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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.
| 50801. |
(A) .^(20)Ne and .^(22)Ne are isotones. (R) Noble gases do not exist as isotopes as they are not reactive. |
| Answer» <html><body><p>Both (<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>) and (A) are true and reason is the <a href="https://interviewquestions.tuteehub.com/tag/correct-409949" style="font-weight:bold;" target="_blank" title="Click to know more about CORRECT">CORRECT</a> <a href="https://interviewquestions.tuteehub.com/tag/explanation-455162" style="font-weight:bold;" target="_blank" title="Click to know more about EXPLANATION">EXPLANATION</a> of assertion.<br/>Both (R) and (A) are true but reason is not correct explanation of assertion<br/>Assertion (A) is true but reason (R) is false<br/>Assertion (A) and reason (R) both are false</p>Answer :D</body></html> | |
| 50802. |
A 20g chunk of dry ice is placed in an empty 0.75 litre wire bottle tightly closed what would be the final pressure in the bottle after all CO_2 has been evaporated and temperature reaches to 25^(@)C? |
| 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> :Pressure <a href="https://interviewquestions.tuteehub.com/tag/inside-1045864" style="font-weight:bold;" target="_blank" title="Click to know more about INSIDE">INSIDE</a> the <a href="https://interviewquestions.tuteehub.com/tag/bottle-390839" style="font-weight:bold;" target="_blank" title="Click to know more about BOTTLE">BOTTLE</a> `= P+` atm pressure `=14.828+1=15.828atm`</body></html> | |
| 50803. |
A 2.0g mixture of Na_(2)CO_(3) and NaHCO_(3) suffered a loss of 0.12g on heating. Percentage of Na_(2)CO_(3) in the mixture- |
| Answer» <html><body><p>83.8<br/>16.2<br/>38.8<br/>61.2</p>Answer :A</body></html> | |
| 50804. |
A 200 mL solution of I_(2) divided into two unequal parts. Part I reacts with pypo solution in acidic medium and required 8 " mL of " 2 M hypo solution for complete neutralisation. Part II was added with 300 " mL of " 0.1 M NaOH solution residual base required 30 " mL of " 0.1 M H_(2)SO_(4) solution for complete neutralisation. Calculate the value of 20 times the initiall concentration I_(2)? |
| Answer» <html><body><p>1<br/>2<br/>3<br/>4</p>Solution :First part: mmoles of `Na_(2)S_(2)O_(3)` used <br/> `=8xx2xx1 ("n-factor")=16` <br/> `I_(2)+2Na_(2)S_(2)O_(3)to2NaI+Na_(2)S_(4)O_(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)` <br/> m" mol of "`I_(2)` used `=(1)/(2)` m" mol of "`Na_(2)S_(2)O_(3)=(16)/(2)=8`..(i) <br/> Second Part: <br/> `3I_(2)+6NaOHto5NaI+NaIO_(3)+3H_(2)O` <br/> mmoles of `H_(2)SO_(4)=` excess `NaOH=30xx0.1xx2` <br/> `xx("n-factor")=6` <br/> m" mol of "total `NaOH=300xx0.1xx1("n-factor")=<a href="https://interviewquestions.tuteehub.com/tag/30-304807" style="font-weight:bold;" target="_blank" title="Click to know more about 30">30</a>` <br/> m" mol of "NaOH `used=30-6=24` <br/> m" mol of "`I_(2)` used `=(1)/(2)` m" mol of "NaOH used <br/> `=(24)/(2)=12 m" mol of "I_(2) used` <br/> Total m" mol of "`I_(2)` used `=` part I`+` part II <br/> `=8+12=20mmol` <br/> `M of I_(2)=(mmol)/(V_(mL))=(20)/(200)=0.1M` <br/> 20 <a href="https://interviewquestions.tuteehub.com/tag/times-1420471" style="font-weight:bold;" target="_blank" title="Click to know more about TIMES">TIMES</a> the initial `M_(I_(2))=0.1xx20=2`</body></html> | |
| 50805. |
A 200 mL sample of a gaseous hydrocarbons has a density of 2.52gL^(-1) at 55^(@)C and 720mm Hg. What Is its formula? |
| Answer» <html><body><p>`C_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)H_(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)`<br/>`C_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)H_(10)`<br/>`C_(5)H_(12)`<br/>`C_(6)H_(6)`</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> | |
| 50806. |
A 200 mL sample of a citrus fruit drinks containing ascorbic acid(vitamin C, mol. We 176.13) was acidified with H_(2)SO_(4) and 10mLof 0.250M I_(2) was added. Some of the iodine was reduced by theascorbicacid to I^(-). The excess of I_(2)required 4.6mL of 0.01M Na_(2)S_(2)O_(3) for reduction. What was the vitaminC contentof the drink in mg vitamin per mL drink? The reactions are: C_(6)H_(8) O_(6) + I_(2) rarrC_(6) H_(6) O_(6) + 2HI 5H_(2)O + S_(2)O_(3)^(2-) + 4I_(2) rarr 2SO_(4)^(2-) + 8I^(-) + 10H^(-) |
| 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> :`0.058mg//mL`</body></html> | |
| 50807. |
A 2.00 litre evacuated container has a mass of 1050.0g. When the container is filled with an unknown gas at 800mm Hg pressure and 25.0^(@)C the mass is 1052.4g. What is the molar mass of the gas (in g"mol"^(-1))? |
| Answer» <html><body><p>28<br/>31<br/>54<br/>56</p>Answer :a</body></html> | |
| 50808. |
A 20 mL sample of a Ba(OH)_(2) solution is titrated with 0.245 M HCl. If 27.15 mL of HCl is required, what is the molarity of the Ba(OH)_(2) solution ? |
| Answer» <html><body><p>0.166 M<br/>0.180 M<br/>0.333 M <br/>0.666 M</p>Solution :N//A</body></html> | |
