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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.
| 51151. |
35ml of dry H_2 gas were collected at 6^@C and 758 mm pressure. What is the volume in ml of H_2 at STP? |
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Answer» 32 |
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| 51152. |
350 cm^3 of oxygen and 275cm^3 of another gas 'A' diffused in same time under similar conditions. Find the molecualr mass of the gas 'A'. |
| Answer» SOLUTION :51.86 G.`MOL^(-1)` | |
| 51153. |
350 cm^3 of CH4and 175cm^3 of an unknown gas 'A' diffused in the same time under similar conditions. The molecular mass of gas A is |
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Answer» 32 |
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| 51154. |
35 mL sample lof hydrogen peroxide gives off 494 mL of O_(2) at 27^(@) C and 1 atm pressure. Volume strength of H_(2)O_(2) sample will be : |
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Answer» 10 V |
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| 51155. |
3.5 grams aluminium bronze was heated in aqueous hydroxhloric solution. The hydrogen liberated measured 4.15L at STP. What is the percentage of copper in the given alloy? |
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Answer» |
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| 51156. |
3.45 g of a metallic carbonate were mixed with 240 mL of N//4 HCl. The excess acid was neutralised by 50 mL of N//5 KOH solution. Calculate the equivalent mass of the metal. |
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Answer» `=(3.45xx4xx1000)/(200)=69` Equivalent mass of metal =69- EQ. mass of carbonate =(69-30)=39 |
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| 51157. |
34.95 ml of phosphorus vapour weighs 0.0625gat 0.1 bar pressure. What is the molar mass of phosphorus ? |
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Answer» Solution :`PV=nRT` `0.1"bar" xx(34.*05)/(10000L)=(0*0625)/(M) xx0*083xx819K` `M=(0*0625xx0*083xx819Kxx1000)/(34*05xx0*1)=(4248*5625)/(3*405)=1247*7 "MOL"^(-1)` |
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| 51158. |
343K = ........""^(@)F. |
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Answer»
`t_(C)^(@)=T_(K)-273 = 343 -273=70""^(@)C` `""^(@)F=9/5 (""^(@)C)+32=9/5(70)+32=158 ""^(@)F` |
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| 51159. |
3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution are |
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Answer» `3.67xx10^(26)` `=(6.02xx10^(23)xx11xx34.2)/(342) = 6.62xx10^(23)` Number of oxygen atoms in 9 g of `H_2O` `=(6.02xx10^(23)xx90)/(18) =30.1xx10^(23)` Total number of oxygen atoms `(30.1 +6.62)10^(23)` `=36.72xx10^(23) = 3.672xx10^(24)` |
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| 51160. |
34.95 mole of phosphorus vapour weighs 0.0625gat 0.1 bar pressure. What is the molar mass of phosphorus ? |
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Answer» Solution :ACCORDING to ideal gas equation. `Pv= nRT` Now, `n=(W)/(M):. Pv=(WRT)/(M) or M=(WRT)/(PV)` Substituting the values of from data `M=(0*0625gxx0*83 "BAR" L K^(-1) "mol"^(-1) xx819*15K)/(1"bar" xx34*04xx10^(-3)L)` `:. M=124*8 g "mol"^(-1)` |
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| 51161. |
34.05 mL of phosphorus vapour weights 0.0625 g at 546^(@)C and 0.1 bar pressure. What is the molar mass of phosphorus ? |
