This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your 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 …………….. |
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Answer» Solution :`20%` Volume percentage of BENZOIN` = ("Volume of the benzoin")/("Volume of the solution in ML") xx 100` `= (10)/(50) xx 100 = 20%` |
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| 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. |
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Answer» 1.12 gm lit = m.eq of Hypo = `40xx0.5=20` meq Now NORMALITY = `(20)/(50)=0.4N` `IMPLIES M=0.2M` `therefore` Volume strength = `Mxx11.2=0.2xx112=2.24gm//lit.` |
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| 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. |
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Answer» Solution :REACTIONS involved may be given as : `2KI+H_(2)O+O_(3) to 2KOH+I_(2)+O_(2) uarr` `I_(2)+2Na_(2)S_(2)O_(3) to 2NaI+Na_(2)S_(4)O_(6)` 1 mole `O_(3)=2 mol e Na_(2)S_(2)O_(3)` No. of MOLES of hypo `=("Mass")/("Molecular mass "(158))` `(ExxNxxV)/(1000xx158)` where, `E_(Na_(2)S_(2)O_(3))=158, N = 0.08, V=15` `therefore`No. of moles of hypo `=(158xx0.08xx15)/(1000xx158)=1.2xx10^(-3)` No. of moles of `O_(3)=(1)/(2)` mole of hypo `=(1)/(2)xx1.2xx10^(-3)` `=6xx10^(-4)` mole Volume of `O_(3)` at NTP =No. of moles`xx 22400` `=6xx10^(-4)xx22400` =13.44 mL at NTP |
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| 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 |
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Answer» 120 PPM |
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| 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. |
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Answer» Solution :` 4 " mL of " (N)/(50) Hcl=(4xx0.02xx50)/(1000)g CaCO_(3)` So, the temporary HARDNESS`=(4xx0.02xx50)/(1000xx50)g CaCO_(3)` `=80ppm` After boiling with `Na_(2)CO_(3)` and filtration, 8 " mL of " `(N)/(50)` HCl `-=(8)/(50)=0.16m" Eq of "(HCl)/(50mL) Na_(2)CO_(3)` FILTRATE `-=(0.16xx200)/(50)m" Eq of "(HCl)/(200mL) of Na_(2)CO_(3)` filtrate `-=0.64 mEq` Initially, total m" Eq of "`Na_(2)CO_(3)-=10xx(1)/(10)=1mEq` 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)` So PERMANENT hardness`=(0.36xx10^(-3)xx50xx10^(6))/(200)=90ppm` Hence, temporary hardness`=80ppm`, permanent hardness `=90ppm` |
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| 51006. |
50 ml of tap water contains 20 mg of dissolved solids. The TDS value in ppm is …………… |
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Answer» Solution :400 ppm The TDS value in ppm `= ("MASS of the dissolved solids")/("Mass of water") XX 10^(6)` `=(20 xx 10^(-3) G)/(50g) xx 10 ^(6) = 400 ppm` |
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| 51007. |
50 ml of oxygen wre collected at 10^(@)C under 750 mm pressure. Calculate volume at STP. |
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Answer» Solution :`v_(1)=50 ml. P_(1) =750 mm, P_(2) =760 mm` `v_(2) ? T=10+273=283, T_(1)=273*15 K` `(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)` `:. V_(2)=47*6247 ml` |
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| 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» ANSWER :D | |
| 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» Solution :64 min | |
| 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? |
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Answer» Solution :Solution is diluted ,` N_1 V_1 =N_2V_2` ` 2 xx 10 ^(-3) xx 50 =N_2 xx 100` `N_2 = 10 ^(-3)rArr [OH^(-) ] =10 ^(-3) ` `POH = 3 thereforepH =11` |
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| 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 |
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Answer» `0.034g` `I_(2)+2S_(2)O_(3)^(-2)rarr2I^(-)+S_(4)O_(6)^(-2)` `W_(H_(2)O_(2))=8XX10^(-3)XX(250)/(1000)xx34=0.068gm` |
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| 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 |
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Answer» `8xx10^(-3)M` `I_(2)+2S_(2)O_(3)^(-2)rarr2I^(-)+S_(4)O_(6)^(-2)` Eqts of `H_(2)O_(2)` = EQ `Na_(2)S_(2)O_(3)` `(2xx50)/(1000)xxM=(1xx20)/(1000)xx0.04impliesM=8xx10^(-3)` |