| 50809. |
A 20 mL mixture of ethane, ethylene, and CO_2 is heated with O_2. After the explosion, there was a contraction of 28 mL and after treatment with KOH, there was a further contraction of 30 mL. What is the composition of the mixture? |
| Answer» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>></p>Solution :Let the volume of ethane, ethylene, and `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>)`, respectively, are a,b and `(20-a-b)mL` <br/> Now contraction after cooling`=28mL` <br/> `thereforeV_(R)-V_(P)=25` <br/> `V_(R)=` Volume of `C_(2)H_(6)+` volume of `C_(2)H_(4)+` volume of `CO_(2)+` volume of `O_(2)` used for combustion <br/> `V_(P)=` volume of `CO_(2)` produced (Volume of `H_(2)O` is not taken <a href="https://interviewquestions.tuteehub.com/tag/since-644476" style="font-weight:bold;" target="_blank" title="Click to know more about SINCE">SINCE</a> it condenses) <br/> Considering the combustion <a href="https://interviewquestions.tuteehub.com/tag/gases-13668" style="font-weight:bold;" target="_blank" title="Click to know more about GASES">GASES</a>. <br/> (a). `C_(2)H_(6)+(7)/(2)O_(2)to2CO_(2)+3H_(2)O` <br/> (b). `C_(2)H_(4)+3O_(2)to2CO_(2)+2H_(2)O` <br/> (c). `CO_(2)to` no <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a> <br/> `V_(R)=(a+(7)/(2)a)+(b+3b)+(20-a-b)` <br/> `V_(P)=(2a+2b)+(20-a-b)` <br/> `V_(P)-V_(R)=(5)/(2)a+2b=28` <br/> `therefore5a+4b=56`..(i) <br/> There is a further contraction of 32 mL on treatment with KOH. Volume of `CO_(2)` produced `+` Volume of `CO_(2)` original <br/> `=32` <br/> `(2a+2b)+(20-a-b)=32` <br/> `therefore a+b=12`...(ii) <br/> On solving (i) and (ii) we get <br/> `a=8mL` (volume of `C_(2)H_(6))` <br/> `b=4mL` (Volume of `C_(2)H_(4))` <br/> Volume of `CO_(2)=8mL`</body></html> | |
| 50810. |
A 20 ml mixture of C_(2)H_(4)"and"C_(2)H_(2) undergoes sparking in gas eudimeter with just sufficient |
| 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/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a></body></html> | |
| 50811. |
A 20 litre container at 400 K contains CO_2(g) at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the container is now decreased by moving the mavable piston fitted in the container. The maximumvolume of the conatiner, when pressure of CO_2 attains its maximum value will be : Give that : SrCO_3 (s)hArr SrO(s) + CO_2(g) K_p = 1.6 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>><a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> litre<br/>5 litre <br/>10 litre<br/>4 litre</p>Solution :Given that `K_p = 1.6 atm` <br/> `V_1 = 20 L, V_2 = ?, T_1 = 400 K , T_2 = <a href="https://interviewquestions.tuteehub.com/tag/400k-315702" style="font-weight:bold;" target="_blank" title="Click to know more about 400K">400K</a>` <br/> `K_P = P_(CO_2)`<br/> `:. P_(CO_2) = 1.6 atm` <br/> `P_1 = 0.4 atm` <br/> `P_2= 1.6 atm` <br/> `(P_1V_1)/T_1= (P_2V_2)/T_2` <br/> `:. V_2 = (P_1V_1)/T_1 xx T_2/P_2 =(0.4 atm xx 20L)/(cancel 400K) xx (cancel400K)/(1.6atm) = <a href="https://interviewquestions.tuteehub.com/tag/5l-326334" style="font-weight:bold;" target="_blank" title="Click to know more about 5L">5L</a>`</body></html> | |
| 50812. |
A 20 litre container at 400 K contains CO_(2) (g)at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container . The maximum volume of the container when pressure of CO_(2) attains its maximum value, will be ("Given that " : SrCO_(3)(s) hArr SrO (s) + CO_(2) (g)K_(p) =1*6 atm) |
| Answer» <html><body><p>5 litre <br/>10 litre<br/>4 litre<br/>2 litre </p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/maximum-556915" style="font-weight:bold;" target="_blank" title="Click to know more about MAXIMUM">MAXIMUM</a> pressure of `CO_(2)` = Pressure of `CO_(2)` at equilibrium <br/> For the reaction <br/> `SrCO_(3) (s) <a href="https://interviewquestions.tuteehub.com/tag/harr-2692945" style="font-weight:bold;" target="_blank" title="Click to know more about HARR">HARR</a> <a href="https://interviewquestions.tuteehub.com/tag/sro-632549" style="font-weight:bold;" target="_blank" title="Click to know more about SRO">SRO</a> (s) + CO_(2) (g) `<br/> `K_(p) = p_(CO_(2))= 1.6 " <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a> = maximum pressure of "CO_(2)`<br/> Volume of container at this stage<br/> `V=nRT/P""` ...(i) <br/> <a href="https://interviewquestions.tuteehub.com/tag/intially-2132834" style="font-weight:bold;" target="_blank" title="Click to know more about INTIALLY">INTIALLY</a>, no . of moles of the gas (n) <br/>` (PV)/(RT)= (0.4 xx20)/(RT)= 8/(RT)` <br/> As no. of moles of the gas will remain constant , therefore, at equilibrium also, <br/> `n=8/(RT)` <br/> Putting this value in eqn. (i), <br/>`V= 8/(RT)*(RT)/P = 8/P= 8/1.6=5L`</body></html> | |
| 50813. |
A 20 liter container at 400 contains CO_(2)(g) at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of CO_(2) attains Its maximum value will be : Given that : SrCO_(3)(S)hArrSrO(S)+CO_(2)(g) K_(p)=1.6 atm |
