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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.
| 51001. |
50 mL tincture of benzoin, an antiseptic solution contains 10 ml of benzoin. The volume percentage of benzoin is …………….. |
| Answer» <html><body><p></p>Solution :`20%` <br/> Volume percentage of <a href="https://interviewquestions.tuteehub.com/tag/benzoin-395725" style="font-weight:bold;" target="_blank" title="Click to know more about BENZOIN">BENZOIN</a>` = ("Volume of the benzoin")/("Volume of the solution in <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a>") xx 100` <br/> `= (10)/(<a href="https://interviewquestions.tuteehub.com/tag/50-322056" style="font-weight:bold;" target="_blank" title="Click to know more about 50">50</a>) xx 100 = 20%`</body></html> | |
| 51002. |
50 ml solution of H_(2)O_(2) was treated with excess KI (s) and the solution was acidified with acetic acid. The liberated I_(2) required 40 ml of 0.5 M Na_(2)S_(2)O_(3) solution for the end point using starch is indicator. Find the molairty and volume strength of the H_(2)O_(2) solution. |
| Answer» <html><body><p>1.12 gm lit<br/><a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>.24 gm/lit<br/>5.6 gm/lit <br/><a href="https://interviewquestions.tuteehub.com/tag/none-580659" style="font-weight:bold;" target="_blank" title="Click to know more about NONE">NONE</a> </p>Solution :Meq of `H_(2)O_(2)` M <a href="https://interviewquestions.tuteehub.com/tag/50-322056" style="font-weight:bold;" target="_blank" title="Click to know more about 50">50</a> ml <br/> = m.eq of Hypo = `40xx0.5=20` meq <br/> Now <a href="https://interviewquestions.tuteehub.com/tag/normality-1124423" style="font-weight:bold;" target="_blank" title="Click to know more about NORMALITY">NORMALITY</a> = `(20)/(50)=0.4N` <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> M=0.2M` <br/> `therefore` Volume strength = `Mxx11.2=0.2xx112=2.24gm//lit.`</body></html> | |
| 51003. |
50 mL sample of ozonised oxygen at NTP was passed through a solution of potassium iodide. The liberated iodine required 15 mL of 0.08N Na_(2)S_(2)O_(3) solution for complete titration. Calculate the volume of ozone at NTP in the given sample. |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/reactions-20919" style="font-weight:bold;" target="_blank" title="Click to know more about REACTIONS">REACTIONS</a> involved may be given as : <br/> `2KI+H_(2)O+O_(3) to 2KOH+I_(2)+O_(2) uarr`<br/> `I_(2)+2Na_(2)S_(2)O_(3) to 2NaI+Na_(2)S_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)O_(6)`<br/> 1 mole `O_(3)=2 mol e Na_(2)S_(2)O_(3)` <br/>No. 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 hypo `=("Mass")/("Molecular mass "(158))` <br/> `(ExxNxxV)/(1000xx158)` <br/> where, `E_(Na_(2)S_(2)O_(3))=158, <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> = 0.08, V=15` <br/> `therefore`No. of moles of hypo `=(158xx0.08xx15)/(1000xx158)=1.2xx10^(-3)` <br/> No. of moles of `O_(3)=(1)/(2)` mole of hypo <br/> `=(1)/(2)xx1.2xx10^(-3)` <br/> `=6xx10^(-4)` mole<br/> Volume of `O_(3)` at NTP =No. of moles`xx 22400` <br/> `=6xx10^(-4)xx22400` <br/> =13.44 mL at NTP</body></html> | |
| 51004. |
50 ml of water sample requires 6 ml of soap solution to produce a good lather. 1 ml of soap solution is equal to 0.001 g of CaCO_3 . Then the degree of hardness of water is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/120-270396" style="font-weight:bold;" target="_blank" title="Click to know more about 120">120</a> <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/>1.2 ppm <br/><a href="https://interviewquestions.tuteehub.com/tag/112-268456" style="font-weight:bold;" target="_blank" title="Click to know more about 112">112</a> ppm<br/> 2.4 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> :A</body></html> | |
| 51005. |
50 " mL of " water required 4 " mL of " (N)/(50)HCl for complete neutralisation. 200 " mL of " this water was then boiled with 10 " mL of " (N)/(10) soda reagent. After filtration, the filtrate and the washing were made up to 200 mL with distilled water 50 " mL of " this solution required 8.0 " mL of " (N)/(50) HCl for complete neutralisation. CAlculate the temporary and permanent hardness in ppm. |
