ml with Methylorange indicaor from the begining. NaHCO_(3)+HCl rarr NaCl +H_(2)O +CO_(2) Na_(2)CO_(3)+2HCl rarr 2NaCl +CO_(2) +H_(2)O What amount of NaOHis required to convertNaHCO_(3) to Na_(2)CO_(3) in 1 litre of solution |
| Answer» <html><body><p>`2g`<br/>`20g`<br/>`1g`<br/>`0.5g`</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> :`NaHCO_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)+NaOH rarr Na_(2)CO_(3) +H_(2)O` <br/> 84g of `NaHCO_(3)` requires 40g of `NaOH` <br/> `2.1g` of `NaHCO_(3)` requires? <br/> `= (40)/(<a href="https://interviewquestions.tuteehub.com/tag/84-339262" style="font-weight:bold;" target="_blank" title="Click to know more about 84">84</a>) xx 2.1 = 1g`</body></html> | |
| 51394. |
20 mL of x M HCl neutralises 5 mL of 0.2 M Na_(2)CO_(3) solution to phenolphthalein end-point. The value of x is |
| Answer» <html><body><p>`0.167M`<br/>`0.133M`<br/>`0.150M`<br/>`0.05M`</p>Solution :With phenolphthalein, <br/> <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a>. `HCl=(1)/(2)` eq. `Na_(2)CO_(<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/20xx-292987" style="font-weight:bold;" target="_blank" title="Click to know more about 20XX">20XX</a> <a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a> xx1=(1)/(2)xx5xx0.2xx2impliesx=0.05`</body></html> | |
| 51395. |
20 ml of the solution containg Na_(2)CO_(3) and NaHCO_(3) is titrated with 0.1MHCI using Phenolphthalein indicator the end point was 10ml. 20ml of the same solution is titrated with 0.1M HCI, the end point was 25ml with Methylorange indicaor from the begining. NaHCO_(3)+HCl rarr NaCl +H_(2)O +CO_(2) Na_(2)CO_(3)+2HCl rarr 2NaCl +CO_(2) +H_(2)O What is amount of NaHCO_(3) present in 1 litre of solution |
| Answer» <html><body><p>`2.1g`<br/>`1.05g`<br/>`8.1g`<br/>`0.844g`</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> :Using <a href="https://interviewquestions.tuteehub.com/tag/methyl-1095247" style="font-weight:bold;" target="_blank" title="Click to know more about METHYL">METHYL</a> orange, <br/> `x +y = 25 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 0.1 = 2.5, y = 2.5 - 2 = 0.5` <br/> Wt of `NaHCO_(3)` in 20ml `=(0.5)/(1000) xx <a href="https://interviewquestions.tuteehub.com/tag/84-339262" style="font-weight:bold;" target="_blank" title="Click to know more about 84">84</a> = 0.042g` <br/> wt of `NaHCO_(3)` in 1 lit `=(0.042)/(20) xx 1000 = 2.1g`</body></html> | |
| 51396. |
20 ml of nitric oxide combines with 10 ml of oxygen at STP to give NO_(2). The final volume will be |
| Answer» <html><body><p>30ml<br/>20ml<br/>10m<br/>40ml</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> | |
| 51397. |
20 mL of HCl having a certain normality neutralises exactly 1.0 g CaCO_(3) . The normality of acid is |
| Answer» <html><body><p>0.5 <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a><br/>0.12 N<br/>0.01 N<br/>1.0 N</p>Solution :`underset((100 g))(CaCO_(3)) + underset((73 g))(2HCl) to CaCl_(2) + H_(2)O + CO_(2)` is<br/> 100 g of `CaCO_(3)` shall be <a href="https://interviewquestions.tuteehub.com/tag/neutralised-2188316" style="font-weight:bold;" target="_blank" title="Click to know more about NEUTRALISED">NEUTRALISED</a> by<br/> `(73)/(100)=0.73 g ` HCl <br/> `therefore 20` mL of HCl has HCl in equivalent `=(0.73)/(36.5)` <br/> Hence Normality `=(0.73)/(36.5)<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>(1000)/(20)=1.0 N`</body></html> | |
| 51398. |
20 mL of hydrogen measured at 15^(@)C are heated to 35^(@)C. What is the new volume at the same pressure ? |