| Answer» <html><body><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> litre<br/>5litre<br/>10litre<br/>4 litre</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> | |
| 50814. |
A 2.0 L container at 25^(@)C contains 1.25 mol of oxygen and 3.3mol of carbon. (a) What is the initial in the flask ? (b) If carbon and oxygen and oxygen react as completely as possible to form CO, what will be the final pressure in the container ? |
| Answer» <html><body><p></p>Solution :(a) V=2.0 L, T=298 <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>, n=1.25 mol (because only <a href="https://interviewquestions.tuteehub.com/tag/gas-1003521" style="font-weight:bold;" target="_blank" title="Click to know more about GAS">GAS</a> will exert pressure). Hence, <br/> PV=nRTor`P=(<a href="https://interviewquestions.tuteehub.com/tag/nrt-1109890" style="font-weight:bold;" target="_blank" title="Click to know more about NRT">NRT</a>)/(V)=((1.25" mol")(0.0821" L atm "K^(-1)mol^(-1))(298 K)/(2.0" L")=15.3" atm"` <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>) The reaction will be :`C(s)+(1)/(2)O_(2)(g) to CO(g)` <br/> As 1 mol of C reacts with `(1)/(2)` mol of `O_(2)`, the limiting reactant will be 1.25 mol of `O_(2)`. Thus, `(1)/(2)` mol of `O_(2)` produces CO=1 mol. <br/> `:.1.25` mol of `O_(2)` will produce CO=2.50 mol. <br/> Now, n(<a href="https://interviewquestions.tuteehub.com/tag/gaseous-467815" style="font-weight:bold;" target="_blank" title="Click to know more about GASEOUS">GASEOUS</a>)=2.50 mol. Hence, <br/> `P=(nRT)/(V)=((2.50" mol")(0.0821" L atm "K^(-1)mol^(-1)(298"K"))/(2.0" L ")=30.6" atm "`.</body></html> | |
| 50815. |
A 2.0 g sample containing Na_2CO_3 and NaHCO_3 loses 0.248 g when heated at 300° C, the temperature at which NaHC03 decomposes to Na_2CO_3, CO_2 and H_2O. What is the percentage of Na_2CO_3 in the mixture ? |
| 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> :`Na_2CO_3` does not decompose at the given temperature. It is only `NaHCO_3` which decomposes at this temperature. The corresponding equation is:<br/> `underset(168.016 g) (2NaHCO_(3)) overset(300^(@) C) to underset(44.01 g)(Na_(2)CO_(3)) + underset(18.016 g)(CO_(2)) + H_(2)O` <br/> `CO_(2)` and `H_(2)O` scape as gases and cause a loss in mass. <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> mass of `CO_2 + H_2O` (loss in mass)<br/> =44.01 + 18.016 = 62.026 g<br/> As according to the above equation,`therefore 0.248` g mass is <a href="https://interviewquestions.tuteehub.com/tag/lost-537630" style="font-weight:bold;" target="_blank" title="Click to know more about LOST">LOST</a> by `NaHCO_3 = 168.016` g <br/> `therefore 0.248`gmasswillbelostby`NaHCO_(3) = (168.016)/(62.026) xx 0.248 = 0.672 g` <br/> Since, the total mass of the sample is 2.0 g, the mass of `Na_2CO_3` <a href="https://interviewquestions.tuteehub.com/tag/present-1163722" style="font-weight:bold;" target="_blank" title="Click to know more about PRESENT">PRESENT</a> in it = `2.0 - 0.672 = 1.328 g` <br/> `therefore` <a href="https://interviewquestions.tuteehub.com/tag/percentage-13406" style="font-weight:bold;" target="_blank" title="Click to know more about PERCENTAGE">PERCENTAGE</a> of `Na_(2)CO_(3) = (1.328)/2 xx 100 = 66.4` <br/> Hence, the given sample contains `66.4% Na_(2)CO_(3)`.</body></html> | |
| 50816. |
A 2 lit vessel is filled by 1 mole of each gas A "&" B. If K_(C) for reaction A(g)hArrB(g) is 1.5 at temp. T. [Atomic mass of A is 40 & B "is" 20]. |
| Answer» <html><body><p>`[A]` <a href="https://interviewquestions.tuteehub.com/tag/vs-724135" style="font-weight:bold;" target="_blank" title="Click to know more about VS">VS</a> <a href="https://interviewquestions.tuteehub.com/tag/time-19467" style="font-weight:bold;" target="_blank" title="Click to know more about TIME">TIME</a> is graph `I`<br/>`[B]` vs time is graph `I`<br/>`[A]` vs time is graph `<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>`<br/>`[B]` vs time is grap `II`</p>Solution :N//A</body></html> | |
| 50817. |
A 2 lit solution contains 0.04 mol of each of [CO(NH_(3))_(5)SO_(4)]Br and [CO(NH_(3))_(5)Br]SO_(4). To 1 lit of this solution, excess of AgNO_(3) is added. To the remaining solution of excen of BaCl_(2) is added. The amounts of precipitated salts, respectively, are |
