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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. |
In which ratio of volumes 0.4 M HCl and 0.9 M HCl are to be mixed such that the concentration of the resultant solution becomes 0.7 M ?A. `4 : 9`B. `2 : 3`C. `3 : 2`D. `1 : 1` |
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Answer» Correct Answer - B Let the volume ratio = x : y a/c to question `0.7=(0.4x +0.9 y)/(x + y) rArr 0.7x + 0.7 y = 0.4 x + 0.9 y` `rArr 0.3x = 0.2 y rArr (x)/(y)=(0.2)/(0.3)rArr =(0.2)/(0.3)rArr (x)/(y)=2:3` |
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| 202. |
The equivalent weight of `KMnO_(4)` in (a) neutral medium, (b) acidic medium and (c ) alkaline medium is `M//..` ( where `M` is mol.wt. of `KMnO_(4)`)A. Molecular weightB. `("Molecular weight")/(2)`C. `("Molecular weight")/(3)`D. `("Molecular weight")/(5)` |
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Answer» Correct Answer - C `KMnO_(4)` in neutral medium changes to `MnO_(2)` `underset((+7))(MnO_(4)^(-))to underset((+4))(MnO_(2))` Thus n-factor = change in O.N. = 3 `therefore "Eq. wt." = ("Mol.wt.")/(3)` |
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| 203. |
V.P. of a solution containing non-volatile solute isA. more than the vapour pressure of a solventB. lass that the vapour pressure of solventC. equal to the vapour pressure of solventD. none |
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Answer» Correct Answer - B Addition of non-volatile solute always lowers the vapour pressure. |
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| 204. |
An aqueous solution of methanol in water has vapour pressure:A. equal to that of waterB. equal to that of methanolC. more than that of waterD. less than that of water |
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Answer» Correct Answer - C Methanol-water system shows positive deviations from Raoult,s law, Hence, vapour pressure of solution is more than that of water. |
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| 205. |
By adding water to the solution, itsA. concentration remains sameB. concentration increasesC. ionisation decreasesD. concentration decreases |
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Answer» Correct Answer - D Addition of water increases dilution,`V prop 1/C` |
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| 206. |
When a substance is dissolved in a solvent the vapour pressure of solvent decreases. This brings:A. an increase in b.bt. Of the solutionB. a decrease in b.pt. of a solutionC. an increase in f.pt. of the solventD. none |
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Answer» Correct Answer - A Lower is the vapour pressure of liquid, higher is its boiling point. |
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| 207. |
When a substance is dissolved in a solvent the vapour pressure of solvent decreases. This brings:A. an increase in the b.p. of the solutionB. a decrease in the b.p. of the solventC. The solution having a higher freezing point than the solventD. the solution having a lower osmotic pressure than the solvent. |
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Answer» Correct Answer - A Greater the lowering in V.P., higher is the b.pt. |
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| 208. |
When the concentration is expressed as the number of moles of a solute per litre of solution it is known asA. NormalityB. MolarityC. Mole fractionD. Mass percentage |
| Answer» Correct Answer - B | |
| 209. |
By dissolving 5 g substance in 50 g of water, the decrease in freezing point is `1.2^(@)C`. The gram molal depression is `1.85^(@)C`. The molecular weight of substance isA. `105.4`B. `118.2`C. `137.2`D. `154.2` |
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Answer» Correct Answer - D We know that, `m=(1000 K_(f)xx w)/(Delta TW)` Here, `Delta T=1.2^(@)C, K_(f)=1.85^(@)` `w=5g, W=50 g` `m=(1000xx1.85xx5)/(1.2xx50)=154.2`. |
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| 210. |
The process of getting fresh water from sea water is known asA. osmosisB. FillrationC. DiffusionD. Reverse osmosis |
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Answer» Correct Answer - D Reverse osmosis involves movement of solvent particles through semipermeable membrane from concentrated solution to dilute solution under pressure. |
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| 211. |
Which of the following cannot be separted from air by the process of fractional distillation?A. Benzene-tolueneB. Water - ethyl alcoholC. Water - nitric acidD. Water - hydrochloric acid |
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Answer» Correct Answer - A Aromatic compound generally separated by fractional distillation e.g. Benzene + Toluene. |
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| 212. |
All form ideal solution exceptA. `C_(2)H_(5)Be` and `C_(2)H_(5)I`B. `C_(6)H_(5)Cl` and `C_(6)H_(5)Cl` and `C_(6)H_(5)Br`C. `C_(6)H_(6)` and `C_(6)H_(5)CH_(3)`D. `C_(2)H_(5)I` and `C_(2)H_(5)OH` |