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Answer» Solution :According to ideal GAS law, `pV=nRT=(w)/(M)RT` `therefore M=(wRT)/(pV)` where, Molecular MASS of phosphorus, (w) = 0.0625 g Gas Constant `(R )=8.314xx10^(-2)` BAR `L mol^(-1)K^(-1)` Pressure of phosphorus vapour (p) = 1.0 bar Absolute TEMPERATURE `(T)=(546+273)K=819 K` Volume of Phosphorus vapour (V) = 34.05 mL = 0.03405 L `therefore M =((0.0625 g)(8.314xx10^(-2)"bar L mol"^(-1)K^(-1))(819K))/(("1 bar")(0.03405 L))` `= 124.98 = 125 g mol^(-1)` Note : If 0.1 bar is corrct for according to text book then answer will be 1249.8. It is not real because, `P_(4)=31xx4=124` |
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| 51162. |
34 . 05 mL of phosphorus vapours weigh 0.0625 g at 546 .^(@)C and 0.1 bar pressure . What is the molart mass of phosphorus ? |
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Answer» <P> Solution :ACCORDING to gas EQUATION,`PV = nRT` or `PV =w/M RT` Given : `P = 0.1 " bar," V =340.5 mL = 340.5 xx 10^(-3) dm^3, W = 0.0625 g, R = 0.0831 " bar " dm^3 K^(-1) mol^(-1)` `T = 546^@C = 819 K, M` = ? `:. M=(W.R.T)/(P.V)= (0.0625 xx0.0831 xx 819)/(0.1 xx 340.5 xx 10^(-3))` = `124.92 g mol^(-1)` Hence, molar MASS of phosphorus is `124.92 g mol^(-1)`. |
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| 51163. |
34.05 mL of phosphorus vapour weigh 0.0625 g at 546^(@)C and 1.0 bar pressure. What is the molar mass of phosphorus ? |
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Answer» <P> Solution :Step 1. Calculation of VOLUME at `0^(@)C` and 1 BAR pressure`(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)), i.e., (1xx34.05)/(546+273)=(1xxV_(2))/(273)"or" V_(2)=11.35 mL` 11.35 mL of vapour at `0^(@)C` and 1 bar pressure weigh `=0.0625" G"` Step 2. Calculation of mass of 22700 mL of `0^(@)C` and 1 bar pressure `:. 22700" mL"` of vapour at `0^(@)C` and 1 bar pressure will weigh `=(0.0625)/(11.35)xx22700=125" g"` `:. """Molar mass" =125" g "mol^(-1)` Alternatively, using `R=0.083" bar "DM^(3)K^(-1)mol^(-1)` PV=nRT, i.e., `n=(PV)/(RT)=(1.0" bar"xx(34.05xx10^(-3)dm^(3))/(0.083" bar " dm^(3)K^(-1)xx819 K)=5xx10^(-4)mol` `:.`Mass of 1 mole `"=(0.0625)/(5xx10^(-4))"g"=125 g` `:. ""` Molar mass`=125 g mol^(-1)` |
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| 51164. |
3.4 gm of H_(2)O_(2) decomposes, the weight of oxygen liberated from it is |
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Answer» 1.6 GM |
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| 51165. |
338 mL clear saturated solution of AgBrO_(3) requires just 30.4mL of H_(2)S_((g)) at 23^(@)C and 748 mm Hg to precipitate all the Ag^(+) ions Ag_(2)S. What will be K_(SP)of AgBrO_(3) ? |
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| 51166. |
3,3-Dimethyl-2-butanol on reaction with HCl yields mainly |
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Answer» 2-Chloro-2,3-dimethlybutane |
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| 51167. |
3.2g of a gas occupies 550 cc of volume at 22^@C and 770 mm of Hg pressure. Find the molecular mass of the gas. |
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Answer» |
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| 51168. |
3.2g of oxygen (At.wt =16) and 0.2g of hydrogen (At.wt = 1) are placed in a 1.12 litre flask at 0^@C. The total pressure of the gas mixture will be atm |
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Answer» <P> Solution :`n_1 = 3./32 = 0.1 , n_2 = (0.2)/2 = 0.1 , n_("TOTAL") = 0.2``P = (0.2 xx 0.0821 xx 273)/(1.12) = 4`. |
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| 51169. |
32.2 gm of an organic compound containing C,H and O when completely combusted produces 61.6gm of CO_(2) and 37.8 gm of H_(2)O. Select the correct option : |
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Answer» % of C in the ORGANIC COMPOUND is 40%. |
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| 51170. |
