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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.
| 30851. |
The fraction of total volume occupied by the atoms in a simple cubic is |
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Answer» `((PI)/(4sqrt2))` |
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| 30852. |
The fraction of the totalvolumeoccupied by the atomsis a fcc is |
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Answer» `PI/6` |
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| 30853. |
The fraction of the total volume occupied by atoms in a simple cube is |
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Answer» `PI/2` |
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| 30854. |
The fraction of chlorine precipitated by AgNO_3 solution from [Cu(NH_3)_5Cl]Cl_2 is: |
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Answer» `1//2` |
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| 30855. |
Thefractionobtainedbetweentemperatures 423-573 Kduringfractionaldistillation of crude petroleum is : |
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Answer» PARAFFIN wax |
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| 30856. |
The fraction by volume of carbon monoxide in producer gas is about : |
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Answer» `1//2` |
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| 30857. |
The f.p. of nitrobenzene is 3^@C . When 1.2 g of chloroform (mol. wt = 120) is dissolved in 100 g of nitrobenzene, the f.p. of the solution is 2.3^@C. When 0.6 g of acetic acid is dissolved in 100 g of nitrobenzene, the f.p. of the solution is 2.64^@C . Calculate the molecular weight of acetic acid. What conclusion can be drawn from |
| Answer» SOLUTION :116.6, DIMER | |
| 30858. |
The f.p constant of benzene is 4.9 and its m.p is 5.51^(@)C. A solution of 0.816g of compound (A) when dissolved in 7.5g of benzene freezes at 1.59^(@)C. The compound (A) has C, 70.58% and H, 5.88%. Determine the molecular weight and molecular formula of (A). Compound (A) on heating with soda lime gives another compound (B) which on oxidation and subsequent acidification gives an acid (C ) of equivalent weight 122. (C ) on heating with soda lime gives benzene. Identify (A), (B) and ( C) and explain the reaction involved. |
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Answer» Solution :We have `Delta T_(f)= K_(f).m` `(5.51-1.59)= 4.90 xx ((0.816)/(M) xx (1000)/(7.5))` M= 136 [M is mol. Wt. of COMPOUND (A)] Moles of `C: H: O= (70.58)/(12): (5.88)/(1): (23.54)/(16)` `=5.88 : 5.88: 1.471` `=4:4:1` `therfore` empirical formula of (A) `=C_(4)H_(4)O` (68) As the molecular weight of (A) is 136, molecular formula of (A) is `C_(8)H_(8)O_(2)` Now, from the GIVENREACTION sequence: 
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| 30859. |
The f.p. of 1% solution of Ca(NO_(3))_(2) in water will be |
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Answer» below `0^(@)C` |
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| 30860. |
The fourth period of the p-block elements contains: |
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Answer» 6 elements |
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| 30861. |
The four colligative properties of the dilute solutions help in calulating the molecular mass of sthe solute which is often called observen molecular mass. It may be same as the theoretical molecular mass (calculated form rthe molecular formula) if the solute behaves normally in solution. In case, it undergoes association or dissociation. the observed molar mass gives different results. The nature of the in solution is expressed in terms of Van't Hoff factor (i) which may be 1 (ifthe solute behave normally), less than 1 (if the solute associates) and more than 1 (if the solute dissociates). The extent of association of dissociation is repreented by lapha which is :alpha=underset(("For association"))((i-1)/(1//n-1))or underset(("for dissoviation"))((i-1)/(n-1)) (16) The molar mass ofd the solute sodium hydrxide obtained form the measurement of osmotic pressure of its aqueous solution at 27^(@)C is 25 g mol^(-1). Therefore, The precentage ionisation of solution is : |
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Answer» 75 = `((40"G mol"^(-1)))/((25"g mol"^(-1)))=1.6` `alpha=(i-1)/(n-1)=(1.6-1)/(2-1)=0.6` % ionisation = 60. |
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| 30862. |
