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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.
| 51651. |
100ml of water at 20^@C and 100ml of water at 40^@C are mixed in calorimeter until constant temperature reached. Now temperature of the mixture is 28^@C. Water equivalent of calorimeter is |
| Answer» <html><body><p>50J<br/>104.5J<br/>` -24.2J`<br/>209J</p>Answer :D</body></html> | |
| 51652. |
100ml of O_2 diffuses in 100 minutes, under similar conditions one litre of H_2 diffuses in |
| Answer» <html><body><p>100min <br/>250min <br/>500min <br/>750min </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> | |
| 51653. |
100mL of H_(2)O_(2) solution having volume strength 11.35 V is mixed with 500mL of 0.5M KI solution to liberate I_(2) gas such that equilibria gets established. All the I_(2) gas liberated is dissolved to form 500mL solution. 200ml of the solution required 50mL of 2/(3)M hypo solution. Calculate volume strength fo remaining H_(2)O_(2) mixture. Round off your answer. |
| 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/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a></body></html> | |
| 51654. |
100mL of each hydrogen and oxygen was mixed and fired. What will be the final volume of the mixture when measured at the original conditions? |
| Answer» <html><body><p></p>Solution :The storichiometric equation is given as <br/> `2H_(2)(g)+O_(2)(g)to2H_(2)O(1)` <br/> 2vol1vol 0 vol Gay-Lussac.s coefficients <br/><a href="https://interviewquestions.tuteehub.com/tag/100ml-265997" style="font-weight:bold;" target="_blank" title="Click to know more about 100ML">100ML</a> 100mL 0mLat <a href="https://interviewquestions.tuteehub.com/tag/start-1224710" style="font-weight:bold;" target="_blank" title="Click to know more about START">START</a> <br/> 0mL <a href="https://interviewquestions.tuteehub.com/tag/50ml-324226" style="font-weight:bold;" target="_blank" title="Click to know more about 50ML">50ML</a> 0mL after the reaction <br/> The composition of the final mixture is that it has only unreacted <a href="https://interviewquestions.tuteehub.com/tag/oxygen-1144542" style="font-weight:bold;" target="_blank" title="Click to know more about OXYGEN">OXYGEN</a> volume =50ml</body></html> | |
| 51655. |
100mL of each 0.2M solutions of H_(2)SO_(4) and HCl are mixed. Calculate the normality of the mixture. |
| Answer» <html><body><p></p>Solution :Milliequivalents of `H_(2)SO_(4)=100xx0.2xx2=40` <br/> Milliequivalents of <a href="https://interviewquestions.tuteehub.com/tag/hcl-479502" style="font-weight:bold;" target="_blank" title="Click to know more about HCL">HCL</a> `=100xx0.2xx1=20` <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> number of milliequivalents of acid in the <a href="https://interviewquestions.tuteehub.com/tag/mixture-1098735" style="font-weight:bold;" target="_blank" title="Click to know more about MIXTURE">MIXTURE</a> =40+2=60 <br/> Total volume of mixture =100+100=200ml <br/> Normality of the mixture `=("M <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a>")/("Volume")=(60)/(200)=0.3 eq L^(-1)`</body></html> | |
| 51656. |
100mL of a gaseou mixture has 20mL ethylence and the remaining oxygen. The mixture Is fired and cooled to room temperature. Determine the final composition of gaseou 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> :<a href="https://interviewquestions.tuteehub.com/tag/20ml-293069" style="font-weight:bold;" target="_blank" title="Click to know more about 20ML">20ML</a> `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>)` + <a href="https://interviewquestions.tuteehub.com/tag/40ml-1875303" style="font-weight:bold;" target="_blank" title="Click to know more about 40ML">40ML</a> `CO_(2)`</body></html> | |
| 51657. |
