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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.
| 201. |
Why in the dissolution of CaCl2, the solubilit increases moderately with high temperature? |
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Answer» Even though the dissolution of CaCI2, is cxothcrmic, the soluhility increases moderately with increase in temperature. Here the entropy factor plays a significant role in deciding the position of equilibrium. |
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| 202. |
Which of the following oxides is soluble in waterA. `Li_(2)O`B. `Na_O`C. `K_(2)O`D. All of these |
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Answer» Correct Answer - 4 Generally oxides `(O^(2-))`, hydroxides `(OH^(-))` are insoluble, except `Li_(2)O, LiOH, Na_(2)O, NaOH, K_(2)O, KOH, BaO, Ba(OH)_(2)` which are soluble and `CaO, Ca(OH)_(2), SrO, Sr(OH)_(2)` which are moderately soluble. Remember these oxides dissolve with evolution of heat and formation of hydroxides. |
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| 203. |
Which of the following bromides is moderately soluble?A. `HgBr_(2)`B. `AgBr`C. `Hg_(2)Br_(2)`D. `PbBr_(2)` |
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Answer» Correct Answer - 1 All common bromides `(Br^(-))` are soluble in water except `AgBr, Hg_(2)Br_(2), PbBr_(2)` which are insoluble and `HgBr_(2)` which is moderately soluble. |
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| 204. |
A colloidal solution of arsenious sulphide is most readily coagulated by the addition of a normal solution?A. NaClB. `CaCl_(2)`C. `Na_(3)PO_(4)`D. `Al_(2)(SO_(4))_(3)` |
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Answer» Correct Answer - D Arsenious sulphide is negatively charged sol more the charge on cation of electrolyte, more the efficiency of electrolyte for coagulation. |
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| 205. |
The freezing point of cyclogexane is 279.65 K. A solution of 14.75 g of a solute in 500 g of cyclohexane has a freezing point of 277.33 K. Calculate the molar mass of the solute. (Given `K_(f)`=20.2 K kg `mol^(-1)`) |
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Answer» Correct Answer - 258.77 g `mol^(-1)` `DeltaT_(f)=T_(f)^(@)-T_(f)=279.65 K-277.33=2.33 K` `K_(f)=20.2" K kg mol"^(-1)=20.2Km^(-1)` `"Molality, m"=(DeltaT_(f))/K_(f)=(2.23K)/(20.2Km^(-1))=0.115 m` Mass of solvent = 500 g = 0.5 kg `"Moles of solute" = "molality"xx"mass of solvent in kg" =0.115" kg mol"^(-1)xx0.5" kg "=0.057" moles"` `"Molar mass of solute"=("Mass of sovent")/("Moles of solute")=(14.75 g)/(0.057" moles")=258.77" g mol"^(-1)`. |
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| 206. |
Which of the following solutions in `H_(2)O` will show maximum depression in freezing point? a.`0.1 M K_(2)[Hgl_(4)]` , b. `0.2 M Ba(NO_(3))_(2)` c.`0.3 M "glucose"` , d.`0.4 M NaCl` |
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Answer» Depression in freezing point `prop` Number of moles Therefore, greater the number of moles greater will be depression in freezing point. `:.` For 0.2 M `Ba (NO)_(3))_(2)`, i=3 |
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| 207. |
Phenol associates in water to double molecules. The values of observed and calculated molecular weight of phenol are `161.84` and `94` , repectively. The degree of association are `161.84` and `94` , repectively. The degree of association of phenol will be a. 60% , b. 84% , c. 45% , d. 80% |
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Answer» b. `i="Calculated molecular weight"/"Observed molecular weight" =94/161.84=0.58` `alpha=(1-i)/(1-n)=(1-0.58)/(1-0.5)=0.42/0.5` `%alpha=0.42/0.5xx100=84%` |
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| 208. |
The freezing point of `0.02 mol` fraction solution of acetic acid (A) in benzene (B) is `277.4K`. Acetic acid exists partly as a dimer `2A = A_(2)`. Calculate equilibrium constant for the dimerisation. Freezing point of benzene is `278.4K` and its heat of fusion `DeltaH_(f)` is `10.042 kJ mol^(-1)`. |