| Answer» <html><body><p>0.01 mol & 0.01 mol <<a href="https://interviewquestions.tuteehub.com/tag/br-390993" style="font-weight:bold;" target="_blank" title="Click to know more about BR">BR</a>>0.01 mol & 0.02 mol <br/>0.02 mol & 0.01 mol <br/>0.02 mol & 0.02 mol</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> :Given that 2L solution contains 0.04 mol each of `[CO(NH_(3))_(6)SO_(4)]Br and [CO(NH_(3))_(5)Br]SO_(4)`. Therefore, the concentration is 0.02 mol `"<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)`. <br/> `underset("0.02 mol")([CO(NH_(3))_(5)SO_(4)]Br)underset("0.02 mol0.02 mol")(rarrBr+[CO(NH_(4))_(5)SO_(4)]^(+))` <br/> `underset("0.02 mol")([CO(NH_(3))_(5)Br]SO_(4))underset("0.02 mol ")(rarrSO_(4)^(-2))+underset("0.02 mol")([CO(NH_(3))_(5)Br]^(+2))` <br/> `underset(("<a href="https://interviewquestions.tuteehub.com/tag/excess-978535" style="font-weight:bold;" target="_blank" title="Click to know more about EXCESS">EXCESS</a>"))(<a href="https://interviewquestions.tuteehub.com/tag/ag-362275" style="font-weight:bold;" target="_blank" title="Click to know more about AG">AG</a>^(+))+underset((0.02))(Br^(-))rarrunderset(0.02)(AgBr)` <br/> `underset(("excess"))(Ba^(+2))+underset((0.02))(SO_(4)^(-2))rarrunderset(0.02)(BaSO_(4))`</body></html> | |
| 50818. |
A 2-L flask contains 1.6 g of methane and 0.5 g of hydrogen at 27^(@)C. Calculate the partial pressure of each gas in the mixture and hence calculate the total pressure. |
| 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_(O_(2))=(70.6)/(32)=2.21, n_(<a href="https://interviewquestions.tuteehub.com/tag/ne-1112263" style="font-weight:bold;" target="_blank" title="Click to know more about NE">NE</a>)=(167.5)/(<a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a>)=8.375, x_(o_(2))=(2.21)/(2.21+8.375)=0.21, x_(Ne)=0.79`</body></html> | |
| 50819. |
A 2 L flask containing nitrogen at 60 cm pressure is connected to a 4 L flask containing carbon monoxide at 48 cm pressure. If the temperature is kept constant, calculate the final pressure of the mixture. |
| 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/52-324553" style="font-weight:bold;" target="_blank" title="Click to know more about 52">52</a> <a href="https://interviewquestions.tuteehub.com/tag/cm-919986" style="font-weight:bold;" target="_blank" title="Click to know more about CM">CM</a></body></html> | |
| 50820. |
(a) 2 g of metal carbonate were dissolved in 50 mL of N HCl. 100 mL of 0.1 N NaOH were required to neutralise the resultant solution. Calculate the equivalent mass of the metal carbonate. (b) How much water should be added to 75 mL of 3 N HCl ot make it a normal solution ? |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :N//A</body></html> | |
| 50821. |
A 1800 gm mixture of anhydrous CuSO_(4)(s) and its hydrated form [CuSO_(4).5H_(2)O(s)] undergoes 20% loss in mass on heating. Mole fractrion of CuSO_(4) in mixture is (Atomic mass of Cu = 64) : |
| Answer» <html><body><p>`(3)/(<a href="https://interviewquestions.tuteehub.com/tag/40-314386" style="font-weight:bold;" target="_blank" title="Click to know more about 40">40</a>)`<br/>`(4)/(9)`<br/>`(5)/(9)`<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>)/(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)`</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> | |
| 50822. |
A 1.50gm samples of type metal (an alloy of S_(n), Pb, Cu and Sb) is dissolved in nitric acid and metastannic acid, H_(2)SnO_(3) precipitate. This is dehydrated by heating to Tin(IV) oxide, which is found to weigh 0.50gm. What percentage of tin was in the original type metal sample? (Sn= 119) |
| Answer» <html><body><p>0.3333<br/>0.2627<br/>0.2938<br/>0.5254</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> | |
| 50823. |
A 15dm^(3) cylinder contains hydroen gas at 5 atm at 27^(@)C, How many balloons of capacity of 1.5 dm^(3) at 1 atm and 27^(@)C can be filled using the gas available from the given cylinder ? |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/40-314386" style="font-weight:bold;" target="_blank" title="Click to know more about 40">40</a></body></html> | |
| 50824. |
A 1.250 g sample of octane(C_(8)H_(18)) is burnedin excess of oxygen in a bomb calorimeter. The temperatre of the calorimeter risesfrom 294.05 K to 300.78K. If heat capacityof the calorimeter is 8.93kJ//K, find the heat transferred to the calorimeter. Also calculate the enthalpy combustion of the sample of octane. |
| 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> :Rise in temperature ` = 300.78 - 294.05K = 6.73K ` <br/> Heattransferredto the calorimeter `= ` Heat capacity of the calorimeter `xx` Rise in <a href="https://interviewquestions.tuteehub.com/tag/temp-1240937" style="font-weight:bold;" target="_blank" title="Click to know more about TEMP">TEMP</a>. <br/> Molar mass of `= ( 8.90kJ K^(-1)) ( 6.73K) = 60.1 kJ` <br/>`C_(8) H_(18) = 8 xx 12 + 18 = 114 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> mol^(-1)` <br/> `:. `<a href="https://interviewquestions.tuteehub.com/tag/enthalpy-15226" style="font-weight:bold;" target="_blank" title="Click to know more about ENTHALPY">ENTHALPY</a> of combustion`= ( 60.1)/( 1.250 ) xx 114 kJ mol^(-1)=5481.1kJ mol^(-1)`</body></html> | |
| 50825. |
A 1.20 gm sample od cocaine, [alpha]_(D)=-16^(@) was dissolvedin 7.50 mL of cholroform and placed in polarimeter tube having a path length of 5.00 am calculate the observed rottion. |
| Answer» <html><body><p>`+1.3^(@)`<br/>`-2.6^(@)`<br/>`+2.6^(@)`<br/>`-1.3^(@)`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :N//A</body></html> | |