| Answer» <html><body><p></p>Solution :` 4 " mL of " (N)/(50) Hcl=(4xx0.02xx50)/(1000)g CaCO_(3)` <br/> So, the temporary <a href="https://interviewquestions.tuteehub.com/tag/hardness-14971" style="font-weight:bold;" target="_blank" title="Click to know more about HARDNESS">HARDNESS</a>`=(4xx0.02xx50)/(1000xx50)g CaCO_(3)` <br/> `=80ppm` <br/> After boiling with `Na_(2)CO_(3)` and filtration, 8 " mL of " `(N)/(50)` HCl <br/> `-=(8)/(50)=0.16m" Eq of "(HCl)/(50mL) Na_(2)CO_(3)` <a href="https://interviewquestions.tuteehub.com/tag/filtrate-461136" style="font-weight:bold;" target="_blank" title="Click to know more about FILTRATE">FILTRATE</a> <br/> `-=(0.16xx200)/(50)m" Eq of "(HCl)/(200mL) of Na_(2)CO_(3)` filtrate <br/> `-=0.64 mEq` <br/> Initially, total m" Eq of "`Na_(2)CO_(3)-=10xx(1)/(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)=1mEq` <br/> Total mEw of `Na_(2)CO_(3)` reacted `=1-0.64=0.36 mEq` of weight of `Na_(2)CO_(3)=0.36xx10^(-3)xx50g CaCO_(3)` <br/> So <a href="https://interviewquestions.tuteehub.com/tag/permanent-590817" style="font-weight:bold;" target="_blank" title="Click to know more about PERMANENT">PERMANENT</a> hardness`=(0.36xx10^(-3)xx50xx10^(6))/(200)=90ppm` <br/> Hence, temporary hardness`=80ppm`, permanent hardness <br/> `=90ppm`</body></html> | |
| 51006. |
50 ml of tap water contains 20 mg of dissolved solids. The TDS value in ppm is …………… |
| Answer» <html><body><p></p>Solution :400 ppm <br/> The <a href="https://interviewquestions.tuteehub.com/tag/tds-659570" style="font-weight:bold;" target="_blank" title="Click to know more about TDS">TDS</a> value in ppm `= ("<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 the dissolved solids")/("Mass of water") <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^(<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/> `=(20 xx 10^(-3) <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>)/(50g) xx 10 ^(6) = 400 ppm`</body></html> | |
| 51007. |
50 ml of oxygen wre collected at 10^(@)C under 750 mm pressure. Calculate volume at STP. |
| Answer» <html><body><p></p>Solution :`v_(1)=50 ml. P_(1) =750 mm, P_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) =<a href="https://interviewquestions.tuteehub.com/tag/760-335611" style="font-weight:bold;" target="_blank" title="Click to know more about 760">760</a> mm` <br/> `v_(2) ? T=10+273=283, T_(1)=273*<a href="https://interviewquestions.tuteehub.com/tag/15-274069" style="font-weight:bold;" target="_blank" title="Click to know more about 15">15</a> K` <br/> `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)) or V_(2) (P_(1)V_(1)T_(2))/(P_(2)T_(1))=(750xx50xx273*15)/(760xx283)` <br/> `:. V_(2)=47*6247 ml`</body></html> | |
| 51008. |
50 ml of oxygen diffuses under certain conditions through a porous membrane. The volume of Hydrogen that diffuses in the same time under the same conditions is |
| Answer» <html><body><p>12.5 <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/><a href="https://interviewquestions.tuteehub.com/tag/25ml-298126" style="font-weight:bold;" target="_blank" title="Click to know more about 25ML">25ML</a> <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> <br/><a href="https://interviewquestions.tuteehub.com/tag/200ml-1820131" style="font-weight:bold;" target="_blank" title="Click to know more about 200ML">200ML</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> :D</body></html> | |
| 51009. |
50 mL of hydrogen diffuses out through a small hole from a vessel in 20 minutes, time needed for 40 mL of oxygen to diffuse out is: 1) 12 min2)64 min3)8 min4)32 min |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a> <a href="https://interviewquestions.tuteehub.com/tag/min-548008" style="font-weight:bold;" target="_blank" title="Click to know more about MIN">MIN</a><br/><a href="https://interviewquestions.tuteehub.com/tag/64-330663" style="font-weight:bold;" target="_blank" title="Click to know more about 64">64</a> min<br/>8 min<br/>32 min.</p>Solution :64 min</body></html> | |
| 51010. |
50 ml of H_(2),O is added to 50 ml of 1xx 10^(-3) M barium hydroxide solution. What is the P^(H) of the resulting solution? |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>.0` <br/>` 3.3` <br/>`<a href="https://interviewquestions.tuteehub.com/tag/11-267621" style="font-weight:bold;" target="_blank" title="Click to know more about 11">11</a>.7` <br/>` 11.0` </p>Solution :Solution is diluted ,` N_1 V_1 =N_2V_2`<br/>` 2 xx 10 ^(-3) xx <a href="https://interviewquestions.tuteehub.com/tag/50-322056" style="font-weight:bold;" target="_blank" title="Click to know more about 50">50</a> =N_2 xx 100` <br/> `N_2 = 10 ^(-3)rArr [OH^(-) ] =10 ^(-3) ` <br/>`<a href="https://interviewquestions.tuteehub.com/tag/poh-1145204" style="font-weight:bold;" target="_blank" title="Click to know more about POH">POH</a> = 3 thereforepH =11`</body></html> | |
| 51011. |
50 ml of given H_(2)O_(2) solution is added to excess KI solution in acidic medium. The liberated I_(2) requires 20 ml of 0.04 M standard Hypo solution. Weight of H_(2)O_(2) present in 250 ml of given solution is |