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Answer» Correct Answer - K=3.22 `"X"_("B")=0.02implies"n"_("B")=0.02" and"" n"_("A")=0.98` `"W"_("A")=0.98xx78"gm"` Molality(m) `=("n"_("B")xx1000)/"W"_("A")=(0.02xx1000)/(0.98xx78)=0.2616` `"K"_("f")=("RT"_("f")^(2)"m"_("A"))/(Delta_("f")"H"xx1000)=(8.314xx(278.4)^(2)78)/(10042xx1000)` `=5.00"K"//"m"` `Delta"T"_("F")="i"xx"K"_("F")"m"` `1="i"xx5xx0.2616` `"i"=1/(5xx0.2616)` `=0.7645` `"i"=1+(1/"n"-1)alpha` As acetic acid dimerise, n=2 `"i"=1-alpha/2` `0.7645=1-alpha/2` `alpha=0.4710` We assume: Molarity=Molality `therefore "Molarity (C)"=0.2616` `2"A "hArr "A"_(2)` `{:("At" t = 0,C,0),("At equi.",C(1-a),(Calpha)/(2)):}` `"K"=(["A"_(2)])/(["A"]^(2))=("C"alpha)/(2"C"^(2)(1-alpha)^(2))=alpha/(2"C"(1-alpha)^(2))=0.471/(2xx0.2616(0.529)^(2))=3.22` |
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| 209. |
Two elements A and B form compounds having molecular formula `AB_(2)` and `AB_(4)`. When dissolved in 20g of benzene, 1g of `AB_(2)` lowers the fpt by 2.3K whereas 1g of `AB_(4)` lowers it by `1.3K (K_(f)` for `C_(6)H_(6) = 5.1 K m^(-1))` At. Mass of A and B areA. `42.64` and `25.58`B. `25.58` and `42.64`C. `22.64` and `55.58`D. `45.64` and `22.58` |
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Answer» Correct Answer - B `2.3 = (5.1 xx 1 xx 1000)/((a+2b)xx 20)` `a +2b = (51 xx 50)/(23)` ..(i) Similary, `a +4b = (51 xx 50)/(13)` ...(ii) Subtracting (i) from (ii) `2b = (51 xx 50 xx 10)/(13 xx 23)` `b = 42.64` Substituting in (i). `a = 25.58` |
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| 210. |
Calculate osmotic pressure of a solutions having `0.1M NaCl & 0.2M Na_(2)SO_(4)` and `0.5MHA`. (Given : Weak acid is `20%` dissociated at `300K`). |
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Answer» `pi=pi_(NaCl)+pi_(Na_(2)SO_(4))+pi_(HA)` `0.1RTxx2+0.2RTxx3+0.5RTxx1.2` `=0.0821xx300(0.2+0.6+0.6)` `=34.482` atm. |
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| 211. |
Mixture of X=0.02 mol of `[Co(NH_(3))_(5)SO_(4)]` Br and 0.02 mol of `[Co(NH_(3))_(5)Br]SO_(4)` was prepared in 2 litre of solution. 1 litre of mixture X + excss of `AgNO_(3)to Y` 1 litre of mixture X + excss of `BgCI_(2)to Z` No. of moles of Y and Z are :A. `0.01, 0.01`B. `0.02, 0.01`C. `0.01, 0.02`D. `0.02, 0.02` |
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Answer» Correct Answer - a Concentration of ions in 1 litre of the solution after mixing will be reduced to half i.e. 0.01 mol. `underset((0.01"mol"))(Br^(-)+AgNO_(3))tounderset((0.01"mol"))(AgBr(Y))` `underset((0.01"mol"))(SO_(4)^(2-)+BaCI_(2))tounderset((0.01"mol"))(BaSO_(4)(Z))` |
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| 212. |
An aqueous soltuion of 6.3 g oxaliec acid digydrate is made upto 250mL. The volume of 0.1 N NaOH required to completely neutralilse 10 mL of this solution is :A. `40 mL`B. `20 mL`C. `10 mL`D. `4 mL` |
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Answer» Correct Answer - a Equivalent mass of oxalic acid `=126/2=63` Normally of exalic acid `=6.3/63xx1000/250=0.4N` `underset(("oxalic acid"))(N_(1)V_(1))=underset((NaOH))(N_(2)V_(2))` `0.4xx10=0.1xxV_(2)` `V_(2)=(0.4xx10)/(0.1)=40mL` |
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| 213. |
Why is the elevation in `b.p. of water different in the following solutions? (i) `0.1` molar `NaCl` solution. (ii) `0.1` molar sugar solution |
| Answer» Correct Answer - It is because number of particles in `0.1M NaCl` solution are doulbel (twice) than that of `0.1M` sugar solution. | |
| 214. |
During depression of freezing point in a solution, the following are in equilibrium:A. liquid solvent-solid solventB. liquid solvent- solid soluteC. liquid solute-solid soluteD. liquid solute-solid solvent |
| Answer» Correct Answer - A | |
| 215. |
The freezing point depression of a 0.1 M aq. Solution of weak acid (HX) is `-0.20^(@)"C".` What is the value of equilibrium constant for the reaction: `"HX(aq)"iff"H"^(+)"(aq)"+"X"^(-1)"(aq)"` [Given: `"K"_("f")" for water=1.8 Kg mol"^(-1)"K. & Molality=Molarity"`]A. `1.46xx10^(-4)`B. `1.35xx10^(-3)`C. `1.21xx10^(-2)`D. `1.35xx10^(-4)` |
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Answer» Correct Answer - B `{:(,HX("aq"),1H^(+)(aq),+, X^(-)(aq)),(t=0,0.1M,0,,0),("At equilibrium",(0.1-"x"),"x",,"x"):}` `Delta"T"_("f")=(0.1+"x")xx"K"_("b")` ` 0.20=(0.1+"x")xx1.8` `"K"=("x"xx"x")/((0.1-"x"))=(0.011xx0.011)/((0.1-0.011))` `"K"=1.35xx10^(-3)` |
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| 216. |