| 50826. |
A 1.2 g mixture of Na_(2)CO_(3) and K_(2)CO_(3) was dissolved in water to form 100 cm^(3) of a solution. 20 cm^(3)of this solution required 40 cm^(3) of 0.1 N HCl for neutralisation. Calculate the mass of Na_(2)CO_(3) and K_(2)CO_(3) in 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> | |
| 50827. |
A 110% sample of oleum contains- |
| Answer» <html><body><p>`44.4%` of `SO_(<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/>55.6% of <a href="https://interviewquestions.tuteehub.com/tag/sulphuric-655420" style="font-weight:bold;" target="_blank" title="Click to know more about SULPHURIC">SULPHURIC</a> acid<br/>55.6%of `SO_(3)`<br/>44.4% of <a href="https://interviewquestions.tuteehub.com/tag/sulphur-1234343" style="font-weight:bold;" target="_blank" title="Click to know more about SULPHUR">SULPHUR</a> acid</p>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> | |
| 50828. |
A 1.10 g sample of copper ore is dissolved and the Cu^(2+)is treated with excess KI. The liberated I_2requires 12.12 mL of 0.10M Na_(2) S_(2)O_3solution for titration. Find the % of copper by mass in ore. |
| Answer» <html><body><p><br/></p>Solution :`Cu^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>+) +erarrCu^(+), 2I^(-) rarrI_(2)+<a href="https://interviewquestions.tuteehub.com/tag/2e-300683" style="font-weight:bold;" target="_blank" title="Click to know more about 2E">2E</a>` <br/> Meq.of `Cu^(+2)` = Meq.of liberated `I_2` <br/> = Meq.of `Na_(2)S_2O_3` <br/> `= 12.12 xx0.1 <a href="https://interviewquestions.tuteehub.com/tag/xx1-1463705" style="font-weight:bold;" target="_blank" title="Click to know more about XX1">XX1</a> =1.212` <br/> `:.(W_(cu^(2+)))/(63.6//1)xx1000=1.212` <br/> ` :. W_(cu) =W_(cu^(2+))=0.077` <br/> `( :. Cuoverset(H_2SO_4)rarrCuSO_(<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/> `:. %Cu=(0.077)/(1.10)<a href="https://interviewquestions.tuteehub.com/tag/xx100-3292680" style="font-weight:bold;" target="_blank" title="Click to know more about XX100">XX100</a> =7%`</body></html> | |
| 50829. |
A 1.0g sample of pyrolusite ore containing MnO_2 (MW = 87)was dissolved in a concentrated HCI solution and liberated Cl_2(g) was passed through a concentrated KI solution releasing I_2 . If the liberated iodine required 16mL of 1.25 M sodium thiosulphate (Na_(2)S_(2)O_3)solution, mass percentage of MnO_2in the given sample is |
| Answer» <html><body><p>` 43.5% `<br/>`87%`<br/>` 21.75 %`<br/>`0.875%`</p>Solution :No.of eqts. `MnO_2` = No.of m.eqts .of `CI_2=` = No.of eqts .KI = No.of m.eqts. hypo No.of m.eqts of `MnO_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)= 16x×1.25 = 16xx5/4=20` <br/>No.of milli <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 `MnO_(2)=<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>` <br/>Weight of `MnO_(2)=10`Weight % of `MnO_(2) =10xx87xx10^(-3)=0.87g` <br/> Weight % of `MnO_(2)=(0.87)/1 xx100 =87%`</body></html> | |
| 50830. |
A 10gmixture of Cu_(2)S and CuS was treated with 200 mL of 075M MnO_(4)^(-)in acid solution producing SO_(2), CU^(2+) and Mn^(2+). The SO_(2) was boiled off and the excess of MnO_(4)^(-). The SO_(2) was boiled off and the excess of MnO_(4)^(-) was titrated with 175 mL of 1M Fe^(2+) solution. Caculaate % fo CuSin original 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> :`57.94%`</body></html> | |
| 50831. |
A 10cm column of air is trapped by a colume of Hg 8cm long is capillary tube horizontally fixed as shown below at 1atm pressure Calculate the length of air column when the tube is fixed at same temperature (a) Vertically with open end up (b) Vertically with open end down (c ) At 45^(@) from horizontal with open end up . |
| 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> :`9.05 <a href="https://interviewquestions.tuteehub.com/tag/cm-919986" style="font-weight:bold;" target="_blank" title="Click to know more about CM">CM</a> 11.18cm 9.3 cm`</body></html> | |
| 50832. |
A 100mL solution containing 0.4 M As_(2)S_(3), 5M NaOH and 6M H_(2)O_(2) are reacted to form AsO_(4)^(3)" and "SO_(4)^(2-) as product. If final solution is allowed to stand for some time, what maximum volume of O_(2) at 1 atm and 273Kcan be obtained by decomposition of H_(2)O_(2)? |
| Answer» <html><body><p>112mL<br/>224mL<br/>336mL<br/>448mL</p>Answer :d</body></html> | |
| 50833. |
A 100mL solution containing 0.4 M As_(2)S_(3), 5M NaOH and 6M H_(2)O_(2) are reacted to form AsO_(4)^(3)" and "SO_(4)^(2-) as product. What may be the correct coefficient of As_(2)S_(3),H_(2)O_(2) and NaOH respectively in a balanced reaction? |
| Answer» <html><body><p>1,14,12<br/>1,12,14<br/>1,28,20<br/>1,28,18</p>Answer :a</body></html> | |
| 50834. |
A 100mL mixture containing 72% of CH_(4) by volume and the rest an unknown gas 'X' was kept in a vessel Due to a very fin e crack the mixture effused out 21mL of the mixture was lost and the remaining mixture contained 68.35% of methane by volume Calculate molecular mass of gas X All the measurement are made at same temperature and pressure . |