The four colligative properties of the dilute solutions help in calulating the molecular mass of sthe solute which is often called observen molecular mass. It may be same as the theoretical molecular mass (calculated form rthe molecular formula) if the solute behaves normally in solution. In case, it undergoes association or dissociation. the observed molar mass gives different results. The nature of the in solution is expressed in terms of Van't Hoff factor (i) which may be 1 (ifthe solute behave normally), less than 1 (if the solute associates) and more than 1 (if the solute dissociates). The extent of association of dissociation is repreented by lapha which is :alpha=underset(("For association"))((i-1)/(1//n-1))or underset(("for dissoviation"))((i-1)/(n-1)) (15) 0.1 M K_(4) [Fe(CN)_(6)] is 60% inized. What will be Van't Hoff factor ? |
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Answer» `1.4` `0.6=(i-1)/(5-1),i=2.4+1=3.4,` |
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| 30863. |
The four colligative properties of the dilute solutions help in calulating the molecular mass of sthe solute which is often called observen molecular mass. It may be same as the theoretical molecular mass (calculated form rthe molecular formula) if the solute behaves normally in solution. In case, it undergoes association or dissociation. the observed molar mass gives different results. The nature of the in solution is expressed in terms of Van't Hoff factor (i) which may be 1 (ifthe solute behave normally), less than 1 (if the solute associates) and more than 1 (if the solute dissociates). The extent of association of dissociation is repreented by lapha which is :alpha=underset(("For association"))((i-1)/(1//n-1))or underset(("for dissoviation"))((i-1)/(n-1)) (14) Benzoic acid undergoes demrisation in benzene solution. The Van't Hoff factor is related to degree of association alpha of the acid as : |
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Answer» `i=1-ALPHA` `(i-1)=(-1)/2alphaor i-1-1/2alpha` |
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| 30864. |
The four colligative properties of the dilute solutions help in calulating the molecular mass of sthe solute which is often called observen molecular mass. It may be same as the theoretical molecular mass (calculated form rthe molecular formula) if the solute behaves normally in solution. In case, it undergoes association or dissociation. the observed molar mass gives different results. The nature of the in solution is expressed in terms of Van't Hoff factor (i) which may be 1 (ifthe solute behave normally), less than 1 (if the solute associates) and more than 1 (if the solute dissociates). The extent of association of dissociation is repreented by lapha which is :alpha=underset(("For association"))((i-1)/(1//n-1))or underset(("for dissoviation"))((i-1)/(n-1)) (13) Which of the following is incorrect ? |
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Answer» Molecular mass of NaCI found by osmotic PRESSURE measurements is half of the theoretial VALUE |
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| 30865. |
The four colligative properties of the dilute solutions help in calulating the molecular mass of sthe solute which is often called observen molecular mass. It may be same as the theoretical molecular mass (calculated form rthe molecular formula) if the solute behaves normally in solution. In case, it undergoes association or dissociation. the observed molar mass gives different results. The nature of the in solution is expressed in terms of Van't Hoff factor (i) which may be 1 (ifthe solute behave normally), less than 1 (if the solute associates) and more than 1 (if the solute dissociates). The extent of association of dissociation is repreented by lapha which is :alpha=underset(("For association"))((i-1)/(1//n-1))or underset(("for dissoviation"))((i-1)/(n-1)) (12) The Van't Hoff factor for diolute solution of glucose is : |
| Answer» Solution :Glucose does not eiother dissociate or ASSOCIATE and van't Hoff FACTOR (i) =1 . | |
| 30866. |
The forward rate constant for the elementary reversible gaseous reaction C_(2)H_(6)ltimplies2CH_(3) "is" 1.57xx10^(-3)s^(-1)at 100 K What is the rate constant for thebackward reaction at this temperature if 10^(-4) moles of CH_(3) and 10 moles of C_(2)H_(6) are present in a 10 litre vessel at equilibrium . |
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Answer» `1.57xx10^(9) L "MOLE"^(-1)s^(-1)` |
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| 30867. |
The formulation of Dettol contains: |
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Answer» CHLOROXYLENOL |
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| 30868. |
The formulation of dettol contains |
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Answer» chloroxylenol |
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| 30869. |
The formulae of Teflon and Freon are respectively |