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| 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) |
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Answer» `Na_(2)S_(2)O_(3)+I_(2)rarrNa_(2)S_(4)O_(6)+2` Nal Total change in O.N of O=2(-1)-2(-2)=2 `therefore` Eq of `H_(2)O_(2)=(2+2xx16)/(2)=17` Let `N_(1)` be the normality of `I_(2)` solution Since one equivalent of `H_(2)O` produces 1 equilvent of `I_(2)` `therefore` 50mL of `N_(1) I_(2)` solution =50mL of `N_(1)H_(2)O_(2)` solution `therefore N_(1)=(20xx0.1)/(50)=0.04` N or STREANGTH of `H_(2)O_(2) "solution" =0.04xx17=0.68 gL^(-1)` |
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| 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. |
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Answer» Solution :`("EFFUSION rate of" CO_(2)) /("Effusion rate of A") = SQRT((M_(A))/(M_(CO_(2))))` `(50/115)/(50/146) = sqrt((M_(A))/(44)) , (1.27)^(2) = (M_(A))/(44)` `M_(A) = 71 therefore` Molecular MASS of A is 71 . |
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| 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 . |
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Answer» SOLUTION :`("Effusion rate of" CO_(2)) /("Effusion rate of A") = SQRT((M_(A))/(M_(CO_(2))))` `(50/115)/(50/146) = sqrt((M_(A))/(44)) , (1.27)^(2) = (M_(A))/(44)` `M_(A) = 71 therefore` MOLECULAR mass of A is 71 . |
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| 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. |
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Answer» |
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| 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 : |
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Answer» Solution :`N_(1)V_(1)+N_(2)V_(2)+N_(3)V_(3)=N_(R )xxV_(R )` `10xx50+12xx25+5xx40=N_(R )=xx1000` `500+300+200=N_(R )=1000` `N_(R )=1` (Resulatnt normality) |
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| 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 |
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Answer» 4.75 |
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| 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) |
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Answer» `13.301` |
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| 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):}] |
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Answer» `6.35` Initial millies MOLES `50 xx 0.05 40 xx 0.1` Final MILLI moles `1.5 , 2.5` `{:(,HCO_(3)^(-)+,H^(+),rarr,H_(2)CO_(3)),("Initial milli mles",2.5,1.5,,"__"),("Final milli moles",1,"___",,1.5):}` `PH = pK_(a_(1)) + "log"([HCO_(3)^(-)])/([H_(2)CO_(3)]) = 6.173` |
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| 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 |
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Answer» 64 `M_(A) = (36 xx 9)/(16) = 20.25` |
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| 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 |
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Answer» 0.68 gm/lit `50xxN=10xx0.2` `N=(0.2)/(5)=(1)/(25),M=(N)/(2)=(1)/(50)` `M=(S)/("M.wt.")impliesS=(1)/(50)xx34=0.68g//L` |
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| 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. |
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Answer» `UNDERSET(162.116 g)(Ca(HCO_(3))_(2) + underset(74.096 g)(Ca(OH)_(2)) to 2CaCO_(3) + 2H_(2)O` `therefore74.096 g` of `Ca(OH)_(2)` react with `Ca(HCO_(3))_(2) = 162.116 g` `therefore 22.2` g of `Ca(OH)_(2)` will react with `Ca(HCO_(3))_(2)` `=(162.116)/(74.096) xx 22.2 = 48.6 g` This much `Ca (HCO_3)_2` is present in 50 litres of hard WATER. `therefore`The amount of `Ca (HCO_3)_2` in one litre of hard water. `=48.06/50 = 0.972 g` |
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| 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. |
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Answer» Solution :The WATER vapours occupies the volume of `N_(2)` gas i.e. 50 litre For `H_(2)O` vapour V=50 litre, w=1.20g, T=300 K, `m=18g MOL^(-1)` `PV=w"/"mRT" or "Pxx50=1.2"/"18xx0.0821xx300` `:. P=0.03284 ATM =24.95 mm`. |