| Answer» <html><body><p>`0.034g`<br/>`0.068g`<br/>`0.136g`<br/>none</p>Solution :`2I^(-)+H_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)O_(2)+2H^(+)rarr2H_(2)O+I_(2)` <br/> `I_(2)+2S_(2)O_(3)^(-2)rarr2I^(-)+S_(4)O_(6)^(-2)` <br/> `W_(H_(2)O_(2))=<a href="https://interviewquestions.tuteehub.com/tag/8xx10-1931283" style="font-weight:bold;" target="_blank" title="Click to know more about 8XX10">8XX10</a>^(-3)<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>(250)/(<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>)xx34=0.068gm`</body></html> | |
| 51012. |
50 ml of given H_(2)O_(2) solution is added to excess KI solution in acidic medium. The liberated I_(2) requires 20 ml of 0.04 M standard Hypo solution. Molarity of H_(2)O_(2) solution is |
| Answer» <html><body><p>`8xx10^(-3)M` <br/>`4xx10^(-3)M`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/5xx10-1901302" style="font-weight:bold;" target="_blank" title="Click to know more about 5XX10">5XX10</a>^(-3)M` <br/>None of these</p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/2i-300397" style="font-weight:bold;" target="_blank" title="Click to know more about 2I">2I</a>^(-)+H_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)O_(2)+<a href="https://interviewquestions.tuteehub.com/tag/2h-300377" style="font-weight:bold;" target="_blank" title="Click to know more about 2H">2H</a>^(+)rarr2H_(2)O+I_(2)` <br/> `I_(2)+2S_(2)O_(3)^(-2)rarr2I^(-)+S_(4)O_(6)^(-2)` <br/> Eqts of `H_(2)O_(2)` = <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a> `Na_(2)S_(2)O_(3)` <br/> `(2xx50)/(1000)xxM=(1xx20)/(1000)xx0.04impliesM=8xx10^(-3)`</body></html> | |
| 51013. |
50 mL of an aqueous solution of H_(2)O_(2) was reacted with excess of KI solution and dilute H_(2)SO_(4) The liberated iodine requried 20 mL 0.1 N Na_(2)S_(2)O_(3) solution for complete interaction calculate the concentration of H_(2)O_(2) in gL^(-1) |
| 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> :`H_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)O_(2)^(-1)KI+H_(2)SO_(4)rarrK_(2)SO_(4)+I_(2)+2H_(2)O^(-2)` <br/> `Na_(2)S_(2)O_(3)+I_(2)rarrNa_(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>)+2` Nal <br/> Total change in O.N of O=2(-1)-2(-2)=2 `therefore` Eq of `H_(2)O_(2)=(2+2xx16)/(2)=<a href="https://interviewquestions.tuteehub.com/tag/17-278001" style="font-weight:bold;" target="_blank" title="Click to know more about 17">17</a>` <br/> Let `N_(1)` be the normality of `I_(2)` solution <br/> Since one equivalent of `H_(2)O` produces 1 equilvent of `I_(2)` <br/> `therefore` 50mL of `N_(1) I_(2)` solution =50mL of `N_(1)H_(2)O_(2)` solution <br/> `therefore N_(1)=(20xx0.1)/(50)=0.04` N or <a href="https://interviewquestions.tuteehub.com/tag/streangth-3079918" style="font-weight:bold;" target="_blank" title="Click to know more about STREANGTH">STREANGTH</a> of `H_(2)O_(2) "solution" =0.04xx17=0.68 gL^(-1)`</body></html> | |
| 51014. |
75 ml of gas A effuses through a pin hole in 73 seconds the same volume of SO_(2) under indential conditions effuses in seconds. Calculate the molecular mass of A. |
| Answer» <html><body><p></p>Solution :`("<a href="https://interviewquestions.tuteehub.com/tag/effusion-966747" style="font-weight:bold;" target="_blank" title="Click to know more about EFFUSION">EFFUSION</a> rate of" CO_(2)) /("Effusion rate of A") = <a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>((M_(A))/(M_(CO_(2))))` <br/> `(50/115)/(50/146) = sqrt((M_(A))/(44)) , (1.27)^(2) = (M_(A))/(44)` <br/> `M_(A) = 71 therefore` Molecular <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 A is 71 .</body></html> | |
| 51015. |
50 ml of gas A effuse through a pin -hole in 146 second . The same volume of CO_2 under identical condition effuse in 115 seconds . Calculate the molecular mass of A . |
| 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> :`("Effusion rate of" 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>)) /("Effusion rate of A") = <a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>((M_(A))/(M_(CO_(2))))` <br/> `(50/115)/(50/146) = sqrt((M_(A))/(44)) , (1.27)^(2) = (M_(A))/(44)` <br/> `M_(A) = 71 therefore` <a href="https://interviewquestions.tuteehub.com/tag/molecular-562994" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULAR">MOLECULAR</a> mass of A is 71 .</body></html> | |
| 51016. |
50 mL of dry ammonial gas was sparked for a long time in an eudiometer tube mercury. After sparking, the volume becomes 97 mL. After washing the gas with water and drying, the volume becomes 94 mL. This was mixed with 60.5 mL of oxygen and the mixture was burnt. After the completion of the combustion of H_(2), the volume of the residual gas was 48.75mL. Derive molecular formula of ammonia. |
| 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> :`NH_(<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> | |
| 51017. |
50 mL of 10 N H_(2)SO_(4), 25 mL of 12 N HCl and 40 mL of 5 N HNO_(3) are mixed and the volume of the mixture is made 1000 mL by adding water. The normality of the resulting solution will be : |