| 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> :87.1</body></html> | |
| 50835. |
A 100 watt bulb emits monochromatic light of wavelength 400 nm.Calculate the number of protons emitted per second by the bulb. |
| 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> :`2.012 <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/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a>) s^(-<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> | |
| 50836. |
A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb. |
| Answer» <html><body><p></p>Solution :Power of the <a href="https://interviewquestions.tuteehub.com/tag/bulb-905352" style="font-weight:bold;" target="_blank" title="Click to know more about BULB">BULB</a> = 100 <a href="https://interviewquestions.tuteehub.com/tag/watt-1450279" style="font-weight:bold;" target="_blank" title="Click to know more about WATT">WATT</a> `= 100 <a href="https://interviewquestions.tuteehub.com/tag/j-520843" style="font-weight:bold;" target="_blank" title="Click to know more about J">J</a> s^(-1)` <br/> Energy of one photon, `E = hv = (hc)/(lamda)((6.626 xx 10^(-34)Js) xx (3 xx 10^(8) ms^(-1)))/(400 xx 10^(-9)m) = 4.969 xx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/19-280618" style="font-weight:bold;" target="_blank" title="Click to know more about 19">19</a>) J` <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 photons emitted `= (100 Js^(-1))/(4.969 xx 10^(-19)J) = 2.012 xx 10^(20) s^(-1)`</body></html> | |
| 50837. |
A 100wartbulbemitsmonochromaticlight ofwavelength 400 bn, Calculatethe numberofphotonsemittedper secondby the bulb. |
| 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> :Powerof the bulb - 100 walt= <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> `js^(-1)` <br/>Energyof one photonE <br/> `hv = (hc )/(lambda)= (6.626xx 10^(34) Js xx3.0xx 10^(<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>) <a href="https://interviewquestions.tuteehub.com/tag/ms-549331" style="font-weight:bold;" target="_blank" title="Click to know more about MS">MS</a>^(-1))/( 400 xx 10^(9) m)` <br/>Noof photon4.9695 `xx 10^(<a href="https://interviewquestions.tuteehub.com/tag/19-280618" style="font-weight:bold;" target="_blank" title="Click to know more about 19">19</a>)` energyfrom 1photon<br/> 100J senergyfromhowmanyphoton<br/> Numberof photonsemited <br/> `=(1xx 100 Js^(-1))/(4.9695 xx 10^(19) J)= 2.0123 xx 10^(20) s^(-1)`</body></html> | |
| 50838. |
A 100 mL solution of 0.1 N HCl was titrated with 0.2 N NaOH solution. The titration was discontinued after adding 30 mL of NaOH solution. The titration was completed by adding 0.25 N KOH solution. The volume of KOH required for completing the titration is : |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/70-333629" style="font-weight:bold;" target="_blank" title="Click to know more about 70">70</a> mL<br/>32 mL<br/>35 mL<br/>16 mL</p>Solution :N//A</body></html> | |
| 50839. |
A 100 ml mixture of Na_(2)CO_(3) and NaHCO_(3) is tittrated against 1 M HCl. If v_(1)L and v_(2)L are consumed when phenolphthalein and methyl orange are used as indicators respectively in two separate titrations, which of the following is true form molarities in the original solution. |
| Answer» <html><body><p>molarity of `Na_(2)CO_(3)=20v_(<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/>molarity of `NaHCO_(3)=10(v_(2)-2v_(1))`<br/>molarity of `Na_(2)CO_(3)=10(v_(2)+v_(1))` <br/>molarity of `NaHCO_(3)=10(v_(2)-v_(1))` </p>Solution :When HPh <a href="https://interviewquestions.tuteehub.com/tag/used-2318798" style="font-weight:bold;" target="_blank" title="Click to know more about USED">USED</a> <br/> `(1)/(2)xx` Eqts of `Na_(2)CO_(3)` = Eq of HCl <br/> `(1)/(2)xx(<a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a>)/(1000)xxM_(Na_(2)CO_(3))xx2=V_(1)xx1` <br/> `M_(NC_(2)CO_(3))=10V_(1)......(1)` <br/> When meOH used <br/> Eqts of `NC_(2)CO_(3)+` Eqts of `NaHCO_(3)` = Eq of HCl <br/> `((100)/(1000)xxM_(Na_(2)CO_(3))xx2)+((100)/(1000)xxM_(NCHCO_(3))xx1)=V_(2)xx1....(2)` <br/> On solving (1), (2) <br/> `M_(NaHCO_(3))=10(V_(2)-2V_(1))`</body></html> | |
| 50840. |
A 1.00 m^2 column of air extending from the earth's surface through the upper atmosphere has a mass of about x producing an atmospheric pressure. Then 'x' is |
| Answer» <html><body><p>10,300 <a href="https://interviewquestions.tuteehub.com/tag/kg-1063886" style="font-weight:bold;" target="_blank" title="Click to know more about KG">KG</a> <br/>10,300 <a href="https://interviewquestions.tuteehub.com/tag/gm-1008640" style="font-weight:bold;" target="_blank" title="Click to know more about GM">GM</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/1030-266406" style="font-weight:bold;" target="_blank" title="Click to know more about 1030">1030</a> gm <br/>1030 <a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a> </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> | |