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| 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 |
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Answer» 50 % of purity of `NaOH = (40)/(50) xx 100 = 80%` |
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| 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 : |
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Answer» 1.5 M |
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| 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. |
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Answer» |
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| 51028. |
50 g of a sample of NaOH required for complete neutralisation, 1 litre N HCl. What is the percentage purity of NaOH ? |
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Answer» 80 |
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| 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 |
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Answer» 25 G , 14 g , 1/22 .4 mol/lit |
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| 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 |
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Answer» 6.48 volumes `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))` Then, Mass of `H_(2)O_(2)` that liberates 0.508 g of `I_(2)` `= (34)/(254)xx0.508 = 0.068 g` This much `H_(2)O_(2)` is present in `5.0 cm^(3)`. Hydrogen PEROXIDE decomposes as follows : `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))` Then 68 g of `H_(2)O_(2)` GIVES 22400 `cm^(3)` of oxygen 1 g of `H_(2)O_(2)` gives `(22400 cm^(3))/(68)` of oxygen 0.068 g of `H_(2)O_(2)` gives `(22400 cm^(3))/(68 g)xx 0.068 g = 22.4 cm^(3)` of oxygen Thus, `5 cm^(3)` of `H_(2)O_(2)` gives `22.4 cm^(3)` of oxygen at STP 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 THEREFORE, strength of the given `H_(2)O_(2)` sample = 4.48 volumes. |
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| 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 |
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Answer» `12 cm^(3)` =10 millieq `50 cm^(3)` of 0.1 N NaOH = 5 millieq Remaining HCl = 10 - 5 millieq = 5 millieq It is neutralized by `V cm^(3)` of 0.5 N KOH i.e. `0.5xxV` millieq = 5 millieq or `V=10 cm^(3)`. |
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| 51032. |
5-Oxohexanal is obtained by ozonolysis of |
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Answer»
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| 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 :- |
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Answer» Solution :`{:(,2SO_(2),+,O_(2),hArr,2SO_(3)),("INITIAL moles",5,,5,,0),("At equilibrium",5-3,,5-(3)/(2),,3(5xx(60)/(1000)=3)):}` Total number of moles in the vessel`=2+3.5+3=8.5` |
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| 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 |
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Answer» `0*52` atm As 60 % `SO_(2)` is used up, no of moles of `SO_(2) "used up" = 60/100 xx 5 =3` ` :. " No. of moles of " SO_(2) " at equilibrium " = 5-3=2` As 2 moles of `SO_(2) " REACT with 1 mole of " O_(2)` `=1/2 xx 3 = 1.5 " moles" ` i.e. No. of moles of `O_(2)` at equilibrium `=5-1*5 = 3.5 ` As 2 moles of `SO_(2) " produce 2 molesof " SO_(3) ` `:. " No . of moles of "SO_(3)" equilibrium = 3 moles"` `= 2 + 3*5 + 3 = 8*5 ` As 2 moles of `SO_(2)"produce 2 moles of"SO_(3)` ` :. " No. of moles of " SO_(3) " at equilibrium" = 3 moles ` `:. "TOTAL no. of moles at equilibrium "` |
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| 51035. |
5 moles of dihydrogen react with 5 moles ofdinitrogen calculate the mole of Ammonia product. |
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Answer» |
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| 51036. |
5 moles of Ba(OH)_(2) are treated with excess of CO_(2). How much Ba(OH_(2) will be formed? |
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Answer» 39.4 g `therefore` 5 moles of `Ba(OH)_(3)` = 5 moles of `BaCO_(3)` `therefore` MASS of `BaCO_(3)` = Moles of BaCO_(3) XX Molecular mass of `BaCO_(3)` = `5 xx 197) = 985 g |