| 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> N <br/>2 N <br/>3 N <br/><a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a> N</p>Solution :`N_(1)V_(1)+N_(2)V_(2)+N_(3)V_(3)=N_(<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a> )xxV_(R )` <br/> `10xx50+12xx25+5xx40=N_(R )=xx1000` <br/> `500+300+200=N_(R )=<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>` <br/> `N_(R )=1` (Resulatnt normality)</body></html> | |
| 51018. |
50 mL of 0.2 M ammonia solution is treated with 25mL of 0.2 M HCl . If pK_(b) of ammonia solution is 4.75 , the pH of the mixture will be |
| Answer» <html><body><p>4.75<br/>3.75<br/>9.25<br/>8.25</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> | |
| 51019. |
50 ml of 0.2 N H_(2) SO_(4) is mixed with 100 ml of 0.4 N KOH solution and 1.85 lit of distilled water is added . The pH of resulting solution is (log 1.5 = 0.176) |
| Answer» <html><body><p>`13.301`<br/>`0.699`<br/>`1.824`<br/>`12.176`</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> | |
| 51020. |
50 ml of 0.05 M Na_(2)CO_(3) is titrated against 0.1 M HCl. On adding 40 ml of HCl, pHof the resulting solution will be [{:(H_(2)CO_(3):pK_(a_(1))=6.35,pK_(a_(2))=10.33),(log3=0.477,log2=0.30):}] |
| Answer» <html><body><p>`6.35`<br/>`6.526`<br/>`8.34`<br/>`6.173`</p>Solution :`CO_(3)^(2-) + <a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>^(-) rArr HCO_(3)^(-)`<br/> Initial millies <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> `50 xx 0.05 40 xx 0.1` <br/> Final <a href="https://interviewquestions.tuteehub.com/tag/milli-560824" style="font-weight:bold;" target="_blank" title="Click to know more about MILLI">MILLI</a> moles `1.5 , 2.5`<br/> `{:(,HCO_(3)^(-)+,H^(+),rarr,H_(2)CO_(3)),("Initial milli mles",2.5,1.5,,"__"),("Final milli moles",1,"___",,1.5):}` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/ph-1145128" style="font-weight:bold;" target="_blank" title="Click to know more about PH">PH</a> = pK_(a_(1)) + "log"([HCO_(3)^(-)])/([H_(2)CO_(3)]) = 6.173`</body></html> | |
| 51021. |
50 mL each of gases A and B take 150 and 200 seconds respectively for effusing through a pin- hole under the similar conditions . If molecular mass of B is 36 , the molecular mass of A will be nearly |
| Answer» <html><body><p>64<br/>96<br/>128<br/>20</p>Solution :`(V_(A))/(t_(A)) xx (t_(<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>))/(V_(B)) = sqrt((M_(B))/(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>))) , (50)/(150) xx (200)/(50) = sqrt((<a href="https://interviewquestions.tuteehub.com/tag/36-309156" style="font-weight:bold;" target="_blank" title="Click to know more about 36">36</a>)/(M_(A)))` <br/> `M_(A) = (36 xx 9)/(16) = 20.25`</body></html> | |
| 51022. |
50 ml H_(2)O_(2) is completely oxidised by 10 ml, 0.2 N KMnO_(4) solution in acidic medium. The strength of hydrogen peroxide is |
| Answer» <html><body><p>0.68 gm/lit<br/>`(M)/(50)`<br/>`(N)/(25)`<br/>`3.4%(w//v)` </p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a>. `H_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)O_(2)` = eq. `KMnO_(4)` <br/> `50xxN=10xx0.2` <br/> `N=(0.2)/(<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)=(1)/(25),M=(N)/(2)=(1)/(50)` <br/> `M=(S)/("M.wt.")impliesS=(1)/(50)xx34=0.68g//L`</body></html> | |
| 51023. |
50 litres of water containing Ca(HCO_3)_2 when converted into soft water required 22.2 g of Ca(OH)_2. Calculate the amount of Ca(HCO_3)_2 per litre of hard water. |
| 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> :The corresponding chemical equation is: <br/> `<a href="https://interviewquestions.tuteehub.com/tag/underset-3243992" style="font-weight:bold;" target="_blank" title="Click to know more about UNDERSET">UNDERSET</a>(162.116 g)(Ca(HCO_(3))_(2) + underset(74.096 g)(Ca(OH)_(2)) to 2CaCO_(3) + 2H_(2)O` <br/> `therefore74.096 g` of `Ca(OH)_(2)` react with `Ca(HCO_(3))_(2) = 162.116 g` <br/> `therefore 22.2` g of `Ca(OH)_(2)` will react with `Ca(HCO_(3))_(2)` <br/> `=(162.116)/(74.096) xx 22.2 = 48.6 g` <br/> This much `Ca (HCO_3)_2` is present in 50 litres of hard <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a>. <br/> `therefore`The amount of `Ca (HCO_3)_2` in one litre of hard water. <br/> `=48.06/50 = 0.972 g`</body></html> | |
| 51024. |
50 litre of dry N_(2) is passed through 36g of H_(2)O" at "27^(@)C. After passage of gas, there is a loss of 1.20g in water. Calculate vapour pressure of water. |
