InterviewSolution
Saved Bookmarks
InterviewSolution
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.
| 92151. |
45 g of ethylene glycol (C_(2)H_(6)O_(2)) is mixed with 600 g water. Calculate (a) freezing point depression (b) freezing point of the solution (K_(f)" for water = 1.86 K kg mol"^(-1)," Atomic masses : C = 12, H = 1, O = 16 amu"). |
|
Answer» SOLUTION :Here, we are given `w_(2)=45g, w_(1)=600g, K_(f)="1.86 K kg mol"^(-1)` `M_(2)"for SOLUTE "C_(2)H_(6)O_(2)=24+6+32="62 g mol"^(-1)` Substituting these value in the formula, `DeltaT_(f)=(1000K_(f)w_(2))/(w_(1)xxM_(2)),` we get `DeltaT_(f)=("1000 g kg"^(-1)xx"1.86 K kgmol"^(-1)xx45g)/(600 gxx"62 g mol"^(-1))="2.25 K"` Freezing POINT of pure water = 273.15 K `therefore""` Freezing point of the solution `=T_(f)^(@)-DeltaT_(f)=273.15-2.25K=270.9K` |
|
| 92152. |
45 g of ethylene glycol(C_(2) H_(6)O_(2)) is mixed with 600 g of water. The freezingpoint of the solution is (K_(f)for water is 1.86 K kgmol^(-1) ) |
|
Answer» 273.95 K |
|
| 92153. |
45 g of ethylene glycol (C_(2)H_(6)O_(2)) is mixed with 600 g of water. Calculate (a) the freezing point depression and (b) the freezing point of the solution. |
|
Answer» Solution :Depression in freezing point is related to the molality, therefore, the molality of the solution with respect to ethylene glycol. `= ("moles of ethylene glycol")/("mass of water in KILOGRAM")` `therefore` Moles of ethylene glycol `= (45 g)/(62 g MOL^(-1))` = 0.73 mol `therefore` Mass of water in `kg = (600 g)/(1000 g kg^(-1))=0.6 kg` `therefore` Hence molality of ethylene glycol `=(0.73 mol)/(0.60 kg)` `= 1.2 mol kg^(-1)` `therefore` Therefore freezing point depression, `Delta T_(F)=1.86K kg mol^(-1)xx 1.2 mol kg^(-1)` = 2.2 K `therefore` Freezing point of the aqueous solution, `= 273.15 K - 2.2 K` = 270.95 K |
|
| 92154. |
4.5 g of aluminium is deposited at cathode from a molten electrolyte containing Al^(3+) ions by a certain quantity of electric charge. The volume of hydrogen produced at STP from H^(+) ions in a solution by the same quantity of electric charge will be: |
|
Answer» `44.8 L` |
|
| 92155. |
4.5 g of aluminium (at mass 27 amu) is deposited at cathode from Al^(3+) solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from H^(+) ions in solution by the same quantity of electric charge will be |
|
Answer» 22.4L `(4.5)/((27)/(3))="eq. of "H_(2),(4.5)/(9)=eq." of "H_(2)` `2H^(+)+2e^(-)toH_(2)` eq. of `H_(2)="number of MOLES"xx"n FACTOR ".5=n_(H_(2))xx2` `V_(H_(2))=(0.5)/(2)xx22.4,V_(H_(2))=5.6` l. |
|
| 92156. |
4.5 g of aluminium (at. Mass 27 amu) is deposited at cathode from a molten electrolyte containing Al^(3+) ionsby a certain quantity of electric charge . The volume of hydrogen produced at STP from ions in a solution by the same quantity of electric charge will be |
|
Answer» 44.8L |
|
| 92157. |
448mL of a hydrocarbon (A) having C (87.80%), H (12.19%) weight 1.64g at NTP. On hydrogenation it gives 2 methyl pentane. Treatment of (A) with acidic HgSO_(4) gives a new compound (B) of molecular weight C_(6)H_(12)O. Compound (A) does not react with ammoniacal AgNO_(3). What is the structure of (A)? |