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Answer» Correct Answer - D `C_(2)H_(5)I` and `C_(2)H_(5)OH` do not form ideakl solution. |
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| 213. |
Assertion: `C Cl_(4)`and `H_(2)O`are immiscible . Reason : `C Cl_(4)` is a polar solvent.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - C The assertion that `C Cl_(4)` & `H_(2)O` are immiscible is true because `C Cl_(4)` is non-polar liquid while water is polar hence assertion is true and reason is false. |
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| 214. |
The molal elevation constant is the ratio of the elevation in B.P. toA. MolarityB. Molality does not change with temperatureC. Mole fraction of soluteD. Mole fraction of solvent |
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Answer» Correct Answer - B `Delta T_(b)=K_(b)xx m` or `K_(b) = Delta T_(b)//m` |
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| 215. |
Mark the correct relationship between the boiling points of very dilute solutions of `BaCl_(2)(t_(1))` and `LCl(t_(2))` ,having the same molarityA. `t_(1)=t_(2)`B. `t_(1)gt t_(2)`C. `t_(2)gt t_(1)`D. `t_(2)` is approximately equal to `t_(1)` |
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Answer» Correct Answer - B `BaCl_(2)` furnishes more ions than KCl and thus shows higher boiling point `T_(1)gt T_(2)`. |
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| 216. |
Elevation in boililng point was `0.52^(@)C` when 6 gm of a compound X was dissolved in 100 gm of water. Molecular weight of X is (`K_(b)` for water is 0.52 per 1000 gm of water)A. 120B. 60C. 180D. 600 |
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Answer» Correct Answer - B `Delta T_(b)=(100xx K_(b)xx w)/(m xx W) therefore 0.52=(100xx5.2xx6)/(m xx 100)` `m=(100xx5.2xx6)/(0.52xx100)=60`. |
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| 217. |
At higher altitudes the boiling point of water lowers becauseA. Atmospheric pressure is lowB. Temperature is lowC. Atmospheric pressure is highD. None of these |
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Answer» Correct Answer - A The boiling occurs at lower temperature if atmospheric pressure is lower than 76 cm Hg. |
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| 218. |
The value of `K_(f)`for water is `1.86^(@)`, calculated from glucose solution, The value of `K_(f)`for water calculated for NaCl solution will be,A. 1.86B. lt 1.86C. gt1.86D. Zero |
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Answer» Correct Answer - A `K_(f)` is characteristic constant for given solvent. |
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| 219. |
Equimolar solutions in the same solvent have-A. Same boiling point but different freezing pointB. Same freezing point but different boiling pointC. Same boiling and same freezing pointsD. Different boilling and different freezing points |
| Answer» Correct Answer - C | |
| 220. |
What does not change on changing temperatureA. Mole fractionB. NormalityC. MolalityD. None of these |
| Answer» Correct Answer - A::C | |
| 221. |
Colligative properties are used for the determination of ________.A. Molar MassB. Equivalent weightC. Arrangement of moleculaesD. Both (a) and (b) |
| Answer» Correct Answer - C | |
| 222. |
Colligative properties of a solution depends uponA. nature of solute onlyB. nature of both solute and solventC. nuber of solute particlesD. number of solvent particles |
| Answer» Correct Answer - c | |
| 223. |
Increasing the temperature of an aqueous solution wil caseA. Decrease in molalityB. Decrease in molarityC. Decrease in mole fractionD. Decrease in % w/w |
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Answer» Correct Answer - B An increase in temperature increases the volume of the solution and thus decreases its molarity. |
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| 224. |
The sum of mole fractions of A,B and C in a solution containing 0.1 mole each of A,B and C is:A. 0.1B. 0.3C. 1D. 1/3. |
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Answer» Correct Answer - C The sum of mole fractions of all the components =1. |
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| 225. |
Among (P) water (Q) ethanol and (R) mercury the correct order of vapour pressure at room temp isA. PgtQgtRB. QgtPgtRC. RgtQgtPD. QgtRgtP |
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Answer» Correct Answer - B If the liquid is more volatile,more is its V.P. |
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| 226. |
92 g. of ethanol is dissolved in 108g. Of water. The mole fraction of water in the solution isA. 0.25B. 0.75C. 0.5D. 0.35 |
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Answer» Correct Answer - A `"Moles of ethanol"="Mass"/("Molar Mass")=92/46=2` and moles of water `= 108/18=6` mole fraction of ethanol `=2/(2+6)=0.25` |
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| 227. |
20 g of hydrogen is present in 5 litre vessel. The molar concentration of hydrogen isA. 4B. 1C. 3D. 2 |
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Answer» Correct Answer - D Molar concentration `[H_(2)]=("Mole")/("V in litre")=(20//2)/(5)=2`. |