1.00litre sample of a mixture of CH_(4(g)) and O_(2(g)) measured at 25^(@)C and 740 t or r was allowed to react at constant pressure in a calorimeter which together with its contents had a heat capacity of 1260cal//K. The complete combustion of the methane toCO_(2) and H_(2)O caused a temperature rise in the calorimeter of 0.667K. What was the mole per cent of CH_(4) in the original mixture ? (DeltaH_(comb.)^(@)(CH_(4))=-215kcal mol^(-1)) |
| 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> :`9.75%;`</body></html> | |
| 51658. |
100kg of a water sample contains 6g of magnesium sulphate, 10.2g of calcium sulphate and 16.2 g of calcium bicarbonate only. Calculate permanent hardness, temporary hardness and total hardness. |
| Answer» <html><body><p></p>Solution :`100 Kg (10^5g)`of given water <a href="https://interviewquestions.tuteehub.com/tag/sample-1194587" style="font-weight:bold;" target="_blank" title="Click to know more about SAMPLE">SAMPLE</a> contains 6 g of `MgSO_4, 10.2 g CaSO_4` and `16.2g <a href="https://interviewquestions.tuteehub.com/tag/ca-375" style="font-weight:bold;" target="_blank" title="Click to know more about CA">CA</a>(HCO_3)_2` .<br/> Weight of `MgSO_4`in `10^6`g water `= (6)/(10^5) xx 10^6= 60g`<br/>120 g of `MgSO_4` = 100 g of `CaCO_3` <br/>60 g of `MgSO_4 =60/120 xx 100 = 50 g` of `CaCO_3`<br/>Degree of <a href="https://interviewquestions.tuteehub.com/tag/hardness-14971" style="font-weight:bold;" target="_blank" title="Click to know more about HARDNESS">HARDNESS</a> due to `MgSO_4` = 50ppm<br/>weight of `CaSO_4` in `10^6g` of water = `10.2 xx 10^6//10^5 = 102g`<br/>136 g of `CaSO_4` = 100 g of `CaCO_3` <br/>102 g of `CaSO_4 = 102 xx 100//136 = <a href="https://interviewquestions.tuteehub.com/tag/75-334971" style="font-weight:bold;" target="_blank" title="Click to know more about 75">75</a>` ppm<br/>Degree of hardness due to `CaSO_4 = 75` ppm<br/>Weight of `Ca(HCO_3)_2` present in `10^6` gm of water ` = 16.2 xx 10^6//10^5 = 162 g `<br/>162 g of `Ca(HCO_3)_2 -= 100 g `of `CaCO_3` <br/>Degree of hardness due to `Ca(HCO_3)_2 = 100` ppm<br/> Temporary hardness of water sample = 100ppm<br/>Permanent hardness of water sample = 50 + 75 = 125 ppm<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> hardness of water sample = 100+125=225 ppm</body></html> | |
| 51659. |
100kg of a sample of water contained 6 of MgSO_(4) The degree of |
| Answer» <html><body><p>6 ppm<br/>60 ppm<br/>5 ppm<br/>500 ppm</p>Solution :`100 kg=10^(5)g rArr "weight" of MgSO_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)=<a href="https://interviewquestions.tuteehub.com/tag/6g-331964" style="font-weight:bold;" target="_blank" title="Click to know more about 6G">6G</a>` <br/> <a href="https://interviewquestions.tuteehub.com/tag/hardness-14971" style="font-weight:bold;" target="_blank" title="Click to know more about HARDNESS">HARDNESS</a> of `MgSO_(4)=(6g)/(10^(5)g)xx10^(6)=60ppm` <br/> hardness in term of <br/> `CaCO_(3)=60xx(100)/(120)=50` ppm</body></html> | |
| 51660. |
100Kg of a sample of water contained 6 g of MgSO_4 . The degree of hardness |
| Answer» <html><body><p> 6 <a href="https://interviewquestions.tuteehub.com/tag/ppm-1162221" style="font-weight:bold;" target="_blank" title="Click to know more about PPM">PPM</a> <br/> 60 ppm<br/> 5 ppm<br/> <a href="https://interviewquestions.tuteehub.com/tag/50-322056" style="font-weight:bold;" target="_blank" title="Click to know more about 50">50</a> ppm </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> | |
| 51661. |
100grams of a hydrocarbon has 93.71 g of carbon. What is its empirical formula of the hydrocarbon? |
| 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> :`C_(<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)H_(<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> | |
| 51662. |
100cm^3of a given sample of H_2O_2gives 1000cm^3ofO_2 at S.T.P. The given sample 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>% H_2O_2`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/90-341351" style="font-weight:bold;" target="_blank" title="Click to know more about 90">90</a>% H_2O_2` <br/>10vol. `H_2O_2` <br/>100vol. `H_2O_2` </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 51663. |