Which statement comparing solutions with pure solvent is not correctA. A solution containing a non-volatile solute has a lower vapour pressure than pure solventB. A solution containing a non-volatile solute has a lower boiling point than pure solventC. A solution containing a non-volatile solute has a lower freezing point than pure solventD. A solution will have a greater mass than an equal volume of pure solvent if the solute has a molar mass greater than the solvent |
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Answer» Correct Answer - B On mixing non-volatile solute, elevation in boiling point takes place. |
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| 217. |
During depression in freezing point in a solution, the following are in equilibrium :A. liquid solvent and solid solventB. liquid solvent and solid soluteC. liquid solute and solid solute.D. none is correct. |
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Answer» Correct Answer - a Conceptural question. |
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| 218. |
During depression of freezing point in a solution, the following are in equilibrium:A. liquid solvent-solid solventB. liquid solvent-solid soluteC. liquid solute-solid soluteD. liquid solute-solid solute |
| Answer» Correct Answer - A | |
| 219. |
During depression of freezing point in a solution, the following are in equilibrium:A. liquid solvent, solid solventB. liquid solvent, solid soluteC. liquid solvent, solid soluteD. liquid solute, solid solvent |
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Answer» Correct Answer - A During depression of freezing point in a solution liquid solvent and solid solvent are in equilibrium. |
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| 220. |
If `0.04 M Na_(2)SO_(4)` solutions at `300K` is found to be isotonic with `0.05M NaCl` (`100%` dissociation) solutions. Calculate degree of dissociation of sodium sulphide. |
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Answer» `i_(l)C_(1)RT=i_(2)C_(2)RT` `i_(l)C_(1)=i_(2)C_(2)` `0.04(1+2alpha)=0.05xx2` `alpha=0.75=75%` |
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| 221. |
What is Van’t Hoff factor? |
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Answer» Van’t Hoff factor is the ratio of normal molar mass/abnormal molar mass or any other suitable formula. |
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| 222. |
During depression of freezing point in a solution, the following are in equilibrium:A. Liquid solvent-solid solventB. Liquid solvent-solid soluteC. Liquid solute-solid soluteD. Liquid solute-solid solvent |
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Answer» Correct Answer - A Only solvent molecule under go solidification. So liquid solvent and solid solvent remain in equilibrium. |
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| 223. |
Why are the equimolar solutions of NaCl and glucose not isotonic? |
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Answer» Isotonic solutions are those having same concentrations and osmotic pressure but NaCl and Glucose have not the same osmotic pressure due to the different Van’t Hoff factor. |
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| 224. |
The molecular mass of a solute cannot be calculated by one of the following relationsA. `M_(B) = (K_(b) xx 1000 xx w_(B))/(Delta T_(b) xx w_(A))`B. `M_(B) =(w_(B) xx RT)/(piV)`C. `M_(B) = (p_(0) xx w_(B) xx M_(A))/((p_(0)-p) xx w_(A))`D. `M_(B) = (Delta T_(b))/(K_(b)) xx 1000 (w_(B))/(w_(A))` |
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Answer» Correct Answer - A The correct expression for determination of molecular mass is `M_(B) = (K_(b) xx w_(B) xx 1000)/(DeltaT_(b) xx w_(A))` |
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| 225. |
If 100mL of `1M KOH` is diluted to 1 litre, the concentration of resulting solution isA. `1.0M`B. `10.0M`C. `M//100`D. `0.1M` |
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Answer» Correct Answer - D `M_(2) = (100 xx 1)/(1000) = 0.1M` |
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| 226. |
If equimolar solutions of urea, NaCl, sucrose and BaCl2 have boiling points A, B, C and D, then(a) A = C < B < D (b) A = D < B < C (c) A > B > C < D (d) A < B < C < D |
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Answer» Option : (a) A = C < B < D |
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| 227. |