| 50841. |
A 10.0 g piece of gallium (m=69.7) at 25.0^(@) c is placed in 10.0 g of H_(2)O"at" 55.0^(@)c what is the final temperature when this system comes to equilbium ? (Assume the specific heat capacity of liquid Ga is the same as that of solid Ga =0.37jg^((-1))"^(@)C^(-1)). |
| Answer» <html><body><p>`35.0^(@)<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>`<br/>`38.1^(@)C`<br/>`41.8^(@)C`<br/>`52.6^(@)`C</p>Answer :d</body></html> | |
| 50842. |
A 100 meter hollow tube of uniform thickness has two open ends X and Y. Ammonia gas is sent into the tube from end X and hydrogen chloride from end Y, simultaneously. At what distance from end 'X', the gases first meet to form a smoke ring? |
| 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> :Two gases diffuse in to the pipe from their respective ends. A, B and <a href="https://interviewquestions.tuteehub.com/tag/meet-1092928" style="font-weight:bold;" target="_blank" title="Click to know more about MEET">MEET</a> at a point .O.<br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_Al_CHE_XI_V01_B_C02_SLV_031_Q01.png" width="80%"/> <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 Grahm.s law <br/> `(r_(1))/(r_(2)) = sqrt((M_(2))/(M_(1))) = sqrt((36.5)/(17)) = 1.47` <br/> The <a href="https://interviewquestions.tuteehub.com/tag/ratio-13379" style="font-weight:bold;" target="_blank" title="Click to know more about RATIO">RATIO</a> of rate of diffusion is also equal to the ratio of distance travelled. <br/> `(1.47)/(1) = (l)/(100-1),` SOlving l = 59.5 <br/> The two gases first meet at a distance from end X = 59.5meter</body></html> | |
| 50843. |
A 10-L vessel filled with O_(2) at 30 K is connected to an open limb manometer containing glycerine. The level in the open limb was found to be higher than the other limb by 50 cm. Calculate the number of moles of the gas in the vessel (Given d_("glycerine")=2.72 g mL^(-1), d_(Hg)=13.6 g mL^(-1)) |
| Answer» <html><body><p></p>Solution :`(hxxdxxg)Hg=(hxxdxxg)_("<a href="https://interviewquestions.tuteehub.com/tag/glycerine-474196" style="font-weight:bold;" target="_blank" title="Click to know more about GLYCERINE">GLYCERINE</a>")` <br/> `hxx13.6=50xx2.72"or"h=10 <a href="https://interviewquestions.tuteehub.com/tag/cm-919986" style="font-weight:bold;" target="_blank" title="Click to know more about CM">CM</a>`<br/> `P_(gas)=(<a href="https://interviewquestions.tuteehub.com/tag/76-335558" style="font-weight:bold;" target="_blank" title="Click to know more about 76">76</a>+10)_(cm)=86 cm` <br/> `PV=nRT :. (86)/(76)xx10=n0.0821xx300"or"n=0.46" mole"`.</body></html> | |
| 50844. |
A 10L cylinder has helium at 8 atm and 32^@C. How many balloons of 2L each at 2atm and 32^@C can be filled with the gas available from the cylinder. |
| Answer» <html><body><p></p>Solution :The PVof balloon ` = <a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> xx 2 = 4` lit-<a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a><br/> Gas from the cylinder is available to fill in the <a href="https://interviewquestions.tuteehub.com/tag/balloons-891914" style="font-weight:bold;" target="_blank" title="Click to know more about BALLOONS">BALLOONS</a>, until pressureof gas in cylinder drops from 8 atm to 2 atm. <br/> The <a href="https://interviewquestions.tuteehub.com/tag/pv-593601" style="font-weight:bold;" target="_blank" title="Click to know more about PV">PV</a> of helium available ` = 10 xx (8-2) = 60` lit-atm <br/> Number of balloons filled ` = ("PV of the gas available")/("PV of the balloon")= (60)/(4) = <a href="https://interviewquestions.tuteehub.com/tag/15-274069" style="font-weight:bold;" target="_blank" title="Click to know more about 15">15</a>`</body></html> | |
| 50845. |
A 10 g sample of oxygen gas is taken in a container of volume 1 litre and is found to exert a pressure of 3 bar. Which of the following options is correct regarding speed of the molecuels? |
| Answer» <html><body><p>All the molecules are <a href="https://interviewquestions.tuteehub.com/tag/moving-548882" style="font-weight:bold;" target="_blank" title="Click to know more about MOVING">MOVING</a> at a same <a href="https://interviewquestions.tuteehub.com/tag/speed-1221896" style="font-weight:bold;" target="_blank" title="Click to know more about SPEED">SPEED</a> which is equal to 310 m/sec.<br/>`U_("avg")=300m//"sec"`<br/>`U_("mps")=300xxsqrt((<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)/(5)` m/sec.<br/>`U_("mps")=310` m/sec.</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> | |
| 50846. |
A 1.0 g sample of KCIO_3 was heated under such conditions that a part of it decomposed according to the equation: (i) 2KCIO_3to 2KCI+3O_2 and the remaining underwent a change according to the equation (ii) 4KCIO_3to 3KClO_4+ KCI. If the amount of oxygen evolved was 146.8 mL at S.T.R, calculate the percentage by weight of KClO_4 in the residue. |