| Answer» <html><body><p></p>Solution :The <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a> vapours occupies the volume of `N_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)` gas i.e. 50 litre <br/> For `H_(2)O` vapour V=50 litre, w=1.20g, T=300 K, `m=18g <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a>^(-1)` <br/> `PV=w"/"mRT" or "Pxx50=1.2"/"18xx0.0821xx300` <br/> `:. P=0.03284 <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a> =24.95 mm`.</body></html> | |
| 51025. |
50 gm of sample of sodium hydroxide required for complete neutralisation, 1 litre 1 N HCl. What is the percentage purity of NaOH is |
| Answer» <html><body><p>50<br/>60<br/>70<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></p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a> = <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</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> GEW xx V` (in <a href="https://interviewquestions.tuteehub.com/tag/lit-537267" style="font-weight:bold;" target="_blank" title="Click to know more about LIT">LIT</a>) <br/> % of purity of `NaOH = (40)/(50) xx 100 = 80%`</body></html> | |
| 51026. |
5.0 g of H_(2)O_(2) is present in 100 mL of the solution. The molecular mass of H_(2)O_(2) is 34. The molarity of the solution is : |
| Answer» <html><body><p>1.5 M<br/>0.15 M <br/>3.0 M<br/>50 M</p>Solution :N//A</body></html> | |
| 51027. |
5.0 g of bleaching powder was suspended in water and volume made up to half a litre. 20 mL of this suspension when acidified with acetic acid and treated with excess of KI solution liberated iodine which required 20 mL of a decinormal hypo solution for titration. Calculate percentage of available chlorine in bleaching powder. |
| 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> | |
| 51028. |
50 g of a sample of NaOH required for complete neutralisation, 1 litre N HCl. What is the percentage purity of NaOH ? |
| Answer» <html><body><p>80<br/>70<br/>60<br/>50</p>Solution :N//A</body></html> | |
| 51029. |
50 g CaCO_(3) is allowed to dissociated in 22.4 lit vessel at 819^(@)C . If 50% of CaCO_(3) is left at equilibrium , active masses of CaCO_(3) , CaO and CO_2 respectively are |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/25-296426" style="font-weight:bold;" target="_blank" title="Click to know more about 25">25</a> <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> , <a href="https://interviewquestions.tuteehub.com/tag/14-272882" style="font-weight:bold;" target="_blank" title="Click to know more about 14">14</a> g , 1/22 .<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a> mol/lit <br/>1, 1, 1/89.6 mol/lit <br/>25, 14 , 1/89 .6 mol/lit <br/>1 , 1, 1</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> | |
| 51030. |
5.0 cm^(3) of H_(2)O_(2) liberates 0.508 g of iodine from an acidified KI solution. The strength of H_(2)O_(2) solution in terms of volume strenth at STP is |
| Answer» <html><body><p>6.48 volumes <br/>4.48 volumes <br/>7.68 volumes <br/>none of these </p>Solution :The reaction between `H_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)O_(2)` and the acidified <a href="https://interviewquestions.tuteehub.com/tag/ki-533104" style="font-weight:bold;" target="_blank" title="Click to know more about KI">KI</a> solution is <br/> `2KI+H_(2)SO_(4)+underse(5.0 cm^(3))underset(34 g)(H_(2)O_(2))rarr K_(2)SO_(4)+2H_(2)O+underset(0.508 g)underset(254 g)(I_(2))` <br/> Then, Mass of `H_(2)O_(2)` that liberates 0.508 g of `I_(2)` <br/> `= (34)/(254)xx0.508 = 0.068 g` <br/> This much `H_(2)O_(2)` is present in `5.0 cm^(3)`. <br/> Hydrogen <a href="https://interviewquestions.tuteehub.com/tag/peroxide-1151683" style="font-weight:bold;" target="_blank" title="Click to know more about PEROXIDE">PEROXIDE</a> decomposes as follows : <br/> `underset(2xx34 g=68 g)underset(2 mol)(2H_(2)O_(2))rarr 2H_(2)O underset(22400 cm^(3)"(at STP)")underset(1 mol)(+O_(2))` <br/> Then 68 g of `H_(2)O_(2)` <a href="https://interviewquestions.tuteehub.com/tag/gives-1007647" style="font-weight:bold;" target="_blank" title="Click to know more about GIVES">GIVES</a> 22400 `cm^(3)` of oxygen <br/> 1 g of `H_(2)O_(2)` gives `(22400 cm^(3))/(68)` of oxygen <br/> 0.068 g of `H_(2)O_(2)` gives `(22400 cm^(3))/(68 g)xx 0.068 g = 22.4 cm^(3)` of oxygen <br/> Thus, `5 cm^(3)` of `H_(2)O_(2)` gives `22.4 cm^(3)` of oxygen at STP<br/> So, `1 cm^(3)` of `H_(2)O_(2)` gives `(22.4)/(5)cm^(3)` of oxygen at STP `= 4.48 cm^(3)` at STP <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>, strength of the given `H_(2)O_(2)` sample = 4.48 volumes.</body></html> | |
| 51031. |
50 cm^(2)of 0.2 N HCl is titrated against 0.1 N NaOH solution. The titration is discontinued after is completed by adding 0.5 NKOH. The volume of KOH required for completing the titration is |