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| 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 |
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Answer» 5 |
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| 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 |
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Answer» 3 MOLES |
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| 51039. |
5 moles gas are introduced in 1 litre container at 47^(@)C. Select the correct option(s). [R=0.08litre-atm/mol-K] |
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Answer» Pressure would be 128 atm if it behaves ideally |
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| 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: |
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Answer» `-15.921kJ` `= -1 xx (5 xx 0.0821 xx 300) ((1)/(1) - (1)/(10))` `= -110.835` L-atm = 11.22 KJ |
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| 51041. |
5 mL water in watch glass and 10 mL in beaker which water will disappear first ? Why ? |
| Answer» Solution :The water of watch GLASS will DISAPPEAR FIRST as the AMOUNT of water is less so RATE of vaporization will be more. | |
| 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 : |
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Answer» `N//5` |
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| 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 |
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Answer» 0.45 `N_(1)V_(1)+N_(2)V_(2)+N_(3)V_(3)=N_(4)(V_(1)+V_(2)+V_(3))` or `1xx5+20xx(1)/(2)+30xx(1)/(3)=N_(4)(5+20+30)` `(HCI) (H_(2)SO_(4)) (HNO_(3))` `therefore` Resulting normality `(N_(4))=(25)/(55)=0.45N`. |
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| 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: |
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Answer» `N/5` `1 XX 5 + 1/2 xx 20 + 1/3 xx 30 = N xx 1000` `N = 1/40 N` |
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| 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. |
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Answer» Solution :mEq of acid mixture `= mEq of HNO_(3) + mEq of HCl + mEq of H_(2) SO_(4)` LET `N` be the normality of the acid mixture and `V mL` be the volume of `H_(2) SO_(4)` added. `N xx 200= 8 xx 5 xx 4.8 + 2 (17) xx V` Now find `N` of carbonate as follows: `implies N = ("Strength")/(Ew)` Strength `= 0.1//10 mL -= 10 g L^(-1)` `Ew = Mw//2 = 286//2 = 143` (`Mw = 106 + 180`, adding the mass of `10 H_(2) O`) `N = (10)/(143)` Now mEq of acid mixture = mEq of `Na_(2)CO_(3)` solution `N xx 30 = 42.9 xx (10)/(143)` `implies N = 0.1 =` Normality of acid mixture Substituting in equation (i) we get: `0.1 xx 2000 = (40 + 24 + 34) V_(mL)` For grams of sulphate ions: mEq of `H_(2) SO_(4) = 2 xx 17 xx V = 136``(VmL = 4)` Now, mEq of `SO_(4)^(2-) = mEq of H_(2) SO_(4)` `("Weight")/(E_(SO_(4)^(2-))) xx 1000 = 136` `("Weight")/(96//2) xx 1000 = 136` Weight = Grams of `SO_(4)^(2-)` ions `= 6.53 g` |
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| 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. |
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Answer» |
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| 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» Answer :C | |
| 51048. |
5% ionization is occur in 0.01 M CH_3COOH solution. Calculate its dissociation constant. |
| Answer» SOLUTION :`2.63xx10^(-5)` | |
| 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)? |
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Answer» |
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| 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 ? |
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Answer» <P> SOLUTION :Initially,`T_(1)=T,P_(1)=P,V_(1)=V `(Volume of the VESSEL), `n_(1)=(5)/(M)`MOLEFinally,`T_(2)=(T+50)K, P_(2)=P, V_(2)=V, n_(2)=(5-1)/(M)=(4)/(M)` mole (Increase of `50^(@)C`=increase of 50 K because size of `1^(@)C`=1 K) ApplyingPV=nRT `n_(1)T_(1)=n_(2)T_(2)` `(5)/(M)xxT=(4)/(M)(T+50)"or"5T=4T+200"orT=200 K` |
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