| Answer» SOLUTION :`[CH_(3)-UNDERSET(underset(CH_(3))(|))(CH)-C-=C-CH_(3)]` | |
| 92158. |
44gofa sample on complete combusion given 88gm CO_(2) and 36gm of H_(2)O. The molecular formula of the complound may be |
|
Answer» `C_(2)H_(6)` `44G""(88G)/44""36/18` `""2""2` `x=2` `y//2=4` `x=2` `y=4` |
|
| 92159. |
44g of a sample on complete combusion given 88 gm CO_2 and 36 gm of H_2O. The molecular formula of the compound may be :- |
|
Answer» `C_2H_6` `{:(44G,"88G"/44,36/18),(,2,2):}` x=2 y/2=4 x=2 y=4 |
|
| 92160. |
4.4 of CO_2 and 2.24 litre of H_2 at STP are mixed in a container. The total number of molecules present in the container will be : |
|
Answer» `6.022 xx 10 ^(23)` |
|
| 92161. |
4.4 gms of a hydrocarbon on complete combustion produced 13.2 gms of CO_(2) and 7.2 gms of H_(2)O. How many moles of oxygen is consumed per mole of hydrocarbon in the combustion? |
|
Answer» |
|
| 92162. |
4.4 g of CO_(2) contains how many litres of CO_(2) at S.T.P |
|
Answer» `2.4 L` `:.` 4.4 G of `CO_(2)` at S.T.P. = 2.24 L |
|
| 92163. |
4.4 g of CO_(2) and 2.24 litre of H_(2) at STP are mixed in a container. The total number of molecules present in the container will be: |
|
Answer» `6.022xx10^(23)` |
|
| 92164. |
4.4 g of an unknowm gas occupies 2.24 litres of volume at STP. The gas may be |
|
Answer» CARBON dioxide |
|
| 92165. |
4,4-Dinitrodiphenyl is obtained when |
|
Answer» 4-Nitrochlorobenzene is heated with Na/ether |
|
| 92166. |
4.35 g of a mixture of NaCl and Na_(2)CO_(3) was dissolved in 100 mL of water , 20 mL of which was exactly neutralised by 75.5mL of N/0 solution of H_(2)SO_(4) . Calculate the percentageof NaCl and na_(2)CO_(3) in the mixture . |
|
Answer» Solution :In this problem ,`Na_(2)CO_(3)`is neutralised by `H_(2)SO_(4)` solution . Let the amount of `Na_(2)CO_(3)` be x g . ` :. ` equivalent of `na_(2)CO_(3)` in 100 ML in solution = `x/53 ` …(Eqn.4i) ` :. ` m.e of `Na_(2)CO_(3)` in10 mLof solution` = x/53 xx1000 "" ...(Eqn.3)` ` :. ` m.e of `na_(2)CO_(3)` in 20 mL of solution ` = (1000 x)/53 xx20/100 = (200 x)/53 ` Now , m.e of `H_(2)SO_(4)` solution ` = 1/10 xx 75.50 "" ...(Eqn.1)` `:. ` m.e of `Na_(2)CO_(3) ` = m.e of `H_(2)SO_(4) "" ...(Eqn.2)` ` :. "" (200x)/53 = 7.55 ` x = 2 ` :. "" % of Na_(2)CO_(3) = 2/(4.35) xx 100= 45.99 % ` ` % of naCl = 100 - 45.99 = 54.01 % ` |
|
| 92167. |
41 g iron and 70 g carbon monoxide are allowed to react to produce iron carbonyl Fe(CO)_(5). Find the excess reactant and how much mass of it will be left at the end of reaction :- |
|
Answer» `CO, 42 gm` |
|
| 92168. |
4.0g of argon ( at. Mass = 40) in a bulb at a temperature of T K has a pressure P atm. When the bulb was placed in a hotter bath at a temperature 50^(@)C more than the first one, 0.8g of gas had to be removed to get the original temperature T which is equal to |
|
Answer» <P>510K Now, `p V = nRT` or `PV = 0.1 RT ` In hot bath, n `= ( 4.0 - 0.8)/( 40) = 0.08` `:. pV = 0.08 R ( T + 50)` `:. 0.1 RT = 0.08 R ( T + 50)` or `T = 200K` |
|
| 92169. |