3200gm sulphur is present in |
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Answer» 9800 GM `H_(2)SO_(4)` a) 100 moles `H_(2)SO_(4)` B) 20 moles `H_(2)SO_(4)` C) 100 moles `H_(2)SO_(4)` d) 100 moles `SO_(2)` |
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| 51171. |
3.2 g of a mixture of calcium carbonate and sodium chloride was dissolved in 100 mL of 1.02 N HCl. After the reaction the solution was filtered and after separating the precipitate the volume was raised to 200 mL . 20 mL of this solution required 25 mL N//5 caustic soda solution for neutralisation. Find out the percentage of calcium carbonate in the mixture. |
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| 51172. |
3.15 g of oxalic acid dihydrate is dissolved in water and the solution was made up to 100 ml using a standard flask. The strength of the solution in normality is …………… |
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Answer» Solution :`0.5N` Normality `= ("Number of gram equivalents")/("Volume of the solution in L")` `= ("Mass of oxalic ACID")/(" EQUIVALENT mass of oxalic acid") + ` Volume of the solution in L `= ((3.15)/(63))/(0.1) = (0.05)/(0.1) =0.5N` |
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| 51173. |
314.000 has how many significant figures? |
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Answer» 6 |
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| 51174. |
30ml of sample of hydrogen peroxide solution after acidification with dil. H_2SO_4required 30ml of 0.02M KMnO_4solution for complete oxidation. Calculate molarity and percentage strength of the solution. |
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Answer» Solution :`KMnO_(4)+5H_(2)O_(2)+3H_(2)SO_(4)toK_(2)SO_(4)+2MnSO_(4)+8H_(2)O+5O_(2)` 2 moles of `KMnO_(4)` = 5 moles `H_(2)O_(2)` The equation for titration `(M_(1)V_(1))/(n_(1))=(M_(2)V_(2))/(n_(2))=(30xx0.02)/(2)=(M_(H_(2)O_(2))xx30)/(5)` MOLARITY of `H_(2)O_(2)` solution `M_(H_(2)O_(2))=(5)/(2)xx0.02M=0.05M` (w/v)% of `H_(2)O_(2)` solution`=Mxx"G.M.W//10=0.05xx34//10=0.17% |
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| 51175. |
30mL of mixture of methane and ethane in x : y ratio of volumes on combustioin gave 40mL of carbon dioxide. If 30mL of mixture is taken in y:x ratio, what is the volume of carbondioxide obtained under similar conditions? |
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Answer» Solution :The COMBUSTION of methane and ethane is given by the equations `CH_(4) overset(O_(2))to CO_(2)+2H_(2)O`: `C_(2)H_(6) overset(O_(2))to 2CO_(2)+3H_(2)O` 1 volume of `CH_(4)="1 volume of "CO_(2)` 1 volume of `C_(2)H_(6)="2 volume of "CO_(2)` Let, Volume of `CH_(4)` in the given mixture=x Let, volume of `C_(2)H_(6)` in the given mixture=y Then, x+y=30 `""` x+2y=40 `"y=10"x=30-10=20` If y mL of `CH_(4)` is taken the volume of `CO_(2)` obtained =y=10mL If x mL of `C_(2)H_(6)` is taken the volume of `CO_(2)` obtained =2x=2 x 20=40mL The volume of CARBONDIOXIDE obtained by the combustion of 30mL of y:x volumes of methane and ethane =10+40=50mL |
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| 51176. |
30ml of hydrogen peroxide solution was added with added with excess potasium iodide. The iodide liberated was titrated using starch indicator with 20ml of 0.3N sodium thiosulphate solution.Calculate the normality of hydrogen peroxide solution. |
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Answer» Solution :NORMALITY equation for iodometry is given as ,`V_(1)N_(1)=V_(2)N_(2)` `N_(1)=(V_(2)N_(2))/(V_(1))=(20xx0.3)/(30)=0.2` Normality `H_(2)O_(2)` solution = 0.2 eq `L^(-1)` |
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| 51177. |