| Answer» <html><body><p>`12 cm^(3)`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> cm^(3)`<br/>`25 cm^(3)`<br/>`10.5cm^(3)`</p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/50-322056" style="font-weight:bold;" target="_blank" title="Click to know more about 50">50</a> cm^(3) ` of 0.2 <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> <a href="https://interviewquestions.tuteehub.com/tag/hcl-479502" style="font-weight:bold;" target="_blank" title="Click to know more about HCL">HCL</a> `=50xx0.2` millieq <br/> =10 millieq <br/> `50 cm^(3)` of 0.1 N NaOH = 5 millieq <br/> Remaining HCl = 10 - 5 millieq = 5 millieq <br/> It is neutralized by `V cm^(3)` of 0.5 N KOH i.e. `0.5xxV` millieq = 5 millieq <br/> or `V=10 cm^(3)`.</body></html> | |
| 51032. |
5-Oxohexanal is obtained by ozonolysis of |
| Answer» <html><body><p><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PR_CHE_02_XI_C13_E20_080_O01.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PR_CHE_02_XI_C13_E20_080_O02.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PR_CHE_02_XI_C13_E20_080_O03.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PR_CHE_02_XI_C13_E20_080_O04.png" width="30%"/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PR_CHE_02_XI_C13_E20_080_S01.png" width="80%"/></body></html> | |
| 51033. |
5 moles of SO_(2) and 5 moles of O_(2) react in a closed vessel. At equilibrium 60% of the SO_(2) is consumed . The total number of gaseous moles(SO_(2),O_(2)andSO_(3))in the vessel is :- |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>.1<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>.9<br/>10.5<br/>8.5</p>Solution :`{:(,2SO_(2),+,O_(2),hArr,2SO_(3)),("<a href="https://interviewquestions.tuteehub.com/tag/initial-499269" style="font-weight:bold;" target="_blank" title="Click to know more about INITIAL">INITIAL</a> moles",5,,5,,<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>),("At equilibrium",5-3,,5-(3)/(2),,3(5xx(60)/(<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>)=3)):}` <br/> Total number of moles in the vessel`=2+3.5+3=8.5`</body></html> | |
| 51034. |
5 moles of SO_(2)and 5 " molesof " O_(2)are allowed to react . At equilibrium , it was found that 60%SO_(2) is used up . If the pressure of the mixture is one atmosphere, the partialpressureof O_(2) is |
| Answer» <html><body><p>`0*52` atm<br/>`0*21` atm<br/>` 0*41`atm<br/>`0*82` atm</p>Solution :` {:(,<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> SO_(2)(g),+,O_(2)(g),hArr,2 SO_(3)(g)),("Intial",<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>,,5,,0):}` <br/> As 60 % `SO_(2)` is used up, no of moles of `SO_(2) "used up" = 60/100 xx 5 =3`<br/> ` :. " No. of moles of " SO_(2) " at equilibrium " = 5-3=2` <br/> As 2 moles of `SO_(2) " <a href="https://interviewquestions.tuteehub.com/tag/react-613674" style="font-weight:bold;" target="_blank" title="Click to know more about REACT">REACT</a> with <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 " O_(2)` <br/>`=1/2 xx 3 = 1.5 " moles" ` <br/> i.e. No. of moles of `O_(2)` at equilibrium<br/> `=5-1*5 = 3.5 ` <br/> As 2 moles of `SO_(2) " produce 2 molesof " SO_(3) ` <br/> `:. " No . of moles of "SO_(3)" equilibrium = 3 moles"`<br/> `= 2 + 3*5 + 3 = 8*5 ` <br/> As 2 moles of `SO_(2)"produce 2 moles of"SO_(3)` <br/> ` :. " No. of moles of " SO_(3) " at equilibrium" = 3 moles ` <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> no. of moles at equilibrium "` <br/></body></html> | |
| 51035. |
5 moles of dihydrogen react with 5 moles ofdinitrogen calculate the mole of Ammonia product. |
| Answer» <html><body><p> <br/> <br/> <br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<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/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a></body></html> | |
| 51036. |
5 moles of Ba(OH)_(2) are treated with excess of CO_(2). How much Ba(OH_(2) will be formed? |
| Answer» <html><body><p>39.4 g<br/>197 g<br/>591 g<br/>985g<br/></p>Solution :d) `<a href="https://interviewquestions.tuteehub.com/tag/ba-389206" style="font-weight:bold;" target="_blank" title="Click to know more about BA">BA</a>(OH)_(2) + CO_(2) to BaCO_(3) + H_(2)O` <br/> `therefore` <a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a> moles of `Ba(OH)_(3)` = 5 moles of `BaCO_(3)` <br/> `therefore` <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 `BaCO_(3)` = Moles of BaCO_(3) <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> Molecular mass of `BaCO_(3)` = `5 xx 197) = 985 g</body></html> | |
| 51037. |
5 moles of a gas in a closed vessel was heated from 300 K to 600 K. The pressure of the gas doubled. The number of moles of the gas will be |
| Answer» <html><body><p>5<br/>2.5<br/>10<br/>20</p>Answer :A</body></html> | |
| 51038. |