4.08g ofa mixtureof BaO and an unkonwn carbonate MCO_(3) was heated strongly .the residence weighted 3.64 g.this was dissolved in 100 mL of 1 N HCl . The excess of acid required 16 mL of 2.5 N NaOH for complete neutralisation .Identify the metal M. |
Answer» Solution : BAO does not change on heating Suppose the weightof `MCO_(3)` is x g and at . Wt of M is y `{:(MCO_(3) to ,MO +,CO_(2)),(" x g",[3.64 -(4.08-x)],0.44 g ),(,=(x - 0.44)g,):}` APPLYING POAC for m and C atoms, we get `x/(y+60) = ((x-0.44))/(y+16) ""...(1)` and `x/(y+60) = (0.44)/44 = 0.01 ""...(2)` From eqns . (1) and (2) , we have , ` (x-0.44)/(y+16) = 0.01""...(3)` Now , m.e of NaOh = `2.5 xx16 = 40 ""...(Eqn.1)` ` :.` m.e of excess acid= 40 ` :. ` m.e of the acid used to neutralise BaO and MO = m.r of total acid = m.e of excess acid ` = 1xx 100 - 40 = 60` ` :. ` eq. of the acid= `60/100 = 0.06` = eq. of BaO + eq. of MO or `(4.08 -x)/(154//2)+ ((x-0.44))/((y+16)2) = 0.06` `("eq. wt . of BaO"= 154/2 , " eq. wt of MO "= (y+16)/2)` Substituting the value of `((x-0.44)/(y+16)) ` from Eqn . (3) in Eqn .(4) we get , x = 1 and y = 40 hence , the metal M must be Ca. |
|
| 92170. |
403K can be expressed in degree celcius as |
|
Answer» `12.98^(@)C` |
|
| 92171. |
400 mL of standard gold sol is just prevented from coagulation by the addition of 0.96 g of starch before adding 1 mL of 10% NaCl solution, the gold number will be |
|
Answer» `2.4` `therefore` 10 mL of glod sol require strach `= (960)/(400) XX 10=24` mg |
|
| 92172. |
400 cm^(3) of water is added to 6% of 600 g (molar mass =36) of a solute. The molarity contain 20 millimoles of the solute is |
|
Answer» 3 `C=W/("Molar mass" xx V(DM^(3)))` `C=36/(36 xx 0.4)=1/(0.4)=0.25` |
|
| 92173. |
4.0 of NaOH are present in one decilitre solution. What is the molarity of solution ? |
|
Answer» Solution :Molarity of solution (M) `((4g)//(40"G MOL"^(-1)))/((0.1L))=1" mol L"=1M` |
|
| 92174. |
40%of a mixture of 0.2 mole of N_2 and o.6 mole of H_2 react to give NH_3 according to the equation:N_2(g)+3H_2(g)⇔2NH_3(g) at constant temperature and pressure. Then what is the ratio of the final volume to the initial volume of gases ? |
|
Answer» 4:5 |
|
| 92176. |
40 ml of 0.1 M ammonia solution is mixed with 20 ml of 0.1 MHCl. What is the pH of the mixture (pK_(b) of ammonia solution is 4.74) |
|
Answer» 4.74 |
|
| 92177. |
40 mL of 0.125 M of NiSO_(4) solution is electrolysed by a current of 0.05 amp for 40 minutes. (i) Write the equation for the reactions occurring at each electrode. (ii) How many coulombs of electricity passed through the solution ? (iii) How many grams of the product deposited on the cathode ? (iv) How long will the same current have to pass through the solution to remove completely the metal ions from the solution ? (v) At the end of electrolysis how many grams of the product would appear at the anode ? |
| Answer» SOLUTION :120, 0.037 G, 19300 s, 0.08 g | |
| 92178. |
40 mg of pure sodium hydroxide is dissolved in 10 litres of distilled water. The pH of the solution is |
|
Answer» `9.0` `= (40 xx 10^(-3))/(40)xx (1)/(10) = 10^(-4) M` `POH = LOG 10. (1)/([OH^(-)]) = log 10. (1)/(10^(-4)) = 4` `PH + pOH = 14, pH + 4 = 14 rArr pH = 10`. |
|
| 92179. |