| Answer» <html><body><p></p>Solution :`{:("Given <a href="https://interviewquestions.tuteehub.com/tag/conditions-424384" style="font-weight:bold;" target="_blank" title="Click to know more about CONDITIONS">CONDITIONS</a>","Final Conditions"),(V_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)=<a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a> mL,V_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)=?mL),(T_(1)=15+273=288 K ,T_(2)=35+273=308 K):}` <br/> By applying Charles' <a href="https://interviewquestions.tuteehub.com/tag/law-184" style="font-weight:bold;" target="_blank" title="Click to know more about LAW">LAW</a>,`(V_(2))/(308)=(20)/(288)" or " V_(2)=(20)/(288)xx308=21.38` <br/> Volume of hydrogen gas at `35^(@)C=21.38 mL`</body></html> | |
| 51399. |
20 ml of H_(2)O_(2) after acidification with dil H_(2)SO_(4), required 30 ml of N/2 KMnO_(4) for complete oxidation. Calculate the % of H_(2)O_(2) in gr/lit. |
| Answer» <html><body><p>10.75 g/lit<br/>11.75 g/lit<br/>12.75 g/lit<br/>13.75 g/lit</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/meq-1093952" style="font-weight:bold;" target="_blank" title="Click to know more about MEQ">MEQ</a>. of `O_(2)="Meq. of "H_(2)O_(2)="Meq. of "KMnO_(4)` <br/> `(W times 1000)/(34//2)=1/2 times 30` <br/> `W=1/2 times (300 times <a href="https://interviewquestions.tuteehub.com/tag/17-278001" style="font-weight:bold;" target="_blank" title="Click to know more about 17">17</a>)/1000=0.255 gr//20ml` <br/> `N_(H_(2)O_(2))=0.255/17 times 1000/20=0.75` <br/> <a href="https://interviewquestions.tuteehub.com/tag/strength-1229153" style="font-weight:bold;" target="_blank" title="Click to know more about STRENGTH">STRENGTH</a> `=0.75 times 17=12.75` gr/lit</body></html> | |
| 51400. |
20 mL of a solution of H_2SO_4 neutralises 21.2 mL of 30% solution (w/v) of Na_2CO_3. How much water should be added to each 100 mL of the solution to bring down its strength to decinormal ? |
| Answer» <html><body><p><br/></p>Solution :The corresponding equation is: <br/> `underset(105.99 g)(Na_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)CO_(3)) + underset(98.076 g)(H_(2)SO_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)) to Na_(2)SO_(4) + H_(2)O + CO_(2)` <br/> The amount of `Na_2CO_3` present in 21.2 mL solution.<br/> `=3/100 xx 21.2 = 0.63 g` <br/> `therefore 105.99 g` of `Na_(2)CO_(3)` react with `H_(2)SO_(4) = 98.076 g` <br/> `therefore 0.636 g` of `Na_(2)CO_(3)` will react with `H_(2)SO_(4)` <br/> `=98.076/105.99 xx 0.636 = 0.588 g` <br/> This much `H_(2)SO_(4)` is present in 20 mL. <br/> `therefore <a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a>=(NEV)/1000` <br/> `therefore 0.588 = (N xx 49.038 xx 20)/1000` <br/> (Eq. wt. of `H_(2)SO_(4) = (98.076)/2 = 49.038`) <br/> `therefore N=0.599 = 0.6` <br/> Therefore, the normality of the given `H_2SO_4` solution is 0.6 N. Suppose, v mL of water are <a href="https://interviewquestions.tuteehub.com/tag/required-1185621" style="font-weight:bold;" target="_blank" title="Click to know more about REQUIRED">REQUIRED</a> to be added to 100 mL of it to make it decinormal `N/10` <br/> `therefore 0.6 xx 100 = 1/10 xx (100 + v)` which <a href="https://interviewquestions.tuteehub.com/tag/gives-1007647" style="font-weight:bold;" target="_blank" title="Click to know more about GIVES">GIVES</a> v = 500 mL<br/> Hence, 500 mL of water should be added to each 100 mL of the solution of `H_2SO_4` to bring down its strength to decinormal.</body></html> | |