5 moles of A, 6 molesof Z are mixed with suffifcient amount of C to finally produce F. Then find the maximum molesof F which can be produced. Assuming that the product formed can also be reused. Reactions are : A + 2Z rightarrow B B + C rightarrow Z + F |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a> <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> <br/>4.<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a> moles <br/>5 moles<br/>6 moles </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> | |
| 51039. |
5 moles gas are introduced in 1 litre container at 47^(@)C. Select the correct option(s). [R=0.08litre-atm/mol-K] |
| Answer» <html><body><p>Pressure would be 128 atm if it behaves ideally <br/>pressure would be <a href="https://interviewquestions.tuteehub.com/tag/28-299271" style="font-weight:bold;" target="_blank" title="Click to know more about 28">28</a> atm if it follows <a href="https://interviewquestions.tuteehub.com/tag/van-723180" style="font-weight:bold;" target="_blank" title="Click to know more about VAN">VAN</a> der <a href="https://interviewquestions.tuteehub.com/tag/waals-732989" style="font-weight:bold;" target="_blank" title="Click to know more about WAALS">WAALS</a>' equation, a=4 atm-`"litre"^(2)//"<a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a>"^(2)` and b=0<br/>Pressure would be 33.33 atm if it follows van der Waals' equation , a=4 atm-`"litre"^(2)//"mol"^(2)` and b=0.04L/mole<br/>Pressure would be 160 atm if it follows van der <a href="https://interviewquestions.tuteehub.com/tag/waal-732986" style="font-weight:bold;" target="_blank" title="Click to know more about WAAL">WAAL</a>'s equation, a=0atm`-"litre"^(2)//"mol"^(2)` and b=0.04L/mole</p>Answer :abd</body></html> | |
| 51040. |
5 mole of an ideal gas expands isothermally and irreversibly from a pressure of 10 atm to 1 atm against a contant external pressure of 1 atm. W_("irr") at 300K is: |
| Answer» <html><body><p>`-15.921kJ`<br/>`-11.224kJ`<br/>`-110.83kJ`<br/>none of these</p>Solution :`W_("irr") = - P_("ext") (<a href="https://interviewquestions.tuteehub.com/tag/delta-947703" style="font-weight:bold;" target="_blank" title="Click to know more about DELTA">DELTA</a> V) = -1 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> ((nRT)/(P_(2)) - (nRT)/(P_(1)))` <br/> `= -1 xx (<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a> xx 0.0821 xx 300) ((1)/(1) - (1)/(10))` <br/> `= -110.835` L-atm = 11.22 <a href="https://interviewquestions.tuteehub.com/tag/kj-1063034" style="font-weight:bold;" target="_blank" title="Click to know more about KJ">KJ</a></body></html> | |
| 51041. |
5 mL water in watch glass and 10 mL in beaker which water will disappear first ? Why ? |
| Answer» <html><body><p></p>Solution :The water of watch <a href="https://interviewquestions.tuteehub.com/tag/glass-14164" style="font-weight:bold;" target="_blank" title="Click to know more about GLASS">GLASS</a> will <a href="https://interviewquestions.tuteehub.com/tag/disappear-440374" style="font-weight:bold;" target="_blank" title="Click to know more about DISAPPEAR">DISAPPEAR</a> <a href="https://interviewquestions.tuteehub.com/tag/first-461760" style="font-weight:bold;" target="_blank" title="Click to know more about FIRST">FIRST</a> as the <a href="https://interviewquestions.tuteehub.com/tag/amount-374803" style="font-weight:bold;" target="_blank" title="Click to know more about AMOUNT">AMOUNT</a> of water is less so <a href="https://interviewquestions.tuteehub.com/tag/rate-1177476" style="font-weight:bold;" target="_blank" title="Click to know more about RATE">RATE</a> of vaporization will be more.</body></html> | |
| 51042. |
5 mL of N HCl, 20 mL of N//20 H_(2)SO_(4) and 30 mL of N//3 HNO_(3) are mixed together and volume made to one litre. The normality of the resulting solution is : |
| Answer» <html><body><p>`N//5`<br/>`N//10`<br/>`N//20`<br/>`N//40`</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> | |
| 51043. |
5 mL of N HCl, 20 mL of N/2 H_(2)SO_(4) and 30 mL of N/3HNO_(3) are mixed together and volume made to 1L. The normality of resulting solution is |
| Answer» <html><body><p>0.45<br/>0.025<br/>0.9<br/>0.05<br/></p>Solution :Normally <a href="https://interviewquestions.tuteehub.com/tag/equtaion-2619042" style="font-weight:bold;" target="_blank" title="Click to know more about EQUTAION">EQUTAION</a> is <br/> `N_(1)V_(1)+N_(<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)+N_(3)V_(3)=N_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)(V_(1)+V_(2)+V_(3))` <br/> or `1xx5+20xx(1)/(2)+30xx(1)/(3)=N_(4)(5+20+30)` <br/> `(HCI) (H_(2)SO_(4)) (HNO_(3))` <br/> `therefore` Resulting normality `(N_(4))=(<a href="https://interviewquestions.tuteehub.com/tag/25-296426" style="font-weight:bold;" target="_blank" title="Click to know more about 25">25</a>)/(55)=0.45N`.</body></html> | |
| 51044. |