100cc of CO_2 gas is diffused in 25 seconds through a porous membrane. How much time does the same volume fo sulphur dioxide take to diffuse ? |
| 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> :30.17sec</body></html> | |
| 51664. |
1000L of air a STP was dissolved in water and required 2.5xx10^(-5) moles of KMnO_(4) for complete reaction of SO_(2) as pollutants. Thus , SO_(2) content in air is |
| Answer» <html><body><p>`1.4ppm`<br/>`14ppm`<br/>`2.8ppm`<br/>`6.25ppm`</p>Solution :`SO_(2)+H_(2)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> H_(2)SO+_(3)` <br/> `underset(+7)(2MnO_(4)^(-))+underset(+4)(2H_(2)SO_(3)) rarr underset(+6)(5H_(2)SO_(4))+underset(+2)(2Mn^(2+))` <br/> `2 mol <a href="https://interviewquestions.tuteehub.com/tag/es-446539" style="font-weight:bold;" target="_blank" title="Click to know more about ES">ES</a> MnO_(4)^(-) = 5 mol e H_(2)SO_(3) or SO_(2)` <br/> `2.5xx10^(-5)` mole `MnO_(4)^(-) = 5/2 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 2.5xx10^(-5) mol es SO_(2)` <br/> `=6.25xx10^(-5)mol es SO_(2)` <br/> At STP, `6.25xx10^(-5) mol es = 6.25xx10^(-5)xx22.4L SO_(2)` <br/> `=1.4xx10^(-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/> In `1000L` air, `SO_(2)=1.4xx10^(-3)L` <br/> In `10^(6)L(ppm), SO_(2) = (1.4xx10^(-3)xx10^(6))/(10^3) = 1.4ppm`.</body></html> | |
| 51665. |
1000g of a water sample contained a calcium salt equivalent to 1g of CaCO_3. The hardness of that water sample is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a> ppm<br/>500 ppm<br/>1000 ppm<br/>20,000 ppm</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> | |
| 51666. |
1000g of a water sample contained a calcium salt equivalent to 18 of CaSO_(4) The hardness of that water sample is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a> ppm<br/>500 ppm<br/>1000 ppm<br/>220,000 ppm</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/hardness-14971" style="font-weight:bold;" target="_blank" title="Click to know more about HARDNESS">HARDNESS</a> `=("solute <a href="https://interviewquestions.tuteehub.com/tag/weight-1451304" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHT">WEIGHT</a>")/("solution weight")xx10^(6)`</body></html> | |
| 51667. |
100ml of KMnO_4 salution is exactly reduced by 100ml of 0.5 M oxalic acid under acidic condition. The molarity of KMnO_4 solution is |
| Answer» <html><body><p>0.1M<br/>0.2M<br/>0.05M<br/>0.5M</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> | |
| 51668. |
1000 mL O_(2) at NTP was passed through Siemen's ozonizer so that the volume is reduced to 888 mL at same condition. Ozonized oxygen is passed through KI solution. Liberated I_(2) was titrated with 0.05 N hypo. Calculate volume of hypo used. |
| 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> | |
| 51669. |
10.00 mol of pure, 1-butyne was taken in a flask with some basic catalyst, sealed and heated to 500K where it isomerised to produce 2-butyne and 1,3-butadiene. At equlibrium 3.0 mole of 2-butyne and 5.5 mole of 1.3-butadiene were found to be present in the flask. If 2.0 mole of 1.3-butadiene wer added further to the above equilibrium mixture. What would be the new equilibrium composition when equilibrium was established again? 1-butyne 2-butyne 1,3-butadiene |
| Answer» <html><body><p>1.80moles 3.60moles 3.60moles<br/>3.60moles 6.60moles 1.80moles<br/>1.80moles 6.6moles` 3.60moles<br/>3.60moles 1.80moles 6.60moles</p>Solution :N//A</body></html> | |
| 51670. |
100 ml solution having 0.2 M HA (weak acid, K_(a), = 1.0x10^(-5) ) and 0.2 N NaA, 200 ml of 0.1 M NaOH has been added. Furthermore, diluted to IL. Which of the following statement is correct? |