If an equimolar solution of `CaCI_(2)` and `AICI_(3)` in water have boiling point of `T_(1)` and `T_(2)` respectively thenA. `T_(1) gt T_(2)`B. `T_(2) gt T_(1)`C. `T_(1) = T_(2)`D. `T_(1) ge T_(2)` |
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Answer» Correct Answer - B Since number of particles of `AICI_(3)` in solution is more than that of `CaCI_(2),DeltaT_(b)` of `AICI_(3)` is more. Hence boiling point of `AICI_(3)` solution is higher. |
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| 228. |
Statement-1 : The difference in the boiling points of equimolar solution of `HCl` and `HF` decreases as their molarity is decreased. Statement-2 : The extent of dissociation decreases steadily with increasing dilution.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Sttement-2 is True. |
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Answer» Correct Answer - C Extent of dissociation increases steadily with increasing dilution. |
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| 229. |
A solution of `x` moles of sucrose in `100`grams of water freeze at `-0.2^(@)C` As. Ice separates the freezing point goes down to `0.25^(@)C`. How many grams of ice would have separated ?A. 18 gramsB. 20 gramsC. 25 gramsD. 23 grams |
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Answer» Correct Answer - B (i) `DeltaT_(f)=mxxK_(f)` `0.2=(Xxx1000)/(100)xx1.86 X=(0.2)/(10xx1.86)` after freezing `DeltaT_(f)=(Xxx1000)/((100-y))xx1.86 DeltaT_(f)=0.25` On solving, Amount of ice `y=20 g` ice |
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| 230. |
Sea water is found to contain `5.85% NaCI` and`9.50% MgCI_(2)` by weight of solution. Calculate its normal boiling point assuming `80%` ionisation for `NaCI` and `50%` ionisation of `MgCI[K_(b)(H_(2)O) =0.51 kg mol^(-1)K]`A. `T_(b)=101.9^(@)c`B. `T_(b)=104.9^(@)C`C. `T_(b)=108.5^(@)C`D. `T_(b)=110.3^(@)C` |
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Answer» Correct Answer - A `DeltaT_(b)=K_(b).m.i` In `100gm` of solution moles of `NaCl=0.1` `(alpha=0.8)` moles of `MgCl_(2)=0.1` `(alpha=0.5)` `NaClrarrNa^(+)+Cl^(-)` `i_(NaCl)=1+(2-1)0.8=1.8` Effective no. of moles of `MgCl_(2)=0.1xx1.8=0.18` `i_(MgCl_(2))=1+(3-1)0.5=2` Effective no. of moles of `MgCl_(2)=0.1xx2=0.2` Total no. of mole `=0.18+0.2=0.38` `DeltaT_(b)=(0.38)/(100)xx100xx0.51=3.8xx0.51=1.938` So, `T_(b)=100+1.938=101.938` |
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| 231. |
Consider following cases: I: `2M CH_(3)COOH` solution is benzene at `27^(@)C` where there is dimer formation to the extent of `100%` II: `0.5M Kci` aq. Solution at `27^(@)C`, which ionises `100%` Which is/are true statements(s):A. both are isotonicB. I is hypertonicC. II is hypotonicD. none is correct |
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Answer» Correct Answer - A `pi_(I)=2Rxx300xx(1+((1)/(2)-1)1)=300R` `pi_(II)=0.5Rxx300xx2=300R` |
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| 232. |
Molecular mass from osmotic pressure: Pepsin is an enzyme present in the human digestive tract. A solution of a `0.500 gram` sample of purified pepsin in `30.0 mL` of aqueous solution exhibits an osmotic pressure of 892 torr at `27^(@)C`. Estimate the molecular mass of pepsin. Strategy: An enzyme is a protein that acts as a biological catalyst. Pepsin catalyzes the metabolic cleavage of amino acid chain (called peptide chains) in other proteins. To determine molecular mass, we must first find the number of moles `(n)` of pepsin represented by the `0.500 g` sample. We can do this by first rearranging the equation `pi=CRT` for osmotic pressure to find the molarity of the pepsin solution and then multiplying it by the volume of the solution to obtain the number of moles of pepsin. |
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Answer» Step 1. Calculate molarity of pepsin Since `pi =CRT`, then `C=pi/(RT)` `C=((8.92 "torr" xx(1 "atm")/(760 "torr")))/((0.082 (L "atm")/(K mol))(300 K))` `=0.000477 mol L^(-1)` `=4.77xx10^(-4) mol L^(-1)` Note that we convert 8.92 torr to atmospheres to be consistent with units of R. Step 2. Calculate moles of pepsin `Molarity (C )=n_("pepsin")/V_(mL)xx(1000 mL)/L` Since the volume of the solution is `30.0 mL`, the number of moles of pepsin is `n_("pepsin")=(4.77xx10^(-4) mol L^(-1))(30.0 mL)((1L)/(1000 mL))` `=143.1xx10^(-7) mol` `1.43xx10^(-5) mol` Step 3. Calculation the molar mass Knowing both the mass and the number of moles of pepsin, we can calculate the molar mass and hence molecular mass `"moles"_("pepsin")=("mass"_("pepsin"))/("molar mass"_("pepsin"))` or `"molar mass"_("pepsin")=("mass"_("pepsin"))/("moles"_("pepsin"))` `=(0.500 g)/(1.43xx10^(-5) mol)` `=0.349xx10^(5) g mol^(-1)` `=3.50xx10^(4) g mol^(-1)` Thus, the molecular mass of pepsin is approximately 35,000 amu. This is typical for medium-sized proteins. Alternatively, we can calculate the molar mass of pepsin through the equation `(2.74)`. |