30mL of a soltuion containg 9.15g//"litre" of an oxalte K_(X) H_(Y) (C_(2)O_(4))_(Z).nH_(2)O are required for titrating 27mL of 0.12N NaOH and 36mL of 0.12N KMnO_(4) separalty. Calculalte X,Y,Z are in the simple ratio of g atoms. |
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Answer» |
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| 51178. |
30.4 KJ is required to melt one mole of sodium chloride. The entropy change during melting is 28.4 JK^(-1) mol^(-1). Calculate the melting point of sodium chloride. |
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Answer» SOLUTION :GIVEN : `delta H_(f) (NaCl) = 30.4kJ = 30400 J MOL^(-1)` `delta S_(f) (NaCl) = 28.4 JK^(-1) mol^(-1)` `T_(f) = ?` `DeltaS_(f) = (Delta H_(f))/(Delta T_(f)) , T_(f) = (Delta H_(f))/(Delta S_(f))` `T_(f) = (30400 J mol^(-1))/(28.4 J K^(-1) mol^(-1)) = 1070.4K` |
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| 51179. |
30.4 kJ is required to melt one mole of sodium chloride. The entropy change during melting is 28.4 JK^(-1)mol^(-1) . Calculate the melting point of sodium chloride. |
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Answer» Solution :GIVEN : `DeltaH_f (NaCl)=30.4 kJ =30400 "J MOL"^(-1)` `DeltaS_f (NaCl)=28.4 JK^(-1) mol^(-1)` `T_f`=? `DeltaS_f =(DeltaH_f)/(DeltaT_f)` `T_f =(DeltaH_f)/(DeltaS_f)` `T_f=(30400 "J mol"^(-1))/(28.4 J K^(-1) "mol"^(-1))` `T_f` =1070.4 K |
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| 51180. |
30.4 KJ is required to melt one mole of sodium chloride . The entropy change during melting is 28.4JK^(-1) mol^(-1) .Calculate the melting point of sodium chloride . |
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Answer» Solution :Heat REQUIRED for 1 mole of NaCl for MELTING (q) =30.4 K J =30.4 x 1000 J `DELTAS` - entropy change =`28.4 J K^(-1) "mol"^(-1)` Melting POINT =`T_m`=? `DeltaS=q/T_m "" therefore T_m =q/(DeltaS)` `therefore T_m =(30.4xx1000)/28.4`=1070.4 K Melting point of NaCl =1070.4 K |
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| 51181. |
3.01xx10^(23)Ca^(+2) and CO_(3)^(-2)ions are present in CaCO_(3).The mass of sample is..... |
| Answer» Answer :B | |
| 51182. |
3.011xx10^(22) molecules are removed from a vessel Containing 1680 cc of nitrogen at STP. How many moles are remaining ? |
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Answer» Solution :Number of MOLES of `N_(2)` originally present =`("volume")/("GMV") = (1680)/(22400) = (3)/(40)` Number of moles of `N_(2)` removed = `("Number of MOLECULES")/("Avogadro number") = (3.011xx10^(22))/(6.022xx10^(23)) = (1)/(20)` Number of moles remaining `= (3)/(40)-(1)/(20) = (1)/(40) = 2.5xx10^(-2) = 0.025` |
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| 51183. |
3.011 xx 10^(22) atoms of an element weighs 1.15 g. The atomic mass of the element is |
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Answer» 23 |
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| 51184. |
300ml of O_2 diffused though a porous pot in 50 Sec. How long will it take for 500ml of CO_2 to diffuse under similar conditions. |
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Answer» 97.72 SECONDS |
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| 51185. |
3*00 " mol of " PCl_(5) kept in 1 L closed reaction vessel was allowed to attain equilibrium at 380 K. Calculate the composititon of the mixture at equilibrium . K_(c) = 1* 80 |