| Answer» <html><body><p>Initially , the <a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> has pH equal to 5, that is before additions of NaOH <br/>In the final solution , the <a href="https://interviewquestions.tuteehub.com/tag/concentration-20558" style="font-weight:bold;" target="_blank" title="Click to know more about CONCENTRATION">CONCENTRATION</a> of `[OH^(-) ] is 10^(-9) ` M. <br/>After the addition of NaOH , the pH of solution increase by four units. <br/>After the addition of base, the solution losses buffering action and can be restored after the addition of acid. </p>Solution :Initially acid buffer is present <br/><br/> ` pH =5+ <a href="https://interviewquestions.tuteehub.com/tag/log-543719" style="font-weight:bold;" target="_blank" title="Click to know more about LOG">LOG</a> ""(0.2)/( 0.2)= 5` <br/> (b)after dilution to1 <a href="https://interviewquestions.tuteehub.com/tag/lt-537906" style="font-weight:bold;" target="_blank" title="Click to know more about LT">LT</a>, <br/> ` [HA] =0.02 N, [NaA ]= 0.02 N` <br/> ` [NaOH ]=0.02 N` <br/> ` {:( NaOH +, HA to , NaA + H_2O) , (0.02 N , 0.02N , -),( -,-,0.02N):}` <br/> ` therefore` Due to addition of NaOH , buffer disappeared & salt is formed <br/> ` pH =7+ ( 5)/(2) +(1)/(2) log (4 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10 ^(-2)) = 8.8 `</body></html> | |
| 51671. |
100 mL solutionof FeC_(2)O_(4) and FeSO_(4) is compleltely oxidized by 60 mL of 0.2M KMnO_(4) in acid medium. The resulting soltuion is thenreduced by Zn and dil. The reduced soltuion is again oxidized completely by 40 mLof0.2M KMnO_(4). Calculate normality of FeC_(2)O_(4) and FeSO_(4) 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> :`FeSO_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>) = 0.03N, FeSO_(4) = 0.03N`</body></html> | |
| 51672. |
100 ml solution contains 12 mg MgSO_(4). The concentration of solution is |
| Answer» <html><body><p>`10^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)M`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/2xx10-1840192" style="font-weight:bold;" target="_blank" title="Click to know more about 2XX10">2XX10</a>^(-3)<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/>120 ppm<br/>`10^(-3)m`</p>Solution :`M=(12xx10^(-3))/(120xx0.1)=10^(-3)M=2xx10^(-3)N` <br/> ppm `(w_(1))/(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>))xx10^(6)=(12xx10^(-3))/(<a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a>)xx10^(6)=120`ppm <br/> For dilute solution, m = M</body></html> | |
| 51673. |
100 mL sample of hard water is passed through a column of the ion exchange resin RH_(2). The water coming off the column requires 15.17 mL of 0.0265 M NaOH for its titration. What is the hardness of water as "ppm" of Ca^(2+)? |
| 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> :`80.40`;</body></html> | |
| 51674. |
100 mL solution consists 4 g caustic soda. The normality of the solution is : |
| Answer» <html><body><p>`1.0`<br/>`0.1`<br/>`0.5`<br/>`4.0`</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> | |
| 51675. |
100 mL of tap water containing Ca(HCO_(3))_(2) was titrated with N/50 HCl with MeOH as indicator . If 30 mL of HCl were required , calculate the temporary hardness as parts of CaCO_(3) per 10^(6) parts of water . If your answer is 'a x 100' , what is the value of 'a'. |
| 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/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a></body></html> | |
| 51676. |
100 ml of sample of water requires 3.92 mg of K_2 Cr_2O_3 in presence of H_2SO_4 for the oxidation of dissolved organic matter present in it. The COD of the water sample in ppm |
| Answer» <html><body><p>3.1<br/>6.4<br/>12.4<br/>8.2</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> | |
| 51677. |
100 ml of sample is removed from an aqueous solution saturated with CaSO_(4) at 25^(0)C. The water is completely evaported from the sample and deposit of 0.24 g of CaSO_(4) is obtained. The Ksp of CaSO_(4) at 25^(0)C will be. |