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| 233. |
Explain why the melting point of a substance gives an indication of the purity of a substance. |
| Answer» Impurities cause depression in freezing point (melting point). More the Impurities, lower is the freezing point. | |
| 234. |
Unit of lowering of vapour pressure isA. atmB. `Nm^(-1)`C. UnitlessD. `Nm^(2)` |
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Answer» Correct Answer - 1 |
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| 235. |
One molar solution of sulphuric acid is equal toA. Normal solutionB. `(N)/(2)` solutionC. 2N solutionD. 4 N solution |
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Answer» Correct Answer - 3 |
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| 236. |
Distinguish between the boiling point of a liquid and the normal boiling point of a liquid. |
| Answer» The boiling point is the temperature at which the vapour pressure of the liquid becomes equal to the surrounding pressure. The normal boiling point is the temperature at which the vapour pressure is `1 atm`. | |
| 237. |
`2 g` each of two solutes ` A` and `B` (molar mass of `A` is greater than that of `B`) are dissolved separately in `50 g` each of the same solvent. Which will show greater elevation in the boiling point? |
| Answer» The solution containing solute `B` will show greater elevation in the boiling point because `DeltaT_(b) prop (1)/(Mw)`. | |
| 238. |
Solvent and solutes are always defined on the basis ofA. Mass compositionB. Molar compositionC. Physical stateD. All of these |
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Answer» Correct Answer - 4 |
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| 239. |
Ionic solutes that do not react with the solvent undergoA. solvolysisB. solvationC. hydrationD. hydrolysis |
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Answer» Correct Answer - 2 This is a kind of physical change in which molecules of solvent are attached in oriented clusters to the solute particles. |
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| 240. |
What will be the van’t Hoff factor for O.1 M ideal solution? |
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Answer» Van't Hoff factor = 1, because ideal solution does not undergo dissociation or association. |
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| 241. |
Why cutting onions taken from the fridge is more comfortable than cutting onions lying at room temperature? |
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Answer» The vapour pressure is low at lower temperature. So, less vapours of tear – producing chemicals are produced. |
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| 242. |
Cutting onions taken from the fridge is more comfortable than cutting those lying a room temperture. Explain why. |
| Answer» At lower temperature, the vapour pressure is low. Less vapours of the tear-producing chemicals are produced. | |
| 243. |
When a graph is pplotted between `log x//m` and log p, it is straight line with an angle `45^(@)` and intercept `0.3010` on y-axis. If initial pressure is `0.3` atm, what will be the amount of gas adsorbed per gm of adsorbent:A. `0.4`B. `0.6`C. `0.8`D. `0.1` |
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Answer» Correct Answer - B `"log"x/M =log k+1/n log P` `1/n=tan 45^(@) log k=0.3010` `n=1 k=2` `x/m=2xx(0.3)^(1)` `x=0.6` |
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| 244. |
Choose the correct reason (s) for the stability of lyophobic colloidal particles.A. Preferential adsorption of ions on their surface from the solutionB. Preferential adsorption of solvent on their surface from the solutionC. Attraction between different particles having opposite charges on their surfaceD. Potential difference between the fixed layer and the diffused layer of opposite charges around the colloidal particles. |