40 mg of gaseous substance (X_(2) ) occupies 4.8 mL of volume at 1 atm and 27^(@)C. Atomic mass of element X is : (R:0.08 atm L/mole- k ) |
|
Answer» |
|
| 92180. |
4.0 grams of caustic soda contain |
|
Answer» `6.02 xx 10^23 ` atoms of H ` = (6.02 xx 10^23 xx 4)/(40) = 6.02 xx 10^22 Na ` atoms |
|
| 92181. |
40 gm of a carbonate of an alkali metal or alkaline earth metal containing some insert impurities was made to react with excess HCl solution. The liberated CO_(2) occupied 12.315 lit. at 1 atm & 300 K . The correct option is |
|
Answer» MASS of impurity is 1 GM and METAL is Be |
|
| 92182. |
4.0 g of NaOH is contained in one decilitre of a solution. Calculate the following in this solution. (i) Mole fraction of NaOH (ii) Molality of NaOH (iii( Molarity of NaOH. (At. Wt. of Na = 23, O = 16 , Density of NaOH solution is 1.038g//cm^(3)) |
|
Answer» |
|
| 92183. |
4.0 g of NaOH dissolved in 500 ml of solution. Calculate molarity of the solution. |
| Answer» Solution :Molarity `=(4//40)/( 500//1000)/(4)/(5) =(1000)/(500)=0.2` | |
| 92184. |
4.0 g of argon (at mass = 40) ina bulb at a temperature of T K had a pressure P atm, when the bulb was placed in hotter bath at a temperature 50^(@) K more than the first one, 0.8 g of gas had to be removed, to get the original pressure. T is equal to |
|
Answer» 510 K |
|
| 92185. |
4.0 g of argonhas pressure P and temeperature T K in a vessel. On keeping the vessel at50@C higher temperature, 0.8 g of argon was given out to maintain the pressure P. The original temperature was : |
|
Answer» 73 K |
|
| 92186. |
4.0 g of a substance, A, dissolved in 100 g of water depressed the f.p. of water by 0.1^@C, while 4.0 g of another substance, B, depressed the f.p. by 0.2^@C . Which of the two substances has the higher molecular weight? |
|
Answer» |
|
| 92187. |
40% by weight solution will contain how much mass of the solute in 1 solution, density of the solution is 1.2 g/mL |
|
Answer» 480 G when M = mol. Weight of the solute Molarity `=(40xx1.2xx10)/(Mxx1000)`.......(i) Molarity `=("weight of the solute / M")/("VOLUME of solution (in litre)")`..........(ii) From Eqs. (i) and (ii) `("Weight of solute")/(M xx 1000)=(40xx1.2xx10)/(M xx 1000)` Weight of solute = 480 g. |
|
| 92188. |
40% aqueous solution of formaldehyde is known as ............... . |
| Answer» SOLUTION :FORMALIN | |
| 92189. |
4% solution of ethyl nitrite in alcohol is known as…………….. |
|
Answer» |
|
| 92190. |
4-Propene underset(778K)overset(Cl_(2))rarr X overset(alc. NH_(3))rarr Y is |
|
Answer» ALLYLAMINE |
|
| 92191. |
4-Nitrotoluene is trated with bromine to get compound 'P'. 'P' is reduced with Sn and HCl to get compound 'Q', 'Q' is diazotised to and the product is treated with phosphinic acid to get compound 'R'. 'R' is oxidised with alkaline KMnO_(4) to get compound 'X'. compound 'S' is |
|
Answer» 2-bromo-4-hydroxybenzoic acid 
|
|
| 92192. |
4-Nitrobenzaldehyde from 4-nitrotoluene. (Friedel-Crafts reaction) |
Answer» Solution :When 4-nitrotoluene is TREATED with chromium oxide inacetic ANHYDRIDE, a diacetate derivative is OBTAINED which on ACID hydrolysis PRODUCES 4-nitrobenzaldehyde.