| Answer» <html><body><p>`3.115 xx 10^(-4)`<br/>`3.115 xx 10^(-5)`<br/>`3.115 xx 10^(-6)`<br/>`3.115 xx 10^(-3)`</p>Solution :`CaSO_(4(s))hArrCa_((aq))^(+2) + SO_(4(aq))^(2-), Ksp = ?` <br/> According to the given <a href="https://interviewquestions.tuteehub.com/tag/data-25577" style="font-weight:bold;" target="_blank" title="Click to know more about DATA">DATA</a>, solubility of `CaSO_(4)` is `0.24 g` <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/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a> ml` <br/> `[<a href="https://interviewquestions.tuteehub.com/tag/ca-375" style="font-weight:bold;" target="_blank" title="Click to know more about CA">CA</a>^(+2)] = [SO_(4)^(2-)] = 0.01765 M` <br/> `Ksp = [Ca^(+2)] [SO_(4)^(2-)]`</body></html> | |
| 51678. |
100 mL of phosphorus pentachloride is totally decomposed to its trichloride at 1 atm and 546^(@)C. How many molecules of chlorine are formed? |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/decomposition-946042" style="font-weight:bold;" target="_blank" title="Click to know more about DECOMPOSITION">DECOMPOSITION</a> of phosphorus pentachloride is given as <br/> `PCl_(5)rarrPCl_(3)+Cl_(2)` <br/> `"1 mole of "PCl_(5)="1 mole of"Cl_(2)` <br/> `"1 vol of "PCl_(5)="1 vol <a href="https://interviewquestions.tuteehub.com/tag/fo-457881" style="font-weight:bold;" target="_blank" title="Click to know more about FO">FO</a> Cl"_(2)` <br/> `"100 mL of "PCl_(5)="100 mL of "Cl_(2)` <br/> `{:("Given conditions","STP conditions"),(P_(1)="1 atm",P_(2)="1 atm"),(T_(1)="819 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"),(V_(1)="100 mL",V_(2)=?):}` <br/> The <a href="https://interviewquestions.tuteehub.com/tag/volue-3265040" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUE">VOLUE</a> of `Cl_(2)` <a href="https://interviewquestions.tuteehub.com/tag/measured-7257656" style="font-weight:bold;" target="_blank" title="Click to know more about MEASURED">MEASURED</a> at STP <br/> `V_(2)=(P_(1)V_(1))/(T_(1))xx(T_(2))/(P_(2))=(1xx100xx273)/(819xx1)="33.33mL"` <br/> `"22400 mL of "Cl_(2)" at STP"=` <br/> `6.022xx10^(23)" molecules"` <br/> `"33.33 mL of "Cl_(2)" at STP"=?` <br/> Number of molecules of chlorine obtained `=(33.33)/(22400)xx6.022xx10^(23)=8.96xx10^(20)`</body></html> | |
| 51679. |
Phosphine on decomposition produces phosphorus and hydrogen. When 100 ml of phosphine are decomposed the change in volume under laboratory conditions is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/50ml-324226" style="font-weight:bold;" target="_blank" title="Click to know more about 50ML">50ML</a> increase<br/>500mL decrease<br/>900mL decrease<br/>None</p>Answer :A</body></html> | |
| 51680. |
100mL of om at STP was passed through 100 ml of 10 volume H_(2)O_(2)solution. What is the volume strength ofH_(2)O_(2) after the reaction? |
| Answer» <html><body><p>9.5<br/>9<br/>4.75<br/>4.5</p>Solution :`H_2O_2+ O_3 to H_2O + 2O_2` <br/> initial `M_(H_2O_2) = 10/11.2 = 10/11.2 M`<br/> <a href="https://interviewquestions.tuteehub.com/tag/number-582134" style="font-weight:bold;" target="_blank" title="Click to know more about NUMBER">NUMBER</a> of <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 = 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>/22.4` <br/> number of moles of `O_2 = 10/11.2 xx 0.1` <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 stoichiometric equation<br/>on mole of `H_2O_2` produce 1 mole of `O_3` number of moles of `H_2O_2` consumed <br/> `=1/224," remains" = 19/224` <br/> onemolarity = 11.2 (W/V)% <br/> `19/224 xx 0.1 --------- ?`</body></html> | |
| 51681. |
100 mL of HCI gas at 25^(@)C and 740 mm pressure were dissolveed in one litre of water. Caculate the pH of a solution. Given, V.P. of H_(2)O at 25^(@)C is 23.7 mm. |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :`2.4142;`</body></html> | |
| 51682. |