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Answer» `C C l_(2) F_(2)` and `( C_(2) F_(4))_(n)` |
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| 30870. |
The formula weight of an acid is 82.0. In a titration, 100cm^(3) of a solution of this acid containing 39.0 g of the acid per litre were completely neutralised by 95.0cm^(3 of aqueous NaOH containing 40.0 of g NaOH per litre. What is the basicity of the acid? |
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Answer» Solution :Suppose the equivalent wt. of the acid = E `therefore"Normality of the acid solution"=(39gL^(-1))/(E)` EQ. wt. of NaOH = 40 `therefore"Normality of NaOH solution"=(40gL^(-1))/(40)=1N` `"Applying"underset("(Acid)")(N_(1)xxV_(1))=underset("(NaOH)")(N_(2)xxV_(2))` `(39)/(E)xx100=1xx95 or E=41.0` `"Basicity of the acid"=("Formula wt.")/("Equivalent wt.")=(82.0)/(41.0)=2` |
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| 30871. |
The formula to calculate paramagnetic moment of a substance is |
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Answer» `mu_(s) = sqrt(4S(S+2)) B.M` |
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| 30872. |
The formula of zinc phosphite is: |
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Answer» `ZnHPO_3` |
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| 30873. |
The formula of zinc chloride with ammonia is: |
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Answer» `[Zn(NH_3)_2]CL` |
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| 30874. |
The formulaof the deep redliquidformedon warming dichromate with KCI in concentrated sulphuric qacid is _____. |
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Answer» Chromyl chloride `K_(2)CrO_(7) + 2H_(2)SO_(4) rarr 2KHSO_(4) + 2CrO_(3) + H_(2)O` `KCI +H_(2)SO_(4) rarr KHSO_(4) + HCI` `CrO_(3) + 2HCI rarrunderset("Orange -red vapours")(CrO_(2)CI_(2)) + H_(2)O` |
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| 30875. |
The formula of the compound which gives violet colour in Lassainge's test for sulphur with sodium nitroprusside is |
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Answer» `Na_(4)[FE(CN)_(5)NOS]` |
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| 30876. |
The formula of the complex tris(ethylenediamine) cobalt (III) sulphate is: |
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Answer» `[CO(EN)_(2)SO_(4)]` |
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| 30877. |
The formula of the complex, tris (ethylenediamine) cobalt(III) sulphate is : |
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Answer» `[CO(en)_(3)SO_(4)]` |
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| 30878. |
The formula of sulphanilic acid is |
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Answer»
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| 30879. |
The formula of sodium nitroprusside is |
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Answer» `Na_(4)[FE(CN)_(5)NO_(2)]` |
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| 30880. |
The formula of sodium nitroprusside is …… and its IUPAC name is …… |
| Answer» Solution :`Na_(2)[FE(CN)_(5)NO]`, sodium nitrosonium pentacyanoferrate (II) | |
| 30881. |
The formula of sodium nitroprusside is: |
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Answer» `Na_4[FE(CN)_5NOS]` |
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| 30882. |
The formula of sodium cobaltinitrite is …… and its IUPAC name is …… . |
| Answer» Solution :`Na_(3)[Co(ONO)_(6)]`, SODIUM hexanitritochobaltate (III) | |
| 30883. |
The formula of siderite ore is |
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Answer» `FeCO_3` |
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| 30886. |
The formula of rust can be represented by Fe_(2)O_(3). How many mole of Fe are present in 16 gm of rust. |
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Answer» 0.1 0.1mole of rust moles of `Fe=2xx("molar of rust")` `=2xx0.1=0.2` |
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| 30887. |
The formula of Rhamnose is ………………. |
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Answer» `C_(6)H_(12)O_(6)` |
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| 30888. |
The formula of product formed, when sodium thiosulphate solution is added to silver bromide is |
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Answer» `Na_(3)[Ag(S_(2)O_(3))_(2)]` |
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| 30889. |
The formula of phosgene is _________. |
| Answer» Answer :A | |
| 30890. |
The formula of pentaaquanitratochromium(III)nitrate is |
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Answer» `[CR(H_(2)O)_(6)](NO_(3))_(3)` |
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| 30891. |