5 mL of N HCI, 20 mL of N/2 H_2SO_4 and 30 mL ofN/3 HNO_3 are mixed together and the volume made to 1 litre. The normality of the resulting solution is: |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>/5`<br/>`N/10`<br/>`N/20`<br/>`N/40` </p>Solution :`N_(1)V_(1) + N_(2)V_(2) + N_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)V_(3) = NV, V = 1000 mL` (<a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a>)<br/> `1 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 5 + 1/2 xx 20 + 1/3 xx 30 = N xx 1000` <br/> `N = 1/40 N`</body></html> | |
| 51045. |
5 mL of 8 N HNO_(3), 4.8 mL of 5 N HCl, and a certain volume of 17 m H_(2) SO_(4) are mixed together and made upto 2 L. 30 mL of the acid mixture exactly neutralises 42.9 mL of Na_(2 CO_(3) solution containing 0.1 g of Na_(2) CO_(3). 10 H_(2) O in 10 mL of water. Calculate: a. The volume of H_(2) SO_(4) added to the mixture. b. The amount (in g) of the sulphate ions in the solution. |
| Answer» <html><body><p></p>Solution :mEq of acid mixture <br/> `= mEq of HNO_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>) + mEq of HCl + mEq of H_(2) SO_(4)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/let-11597" style="font-weight:bold;" target="_blank" title="Click to know more about LET">LET</a> `N` be the normality of the acid mixture and `<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a> mL` be the volume of `H_(2) SO_(4)` added. <br/> `N xx 200= 8 xx 5 xx 4.8 + 2 (17) xx V` <br/> Now find `N` of carbonate as follows: <br/> `implies N = ("Strength")/(Ew)` <br/> Strength `= 0.1//10 mL -= 10 g L^(-1)` <br/> `Ew = Mw//2 = 286//2 = 143` <br/> (`Mw = <a href="https://interviewquestions.tuteehub.com/tag/106-266627" style="font-weight:bold;" target="_blank" title="Click to know more about 106">106</a> + 180`, adding the mass of `10 H_(2) O`) <br/> `N = (10)/(143)` <br/> Now mEq of acid mixture = mEq of `Na_(2)CO_(3)` solution `N xx 30 = 42.9 xx (10)/(143)` <br/> `implies N = 0.1 =` Normality of acid mixture <br/> Substituting in equation (i) we get: <br/> `0.1 xx 2000 = (<a href="https://interviewquestions.tuteehub.com/tag/40-314386" style="font-weight:bold;" target="_blank" title="Click to know more about 40">40</a> + 24 + 34) V_(mL)` <br/> For grams of sulphate ions: <br/> mEq of `H_(2) SO_(4) = 2 xx 17 xx V = 136``(VmL = 4)` <br/> Now, mEq of `SO_(4)^(2-) = mEq of H_(2) SO_(4)` <br/> `("Weight")/(E_(SO_(4)^(2-))) xx 1000 = 136` <br/> `("Weight")/(96//2) xx 1000 = 136` <br/> Weight = Grams of `SO_(4)^(2-)` ions `= 6.53 g`</body></html> | |
| 51046. |
5 mL of a gaseous hydrocarbon was exposed to 30 mL measure 25 mL of which 10 mL are absorbed by NaOH and the remainder by pyrogallol. Determine molecular formula of hydrocarbon. All measurements are made at constant pressure and temperature. |
| 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/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/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)`;</body></html> | |
| 51047. |
5 kg of water contain 24 mg of MgSO_4and 32.4 mg of Ca(HCO_3)_2 . The total hardness of water is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a> PPM<br/> 6 PPM <br/> 8 PPM<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> 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> | |
| 51048. |
5% ionization is occur in 0.01 M CH_3COOH solution. Calculate its dissociation constant. |
| 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> :`2.63xx10^(-<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> | |
| 51049. |
5 g of CuSO_(4). 5H_(2)O is intendedto be prepared by using CuO and four times the stoichiometric amount of H_(2)SO_(4). Assuming that 10% of the material is lost in crystallisation, what weight of oxide should be taken and how many litre of mL of a 5 M H_(2)SO_(4)? |
| 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> :`w=1.753g, 4.41 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a>`;</body></html> | |
| 51050. |
5 g of an unknown hgas has pressure P at temperature TK in a vessel. On increasing the temperature by 50^(@)C, 1g of the was given out to maintain the pressure P. What was the original temperature of the gas ? |
| 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> :Initially,`T_(1)=T,P_(1)=P,V_(1)=<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a> `(Volume of the <a href="https://interviewquestions.tuteehub.com/tag/vessel-1445872" style="font-weight:bold;" target="_blank" title="Click to know more about VESSEL">VESSEL</a>), `n_(1)=(5)/(M)`<a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a><br/> Finally,`T_(2)=(T+50)K, P_(2)=P, V_(2)=V, n_(2)=(5-1)/(M)=(4)/(M)` mole <br/> (Increase of `50^(@)C`=increase of 50 K because size of `1^(@)C`=1 K) <br/> ApplyingPV=nRT <br/> `n_(1)T_(1)=n_(2)T_(2)` <br/> `(5)/(M)xxT=(4)/(M)(T+50)"or"5T=4T+200"orT=200 K`</body></html> | |