100 mL of H_(2)O_(2) is oxidized by 100 mL of 1M KMnO_(4) in acidic medium (MnO_(4)^(-) reduced to Mn^(+2) ) 100 mL of same H_(2)O_(2) is oxidized by v mL of 1M KMnO_(4) in basic medium (MnO_(4)^(-)reduced to MnO_(2)). Findthe value of v: |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/500-323288" style="font-weight:bold;" target="_blank" title="Click to know more about 500">500</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/>`100//3`<br/>`500//3`</p>Solution :In acidic medium <br/> Eqts of `H_(2)O_(2)` = Eqts of `KMnO_(<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/> `(100)/(<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>)xxMxx2=(100)/(1000)xx1xx5impliesM=2.5` <br/> In basic medium <br/> Eq. of `H_(2)O_(2)` = Eqts of `KMnO_(4)` <br/> `(100)/(1000)xx2.5xx2=(V)/(1000)xx1xx3impliesV=(500)/(3)`</body></html> | |
| 51683. |
100 mL of each pH = 3 and pH = 5 solutions of a strong acid are mixed. What is the resultant pH ? |
| Answer» <html><body><p></p>Solution : <a href="https://interviewquestions.tuteehub.com/tag/ph-1145128" style="font-weight:bold;" target="_blank" title="Click to know more about PH">PH</a> of first solution `= <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^(+) ]=10^(-3)=N_1`<br/>pH of second solution `= 5, [H^(+) ]=10^(-5) = N_2`<br/> Final <a href="https://interviewquestions.tuteehub.com/tag/proton-1170983" style="font-weight:bold;" target="_blank" title="Click to know more about PROTON">PROTON</a> concentration is <a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> by the equation:`[H^+]=(V_1 N_1+V_2 N_2)/(V_1 +V_2)`<br/> Substituting the values` [H^(+)] =(100 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10^(-3) xx 10^(-5))/( 100+100 ) = 5.05 xx 10^(-4)` M <br/> pH of the resultant mixture `= -log (5.05 xx 10^(-4) ) = 4 - log 5.05 = 3.3`</body></html> | |
| 51684. |
100 mL of ethano (d = 0.78g cc^(-1)) I made upto a litre with pure water. Caculate the molality. |
| 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/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.88 <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/kg-1063886" style="font-weight:bold;" target="_blank" title="Click to know more about KG">KG</a>^(-1)`</body></html> | |
| 51685. |
100 mL of each acetylene and oxygen are mixed. The mixture is strongly heated to complete the reaction and colled back to room temperature. What is the maximum volume of carbondioxide obtained. Why? |
| Answer» <html><body><p></p>Solution :The stoichiometric equation for burning of acetylene is given as <br/> `2C_(2)H_(2)+5O_(2)rarr4CO_(2)+2H_(2)O` <br/> `"2 moles of "C_(2)H_(2)="5 moles of "O_(2)` <br/> `"2 volumes of "C_(2)H_(2)=5" volumes of "O_(2)` <br/> `"100 mL of "C_(2)H_(2)=<a href="https://interviewquestions.tuteehub.com/tag/250ml-297852" style="font-weight:bold;" target="_blank" title="Click to know more about 250ML">250ML</a>" of "O_(2)` <br/> `"40 mL of "C_(2)H_(2)=100 mL " of "O_(2)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a> the 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> is oxygen <br/> `"5 moles of "O_(2)="4 moles of "CO_(2)` <br/> `"5 volumes of "O_(2)="4 volumes of "CO_(2)` <br/> `"100 mL of "O_(2)=?` <br/> The maximum <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)` formed `=100xx(4)/(5)="80 mL"` <br/> This is because no <a href="https://interviewquestions.tuteehub.com/tag/sufficient-1233547" style="font-weight:bold;" target="_blank" title="Click to know more about SUFFICIENT">SUFFICIENT</a> `O_(2)` is available.</body></html> | |
| 51686. |
100 ml of a sample of water requires 1.96 mg of potassium dichromate in the presence of 50% H_2SO_4 for the oxidation of dissolved organic matter in it. Calculate the chemical oxygen demand. |