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Answer» Solution :` {:(,PCl_(5)(g),hArr,PCl_(3)(g),+,Cl_(2)(g)),("Initial",3*00 " mol ",,0,,0),("At. eqm.",(3*0-X)"mol",,x " mol",,x" mol "),("Molar conc",(3*0-x),,x,,x):}` (as volume of VESSEL = 1 L) ` K_(c) = ([PC_(3)] [Cl_(2)])/([PC_(5)])` ` 1*80 = (x xxx)/(3-x)=x^(2)/(3-x)or x^(2) = 5*40 - 1*80 x or x^(2) + 1*80 x - 5* 40=0` ` :. x= (-bpm sqrt(b^(2) - 4ac))/(2a)=(-1*80 pm sqrt((1*80)^(2) - 4 (-5*40)))/2=1*59`(neglecting - ve value) `:. " AT. equilibrium",[PCl_(5)]=3- 1*59 = 1*41 M` ` [PCl_(3)] = [Cl_(2)] = 1*59 M ` |
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| 51186. |
3.00 mol of PCl_5 kept in 1 L closed reaction vessel was allowed to attain equilibrium at 380 K. Calculate composition of the mixture at equilibrium. K_c=1.80 PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g)) |
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Answer» Solution :`{:("Equilibrium reaction :",PCl_(5(g)) hArr , PCl_(3(g)) + , Cl_(2(g))),("Initial mol :", 3.0,0,0),("Change of mol :", -x,x,+x),("Mol at equilibrium :",(3-x),x,x),("mol" L^(-1) "at equilibrium :",(3-x)/1=(3-x),x/1=x, x/1=x):}` where , x=(MOLE of `PCl_5`) = (Mole of `PCl_3` and `Cl_2` ) So, `K_c=([PCl_3][Cl_2])/[[PCl_5]]` `therefore 1.80 = ((x)(x))/(3-x)=x^2/(3-x)` `therefore x^2`=1.80 (3-x)=5.4 -1.80x `therefore x^2+1.80x-5.4=0` `therefore` a=1 |b=1.80|c=-5.4 `x=(-bpmsqrt(b^2-Delta4ac))/(2A)` `=(-1.80 pm sqrt((-1.8)^2-4(1)(-5.4)))/(2(1))` `=(-1.80 pm sqrt(3.24+21.6))/2` `=(-1.80 pm sqrt(24.84))/2=(-1.8 pm 4.984)/2` `therefore x=3.184/2` OR `(-6.084)/2` =1.592 M OR -3.042 M So, x=1.592and -3.42 is IMPOSSIBLE At equilibrium `(PCl_5)`=(3-x)=(3-1.592) =1.408 M `approx` 1.41 M equilibrium `[PCl_3]=[Cl_2]`= x= 1.592 M `approx` 1.59 M |
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| 51187. |
300 litres of ammonia gas at 20^(@)C and 20 atmosphere pressure are allowed to expend in a space of 600litres capacity and to a pressure of one atmosphere. Calculate the drop in temperature. |
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Answer» DROP intemperature =293-29.3=263.7 K |
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| 51188. |
300 K temperature at 2.0xx10^(4) Pa pressure volume of 0.09 mole CO_(2) gas is 112.0xx10^(-3)m^(3)then calculate volume at 4.0xx10^(4) Pa ? |
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Answer» <P> Solution :`64.2xx10^(-3)m^(3)``p_(1)V_(1)=p_(2)V_(2)` `therefore V_(2)=(p_(1)V_(1))/(p_(2))=(2.0xx10^(4)Pa xx112.0xx10^(-3)m^(3))/(4.0xx10^(4)Pa)` `= 64.2xx10^(-3)m^(3)` (Note : As the pressure divided the volume become half.) |
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| 51189. |
300 पर एक अभिक्रिया के लिए साम्य स्थिरांक 10 है| DeltaG^(ө) का मान क्या होगा ? R=8.314 JK^(-1) "mol"^(-1) |
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Answer» `-5.74kJ` `=-2.303xx8.314xx300 log 10` `=-5744.1 J RARR -5.74 KJ` |
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| 51190. |
30 Volumes H_(2)O_(2)means |
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Answer» `30% H_(2)O_(2)` |
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| 51191. |
30 volumes H_2O_2 means.... |
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Answer» 30% `H_2O_2` |
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| 51192. |
30 volume H_(2)O_(2) means _________. |
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Answer» 30% (w`//`v) `H_(2)O_(2)` solution |
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| 51193. |
30% solution of hydrogen peroxide is well known as |
| Answer» Answer :A | |
| 51194. |
30 mL of N//10 HCl are required to neutralise 50 mL of a sodium carbonate solution. How many mL of water must be added to 30 mL of this solution so that the solution obtained may have a concentration equal to N//50 ? |
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Answer» |
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| 51195. |