|
|
| 92193. |
4 moles of an ideal gas (C_(v)=15J) is subjected to the following process represented on P – T graph. From the given data find out whether the process is isochoric or not ? also calculate q, w, DeltaU, DeltaH, |
Answer» Solution : PV = NRT `4V=4Rxx400` `V=400R"".....(1)` `3V=4Rxx300` `V=400R" ".....(2)` i.e., V is CONSTANT w = 0 `DeltaU=nC_(V)DeltaTrArr4xx15xx100=6000J` `DeltaH=nC_(P)DeltaTrArrn(C_(V)+R)DeltaT` `rArr4xx(15+8.3)xx100` `rArr9320J` `q=DeltaU=6kj` |
|
| 92194. |
4 moles of NaBH_(4) react completely with I_(2). calculate volume of gaseous products at STP from above process. (Give you answer by dividing 4.48) |
|
Answer» |
|
| 92195. |
4 moles of A are mixed with 4 moles of B. At equilibreium for the reaction A+B hArr C+D.2 moles of C and D are formed. The equilibrium constant for the reaction will be |
|
Answer» `1/4` `{:("Initial conc",4,4, 0,0),("After T TIME conc",(4-2),(4-2),0,0):}` Equlibrium constant `=([C][D])/([A][B])=(2XX2)/(2xx2)=1` |
|
| 92196. |
4 moles each of SO_(2) and O_(2) gases are allowed to react to from SO_(3) in a closed vassel. At equlibrium 25% of O_(2) is used up. The total number of moles of all the gases at equlibrium is |
|
Answer» `6.5` `{:("INITIAL moles",4,4,0),("Moles at EQM.",4(1-0.25),4(1-0.25),0.25xx2),("TOTAL NUMBER of moles",=41,(1-0.25)+4,(1-0.25)+0.5):}` `=4-1+4-1+0.5=6.5.` |
|
| 92197. |
4 moles each of SO_(2) and O_(2) gases are allowed to react to form SO_(3) in a closed vessel. At equilibrium 25% of O_(2) is used up. The total number of moles of all the gases at equilibrium is |
|
Answer» `6.5` Total number of moles `=4(1-0.25)+4(1-0.25)+0.5` `=4-1+4-1+0.5=6.5` |
|
| 92198. |
4 moles each of SO_(2) and O_(2) gases are allowed to react to form SO_(3) in a closed vessel. At equilibrium, 25% of O_(2) is used up. The total number of moles of all the gases at equilibrium is |
|
Answer» Solution :Moles of `O_(2)` used up = 25 % of 4 = 1 mol. `{:(,2SO_(2),+,O_(2),hArr,2SO_(3)),("Initial moles",4,,4,,0),("Moles at eqm.",4-2,,4-1,,),(,=2,,=3,,2):}` Therefore total NUMBER of moels of all gases at eqm. = 2+3+2 = 7 |
|
| 92199. |
4 mole of pure liquid A (P_(A)^(0)=80mm of Hg) and 6 mole of pure liquid B (P_(B)^(0)=100mm of Hg) are mixed . The vapour pressure of the resulting solution is found to be 90mm of Hg then |
|
Answer» It is an ideal solution |
|
| 92200. |
4 moles of A are mixed with 4 moles of B. When 2 moles of C are formed at equilibrium accordingly to the reaction A+BrarrC+D.K_c is: |
| Answer» ANSWER :B | |