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Answer» Correct Answer - A::D (A) due to preferential adsorption of common ions (C) due to repulsion not due to attraction (D) The layer of oppositely charged particles around any colloidal particles will decrease the potential energy of system as a whole. |
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| 245. |
A negative catalyst will :A. raise the energy of activation for a given reactionB. take away the energy of reactants and deactivate themC. catalyse the backward reaction more than the forward one, thereby shifting equilibrium backward.D. None of these |
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Answer» Correct Answer - A Negative catalyst provides a path of higher activation energy |
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| 246. |
The given graph/data I,II,III and IV represent general trends observed for different physisorption and chemisorption processes under mild conditions of temperature and pressure . Which of the following choice(s) about I,II,III, and IV is/are correct ? A. I is physisorption and II is chemisorptionB. I is physisorption and III is chemisorptionC. IV is chemisorption and II is chemisorptionD. IV is chemisorption and III is chemisorption |
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Answer» Correct Answer - A::C In physisorption on increasing temperature at constant pressure, adsorption decreases while in chemical adsorption on increasing temperature due to requirement of activation energy adsorption will increase at same pressure. SO, I is physisorption while II is chemisorption. III is physical adsorption as on increasing temperature, extent of adsorption is decreasing. IV is representing enthalpy change (which is high) during chemical adsorption (due to bond formation) So, is valid for chemical adsorption. So, answer is (A) and (C) |
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| 247. |
The dispersed phase in colloidal iron (III) hydroxide and collodial gold is positively and negtively charged respectively with of the following statement is not correct ?A. Coagulation in both sols can be throught about by electrophoresisB. Mixing the sols has no effectC. Sodium sulphate solution causes coangulation in both solsD. Magnesium chloride solution coagulates, the gold sol more readily than the ion (III) hydroxide sol. |
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Answer» Correct Answer - B On mixing, they will coagulate each other being +ve and -ve charged. |
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| 248. |
(i) At 298 K the volume of `NH_(3)` adsorbed by 1 g of charcoal is higher than that of `H_(2)` under similar conditions. (ii) The movement of colloidal particles towards the oppositely charged electrodes on passing current is known as Brownian movement. If T for true and F for false then correct option is :A. T, TB. T, FC. F, TD. F, F |
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Answer» Correct Answer - B (i) `NH_(3)` is easi,y liquifiable, so ordered more than `H_(2)`. (ii) This phenomenon is known as electrophoresis, not Brownian movement. |
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| 249. |
Methylene blue, from its aqueous solution, is adsorbed on activated charcoal at `25^(@)C` . For process, the correct statement isA. The adsorption required activation at `25^(@)C`B. The adsorption is accompanied by a decreases in enthalpyC. The adsorption increases with increase of temperatureD. The adsorption is irreversible |
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Answer» Correct Answer - B As the adsorption of methylene blue over activated characoal is physisorption (Reference: NCERT), it is accompanied by decrease in enthalpy. |
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| 250. |
Which of the following gases, will be adsorbed maximum on a solid surface?A. `CO_(2)`B. `O_(2)`C. `N_(2)`D. `H_(2)` |
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Answer» Correct Answer - A Easily liquefiable gases like `CO_(2)` are adsorbed to greater extent than gases like `O_(2), N_(2)` and `H_(2)` |
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