The density and edge length values for a crystalline element with fcc lattice are 10 g cm^(-3) and 400 pm respectively. The number of unit cells in 32 g of this crystal is |
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Answer» `8xx10^23` `=(400xx10^(-10)cm)^3 =64 XX 10^(-24)cm^3` Density of the unit cell=`10g//cm^3` Mass of unit cell=Volumex density `=64xx10^(-24)xx10=640xx10^(-24) g` Now, `640xx10^(-24) g -=` 1 unit cell `therefore 32 g -= 1/(640xx10^(-24))xx32=0.05xx10^24=5xx10^22` |
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| 30892. |
The densities of water and isopropyl alcohol are 0.9982 and 0.7887 g/cc. At 20°C, isopropyl alcohol flowed through a viscometer in 624 s and an equal volume of water flowed through the same viscometer in 200 s. If eta (water) = 1.009 xx 10^(-3) Nm^(-2)s,calculate eta (isopropyl alcohol) at 20°C. |
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Answer» Solution :We have, `eta_(1)/eta_(2) = (t_(1)d_(1))/(t_(2)d_(2))` Suppose that .Y and .2. STAND for alcohol and WATER respectively, `therefore eta_(1) = (624 XX 0.7887)/(200 xx 0.9982) xx (1.009 xx 10^(-3))` `=2.487 xx 10^(-3) Nm^(-2) s` |
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| 30893. |
The density (in g mL^(-1)) of a 3.60M sulphuric acid solution that is 29% H_(2)SO_(4) (Molar mass = 98 g mol^(-1)) by mass will be |
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Answer» 1.45 29.0 g of acid are present in solution=100 g 325.8 of acid are present in soluion `((100g))/((29.0 g))xx(352.8 g)` =1216 g `"Density of solution"=("Mass of solution")/("Volume of solution")` `((1216g))/((1000ML))=`1.22 g `mol^(-1)` |
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| 30894. |
The densities of three gases O_2,SO_2 and H_2 are measured at 400 K and 1 atm.They are 0.8g//L,32/15g//L and 1/25g//L respectively.Which of the following statements is/are correct for these gases ? (Use R=1/12L atm//mol-K). |
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Answer» `SO_2` is SHOWING negative deviation from IDEAL behaviour `implies Z_(O_2)=(1xx32)/(0.8xx1/12xx400)=1.2` (positive deviation, so less compressible than ideal gas) ` Z_(SO_2)=(1xx64)/(32/15xx1/12xx400)=0.9` (negative deviation) `Z_(H_2)=(1xx2)/(1/25xx1/12xx400)=1.5` (positive deviation from ideal gas) 1 atm is a low pressure, and at 400K , `Z_(O_2)` is greater than 1.Hence `Z_(O_2)` will be equal to 1 at temperature lower than 400K. Hence Boyle's temperature of `O_2` will be less than 400 K. |
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| 30895. |
The densities of graphite and diamond at 298 K are 3.31 g//cm^(3), respectively. If standard free entropy difference (DeltaG^(@)) is equal to 1895 J"mol"^(-1), the pressure at which graphite will be transformed into diamond at 298 K is |
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Answer» `9.92 xx 10^(7)` P.a Molar volume of diamond `=12/(3.31) = 3.625 cm^(3)` `therefore` Change in molar volume in the CONVERSION of graphite to diamond `=3.625 - 5.33 = -1.704 cm^(3) = -1.704 xx 10^(-3)` litre. Work done during conversion of graphite into diamond `=-PDeltaV` `=-P xx (-1.704 xx 10^(-3))` lit. atm `=P xx 1.704 xx 10^(-3) xx 101.3` Joules. Standard free energy difference `(DeltaG^(@))` is a measure of work done. `therefore 1895 J = P xx 1.704 xx 10^(-3) xx 101.3` J `therefore P=10.97` atm `=10.97 atm xx 1.013 xx 10^(5) Pa = 11.12 xx 10^(8)` Pa Hence, (D) is the CORRECT answer. |
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| 30896. |
The densities of hydrogen and oxygen are 0.09 and 1.44 g L^(-1). If the rate of diffusion of hydrogen is 1 then that of oxygen in the same units will be |
| Answer» Solution :`r_(O)=r_(H)SQRT((d_(H))/(d_(O)))=1 sqrt((0.09)/(1.44))=sqrt((1)/(16))=(1)/(4)` | |
| 30897. |
The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm^(-3), respectively. If the standard free energy difference (DeltaG^(@)) is equal to 1895 J mol^(-1), the pressure at which graphite will be transformed diamond at 298 K is |
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Answer» `9.92xx10^(5)Pa` |
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| 30898. |
The dencity of the neon will be highest at |
| Answer» Answer :C | |
| 30899. |
The denaturation of proteins can be carried out easily by …………………. |
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Answer» Addition of SOLVENTS such as acetone |
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| 30900. |
The DeltaS for the vaporisation of 1 mol of water is 88.3 J/mole K. The value of DeltaS for the condensation of 1 mol of vapour will be |
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Answer» 88.3 J/mol K |
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