30 mL of mixture of methane and ethane in x:y ratio by volumes on combustion gave 40 mL of carbon dioxide. If 30 mL of mixture is taken in y: x ratio, what is the volume of carbondioxide obtained under similar conditions? |
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Answer» Solution :The combustion of methane and ethane is given by the equations `CH_(4)overset(O_(2))rarrCO_(2)+2H_(2)O, C_(2)H_(6)overset(O_(2))rarr 2CO_(2)+3H_(2)O` 1 volume of `CH_(4) = 1` volume of `CO_(2)` 1 volume of `C_(2)H_(6) = 2` volume of `CO_(2)` Let, volume of `CH_(4)` in the given mixture `= X` Let, volume of `C_(2)H_(6)` in the given mixture `= y` Then, `x + y = 30""x + 2Y = 40` `y = 10""x = 30 - 10 = 20` If y mL of `CH_(4)` is taken, the volume of `CO_(2)` obtained = y = 10 mL If x mL of `C_(2)H_(6)` is taken, the volume of `CO_(2)` obtained `= 2x = 2 xx 20 = 40 mL` The volume of carbondioxide obtained by the combustion of 30 mL of `y: x` volumes of methane and ethane `= 10 + 40 = 50 mL` |
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| 51196. |
30 mL of K_(2)Cr_(2)O_(7) liberated iodine from KI solution when the iodine was titrated with hypo solution (N//20), the titre value was 45 mL. Find the concentration of K_(2)Cr_(2)O_(7) in g per litre. |
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Answer» |
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| 51197. |
30 mL of a H_(2)O_(2) solution after acidification required 30 mL of N/10 KMnO_(4)solution for complete oxidation . Calculate the percentage and volume strength of H_(2)O_(2) solution. |
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Answer» Solution :To determine the normality of `H_(2)O_(2)` solution . From the given DATA, for `H_(2)O_(2), V_(1)=30 mL, N_(1)=?` For `KMnO_(4), V_(2)=30 mL, N_(2)=N//10` Applying normallity equation , `N_(1)V_(1)=N_(2)V_(2) , i.e., 30xxN_(1)=30xx1//10 therefore N_(1)=0.1 N` Thus, the normality of `H_(2)O_(2)` solution =0.1 N Step 2. To determine the percentage STRENGTH of `H_(2)O_(2)` solution, We know that, `H_(2)O_(2) to 2H^(+) + O_(2) + 2e^(-)""therefore"` Eq. wt. of `H_(2)O_(2) =34//2=17` Hence, strength of `H_(2)O_(2)=(1.7xx100)/(1000)=0.17%` Step 3. To determinethe VOLUME strength of `H_(2)O_(2)` solution. Consider the chemical equation, `underset(68 g)(2H_(2)O_(2)) to 2H_(2)O + underset(22400 mL at N.T.P.)O_(2)` Now 68 g of `H_(2)O_(2)` give `O_(2)` at N.T.P. =22400 mL `therefore 1.7 g ` of `H_(2)O_(2)` will give `O_(2)=(22400)/(68)xx1.7=560` mL But 1.7 g of `H_(2)O_(2)` are present in 1000 mL of `H_(2)O_(2)` solution Hence, 1000 mL of `H_(2)O_(2)` solution gives 560 mL of `O_(2)` at N.T.P. `therefore ` 1 mL of `H_(2)O_(2)` solution will give `=(560)/(1000)=0.56 ` mL of `O_(2)` at N.T.P. `therefore` Volume strength of `H_(2)O_(2)` solution =0.56 |
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| 51198. |
30 ml of 20 Vol of H_(2)O_(2) 40 ml of 20Vol H_(2)O_(2) and 30 ml10 Vol of H_(2)O_(2) dissolvedin all proprotios. The normality of the resulting solution is .... |
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Answer» |
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| 51199. |
30 mL H_(2)and 20 L O_(2)reacts to form water then what will left at the end of reaction ? |
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Answer» 5 ML `H_(2)` |
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| 51200. |
30 lit of oxygen is liberated at STPby the complete decompostion of a sample of 1 lit of H_(2)O_(2). The molarity of that H_(2)O_(2) solution is nearly |
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Answer» `0.893M` |
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