| Answer» <html><body><p></p>Solution :One gram equivalent of any <a href="https://interviewquestions.tuteehub.com/tag/oxidant-1144439" style="font-weight:bold;" target="_blank" title="Click to know more about OXIDANT">OXIDANT</a> can give, one equivalent weight of <a href="https://interviewquestions.tuteehub.com/tag/oxygen-1144542" style="font-weight:bold;" target="_blank" title="Click to know more about OXYGEN">OXYGEN</a> (8 <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/> `49 g K_2Cr_2O_7 to 8g " Oxygen"` <br/> ` 1.96 xx 10^(-3) g K_2Cr_2O_7 to (8 xx 1.96 xx 10^(-3))/(49)=0.32 mg ` <br/> Chemical oxygen demand of the given <a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a> ml <a href="https://interviewquestions.tuteehub.com/tag/sample-1194587" style="font-weight:bold;" target="_blank" title="Click to know more about SAMPLE">SAMPLE</a> of water `=(0.32)/(0.1) = 3.2 ppm`</body></html> | |
| 51687. |
100 ml of a sample of water requires 1.96 mg of potassium dichromate in the presence of 50% H_(2) SO_(4) for the oxidation of dissolved organic matter in it. The chemical oxygen demand is 8x xx 10^(-1) ppm. 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/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a></body></html> | |
| 51688. |
100 ml of a sample of water requires 0.98mg of K_(2)(M.W.= 294 )in presence of the H_(2)SO_(4) for the oxidation of dissolved organic matter in it. The COD of the water sample is |
| Answer» <html><body><p>78.4 <a href="https://interviewquestions.tuteehub.com/tag/ppm-1162221" style="font-weight:bold;" target="_blank" title="Click to know more about PPM">PPM</a> <br/>1.6 ppm<br/>3.2 ppm <br/>6.4 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> | |
| 51689. |
100 mL of a gas is stored over mercury in mercury manometer at 27^(@)C radius of inside column is r and that of outside is R and initially mercury level is equal in both column (a) If R=r. Find the new temperature (in K) if due to change in temperature outside level of mercury is raised by 20 mm (i) Assume volume of gas remain constant by some experimental means rArr T_(1) (ii) Volume does not remain constant (pi r^(2) = 10 cm^(2)) rArr T_(2) (b) If R = 2r and inside level falls by 80 mm rArr T_(3) Give the answers by adding T_(1),T_(2),T_(3) to the nearest integer. |
| 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> :1298</body></html> | |
| 51690. |
100 ml of 2M HCl solution completely neutralises 10 g, of a metal carbonate. Then the equivalent weight of the metal is |
| Answer» <html><body><p>50<br/>20<br/>12<br/>100</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> | |
| 51691. |
100 mL of 0.3 N HCl were mixed with 200 mL of 0.6 N H_(2)SO_(4) solution. The final normality of acid was : |
| Answer» <html><body><p>0.4 <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/>0.5 N<br/>0.6 N<br/>0.9 N</p>Solution :N//A</body></html> | |
| 51692. |
100 mL of 0.2 N HCl solution is added to 100 mL of 0.2 N AgNO_(3) solution. The molarity of nitrate ions in the resulting mixture will be : |
| Answer» <html><body><p>0.05 M<br/>0.5 M<br/>0.1 M<br/>0.2 M</p>Solution :N//A</body></html> | |
| 51693. |
100 ml of 0.1M aqueous ferric alum solution is completely reduced with Iron to give Ferrous Ions. The volume of 0.1M acidified KMnO_4solution required for complete reaction with ferrous Ions in the solution will be |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/60-328817" style="font-weight:bold;" target="_blank" title="Click to know more about 60">60</a> ml <br/>30 ml <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<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> ml </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> :`Fe^(+<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)+Ferarr2Fe^(+2)` <br/> `100xx0.1 xx2=0.5xxV`<br/> `V=(100xx0.2)/(0.5)=200/5=40` ml</body></html> | |
| 51694. |
100 ml of 0.1 M HCl and 100 ml of 0.1 M HOCN are mixed . What is the concentrationof .. OCN ^(-)..,and pH of the solution ?(K_a=1.2 xx 10^(-6) ) |