| Answer» <html><body><p><br/></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/suppose-656311" style="font-weight:bold;" target="_blank" title="Click to know more about SUPPOSE">SUPPOSE</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> g of `KClO_3` decomposed according to the equation: <br/> `underset(2 xx 122.55 g)(2KClO_(3)) to underset(2 xx 74.55 g)(2KCl) + underset(3 xx 22.4 L "at S.T.P.")(3O_(2))` <br/> Therefore, `1-x`g of `KClO_(3)` decomposed according to the equation, <br/> `underset(4 xx 122.55 g)(4KClO_(3)) to underset(3 xx 138.55 g)(3KClO_(4)) + underset(74.55 g)(KCl)` <br/> From these equations, it is <a href="https://interviewquestions.tuteehub.com/tag/clear-408920" style="font-weight:bold;" target="_blank" title="Click to know more about CLEAR">CLEAR</a> that oxygen is evolved only in the first type of decomposition. <br/> `therefore 2 xx 122.55 g` of `KCIO_3` on decomposition gives `O_2 = 3 xx 22.4` L <br/> `therefore` x gm of `KCIO_3` on decomposition will give: <br/> `O_(2) = (3 xx 22.4)/(2 xx 122.55) xx x` L at S.T.P. <br/> `therefore` The percentage of `KClO_(4)` in the residue <br/> `=0.394/0.790 xx 100 = 49.9`</body></html> | |
| 50847. |
A 1.0 g sample of H_(2)O_(2) solution containing 'x' per cent by weight requires x mL of KMnO_(4) solution for complete oxidation under acidic conditions. Calculate the normality of the KMnO_(4) solution. |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :N//A</body></html> | |
| 50848. |
A 10 cm column of air is trapped by a column of mercury, 8 cm long, in a capillary tubehorizontally fixed as shown below, at 1 atm pressure. When the tube is held atcertain angle theta^@ with open end up, the weight of Hg is borne partially by the gas. Vertical Height of Hg is a measure of additional pressure on gas. When the tube's held at 45^@ with the horizontal with open end up the length of air column is |
| Answer» <html><body><p>`(76 xx 10)/(76 + 8/(sqrt2)) cm`<br/>`(76 xx 8/(sqrt2))/(76 + 10) cm`<br/>`76 xx 8/(sqrt2)`<br/>`10 xx 8/(sqrt2)` </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> :At `<a href="https://interviewquestions.tuteehub.com/tag/45-316951" style="font-weight:bold;" target="_blank" title="Click to know more about 45">45</a>^@C`, pressure on <a href="https://interviewquestions.tuteehub.com/tag/gas-1003521" style="font-weight:bold;" target="_blank" title="Click to know more about GAS">GAS</a> = `76 + h <a href="https://interviewquestions.tuteehub.com/tag/cos-935872" style="font-weight:bold;" target="_blank" title="Click to know more about COS">COS</a> <a href="https://interviewquestions.tuteehub.com/tag/theta-1412757" style="font-weight:bold;" target="_blank" title="Click to know more about THETA">THETA</a>` <br/> `= 76 + 8/(sqrt(2))` <br/> `P_1l_1 = P_2l_2 , 76 xx 10 = (76 + 8/(sqrt2))l_2`.</body></html> | |
| 50849. |
A 10 cm column of air is trapped by a column of mercury, 8 cm long, in a capillary tubehorizontally fixed as shown below, at 1 atm pressure. When the tube is held atcertain angle theta^@ with open end up, the weight of Hg is borne partially by the gas. Vertical Height of Hg is a measure of additional pressure on gas. The length of air column, when the tube is fixed vertically at the same temperature with open end up is . |
| Answer» <html><body><p>`(<a href="https://interviewquestions.tuteehub.com/tag/76-335558" style="font-weight:bold;" target="_blank" title="Click to know more about 76">76</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> 84)/10 cm`<br/>`(76 xx 10)/84 cm`<br/>`(84 xx 10)/76 cm`<br/>`(76 xx 10)/(84) cm` </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> :Boyle.s <a href="https://interviewquestions.tuteehub.com/tag/law-184" style="font-weight:bold;" target="_blank" title="Click to know more about LAW">LAW</a> , `P_1 V_1 = P_LV_L (or) P_1L_1 = P_2l_2` <br/> `implies 76 xx 10 - (76_1 + 8) l_2 implies l_2 = (76 xx 10)/(84) cm`</body></html> | |
| 50850. |
A 10 cm column of air is trapped by a column of mercury, 8 cm long, in a capillary tube horizantally fixed as shown below, at 1 atm pressure. When the tube is held at certain angle theta^(@), with open end up, the weight of Hg is borne partially by the gas. Vertical Height of Hg is a measure of additional pressure on gas: The length of air coloumn.when the tube is fixed vertically with opend end down at same temperature is |
| Answer» <html><body><p>`(76 <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/68-331600" style="font-weight:bold;" target="_blank" title="Click to know more about 68">68</a>)/10 `<br/>0 cm <br/>`(76 xx 10)/(84) `<br/>`(76 xx 10)/84 `</p>Solution :`underset(("horrizontal"))(p_1l_1) = underset(("<a href="https://interviewquestions.tuteehub.com/tag/vertically-3260386" style="font-weight:bold;" target="_blank" title="Click to know more about VERTICALLY">VERTICALLY</a> down"))(p_2l_2)` <br/> `76 xx 10 = (76 - 8) l_2 implies l_2 = (76 xx 10)/(68) cm`.</body></html> | |