| 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>.2 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>^(-6), 1.3 ` <br/>` 1.2 xx 10^(-6),1` <br/>` 2.6 xx 10^(-8),1.6` <br/>` 2.4 xx 10^(-6), 2.6` </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^(+)]= (100 xx 0.1 )/(200)=0.05,pH =2- log 5 =13` <br/>`[HOCN]=( 100xx 0.1 )/( 200) = 0.05` <br/>` K_a =([H^(+) ][OCN^(-) ])/( [HOCN])=(1.2 xx 10 ^(-6) xx 0.05 )/( 0.05 ) ` <br/> ` = [OCN^(-) ]=1.2 xx 10^(-6) `</body></html> | |
| 51695. |
100 ml of 0.1 M HCl and 100 ml of 0.1 M HOCN are mixed then(K_a= 1.2 xx 10^(-6)) |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/ocn-2202269" style="font-weight:bold;" target="_blank" title="Click to know more about OCN">OCN</a>^(-) `concentration in the solution is `1.2 xx10^(-6) ` <br/>pH of the solution is 1.3<br/>solution is a buffer <br/>`H^(+) ` in the solution is `10^(6) ` </p>Solution :`HOCNhArr H^(+) +OCN^(-) ` <br/> ` K_a <a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> ([H^(+)][OCN^(-)])/( [HOCN]) `<br/> ` 1.2 xx 10 ^(-6)= (5xx 10 ^(_2)[OCN^(-) ])/( 5 xx10 ^(-2)) ` <br/>`[ OCN^(-)]= 1.2 xx 10 ^(-6) ` <br/> Approximation method, <a href="https://interviewquestions.tuteehub.com/tag/weak-729638" style="font-weight:bold;" target="_blank" title="Click to know more about WEAK">WEAK</a> acid conribution is neglected. <br/>` [H^(+) ] =(100 xx <a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>.1 +0)/( <a href="https://interviewquestions.tuteehub.com/tag/200-288914" style="font-weight:bold;" target="_blank" title="Click to know more about 200">200</a>) = 0.05 therefore pH =1.3 `</body></html> | |
| 51696. |
100 mL of 0.01 M XO_(4)^(-) is reduced to X^(n+) by 100 mL of 0.05 M Fe^(2+) in acidic medium. Thusoxidation state of X in X^(n+) is ______________ |
| 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> :`100xx0.05xx1=100xx0.01xx(7-n)n=2`</body></html> | |
| 51697. |
100 ml of 0.01 M KMnO_(4) oxidizes 10 ml of H_(2)O_(2) sample in acidic medium. The volume strength of H_(2)O_(2) sample is |
| Answer» <html><body><p>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> Vol<br/>5.6 Vol <br/>2.80 Vol<br/>none of these</p>Solution :eq. `KMnO_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)` = eq. `H_(2)O_(2)` <br/> `100xx0.01xx5=10xxNimpliesN=0.5` <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> V.S. = `Nxx5.6=2.8 vol`</body></html> | |
| 51698. |
100 mL NaOH solution pH = 10, So calculate [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/1xx10-1804645" style="font-weight:bold;" target="_blank" title="Click to know more about 1XX10">1XX10</a>^(-<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)` M</body></html> | |
| 51699. |
100 mL each of 1NH_(2)O_(2) " and " 11.2 V H_(2)O_(2) solution are mixed, then the resultant solution will be : |
| Answer» <html><body><p>`3M H_(2)O_(2)`<br/>`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> H_(2)O_(2)`<br/>`25.5 g//L H_(2)O_(2)`<br/>`2.55 g//L H_(2)O_(2)`</p>Solution :`11.2V H_(2)O_(2)-=(11.2)/(5.6)N H_(2)O_(2) i.e., 2N H_(2)O_(2)` <br/> `N_(1)V_(1)+N_(2)V_(2)=N_(R )(V_(1)+V_(2))` <br/> `1xx100+2xx100=N_(R )xx200` <br/> `N_(R )=(3)/(2)=1.5` <br/> Strength `=N xx " Equivalent <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/> `=1.5xx17=25.5 g//L`</body></html> | |
| 51700. |
100 mL , each of 0.25 M NaF and 0.01 5 Ba (NO)_3 are mixed K_(sp)of BaF_2= 1.7 xx 10^(-6) |
| Answer» <html><body><p>A ppt is formed <br/>No ppt is formed <br/>Cannot <a href="https://interviewquestions.tuteehub.com/tag/say-1195451" style="font-weight:bold;" target="_blank" title="Click to know more about SAY">SAY</a> <br/>Some more <a href="https://interviewquestions.tuteehub.com/tag/data-25577" style="font-weight:bold;" target="_blank" title="Click to know more about DATA">DATA</a> are <a href="https://interviewquestions.tuteehub.com/tag/needed-1112853" style="font-weight:bold;" target="_blank" title="Click to know more about NEEDED">NEEDED</a> </p>Solution :`IP =((0.25 xx 100)/(200))^(2) ((0.015 xx 100)/( 200)) gt KsP ` <br/> ` therefore PPt `is formed</body></html> | |