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| 228. |
For a solution of dibasic acid the molarity (M) and normality (N) are related asA. N=M/2B. 2 M = NC. M =ND. M gt N |
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Answer» Correct Answer - B For dibasic acid there are 2 H, groups Hence basicity=2 and normality = 2 M. |
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| 229. |
The amount of anhydrous `Na_(2)CO_(3)` present in 250 ml of 0.25 M solution isA. 6.225 gB. 66.25 gC. 6.0 gD. 6.625 g |
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Answer» Correct Answer - D `M=(w)/(m xxV(l)) , 0.25 = (w)/(106xx0.25), w=6.625 gm` |
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| 230. |
How many grams of `H_(2)SO_(4)` is/are to be dissolved to prepare 200 mL aqueous solution having concentration of `[H_(3)O^(+)]` ions is 1 M at `25^(@)C` temperature?A. 19.6 gB. 0.98 gC. 4.9 gD. 9.8 g |
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Answer» Correct Answer - d Molarity, `M = (w_(2)xx1000)/(M_(2)xxV("in" mL))` where, `w_(2)` is the mass of `H_(2)SO_(4)` in g and `M_(2)` is the molar mass of `H_(2)SO_(4)`. `w_(2) = (1xx98xx200)/(1000) = 19.6 g` `H_(2)SO_(4) + 2H_(2)O hArr 2H_(3)O^(+) + SO_(4)^(2-)` But according to equation, 1 mole of `H_(2)SO_(4)` gives 2 moles of `[H_(3)O^(+)]` ions. Thus, the amount of `H_(2)SO_(4)` to prepare 200 mL solution having 1 M concentration of `H_(3)O^(+)` ions is 19.6/2 = 9.8 g. |
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| 231. |
If 18g of glucose `(C_(6)H_(12)O_(6))` is present in 1018 g of an aqueous solution of glucose, it is said to beA. 1 molalB. 1.1 molalC. 0.5 molalD. 0.1 molal |
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Answer» Correct Answer - D 18 g glucose `=18/180 ` mol =0.1 mol. As it is present in (1018-18) = 1000 g water, the solution is 0.1 molal. |
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| 232. |
If 18 g of glucose is present in 1000 g of solvent, the solution is said to beA. 1 molalB. 1.2 molalC. 0.5 molalD. 0.1 molal |
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Answer» Correct Answer - D `(18)/(180xx1)=(1)/(10)=0.1` molal. |
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| 233. |
58.5 g of NaCl and 180 g of glucose were separately dissolved in 1000 mL of water. Identify the correct statement regarding the elevation of boiling point (b.pt.) of the resulting solutionsA. NaCl solution will show higher elevation of b.pt.B. Glucose solution will show higher elevation of b.pt.C. Both the soltuions will show equal elevation of b.pt.D. The b.pt. of elevation will be shown by neither of the solutions |
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Answer» Correct Answer - A `Delta T_(b) = i K_(b) m` For water `rarr 1000 mL = 1000 g` Molality of `NaCl=(w//M.W.)/("W(solvent)")xx1000` `=(58.5//58.5)/(1000)xx1000=1m` Molality of glucose `=(180//180)/(1000)xx1000=1m` i for NaCl = 2, i for glucose = 1 `Delta T_(b)` for `NaCl gt Delta T_(b)` for glucose |
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| 234. |
What weight of ferrous ammonium sulphate is needed to prepare 100 ml of 0.1 normal solution (mol. wt. 392)A. 39.2 gmB. 3.92 gmC. 1.96 gmD. 19.6 gm |
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Answer» Correct Answer - B `w=(0.1xx100xx392)/(1000)=3.92 g` |
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| 235. |
1000 gram aqueous solution of `CaCO_(3)` contains 10 gram of carbonate. Concentration of solution is:A. 10 ppmB. 100 ppmC. 1000 ppmD. 10000 ppm |
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Answer» Correct Answer - D `10^(3)` parts of `CaCO_(3)` has number of parts = 10 `10^(6)` parts of `CaCO_(3)` has number of parts `=(10)/(10^(3))xx10^(6)=10,000` ppm. |
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| 236. |
The vapour pressure will be lowest forA. 0.1 M sugar solutionB. 0.1 M KCl solutionC. `0.1 M Cu(NO_(3))_(2)` solutionD. `0.1 M AgNO_(3)` solution |
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Answer» Correct Answer - C Vapour pressire of a solvent is lowered by the presence of solute in it. Lowering in vapour pressure is a colligative property i.e., it depends on the no. of particles present in the solution. `Cu(NO_(3))_(2)` give the maximum no. of ions. (i.e., 3) so it causes the greatest lowering in vapour pressure of water. |
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| 237. |
Lowering of vapour pressure is highest forA. UreaB. 0.1 M glucoseC. `0.1 M MgSO_(4)`D. `0.1 M BaCl_(2)` |
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Answer» Correct Answer - D `(P^(@)-P_(s))/(P^(@))` = molality `xx(1-alpha + x alpha + y alpha)` the value of `P^(@)-P_(s)` is maximum for `BaCl_(2)`. |
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| 238. |
The aqueous solution that has the lowest vapour pressure at a given temperature, isA. 0.1 M sodium phosphateB. 0.1 M barium chlorideC. 0.1 M sodium chlorideD. 0.1 M glucose |
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Answer» Correct Answer - a `Na_(3)PO_(4) hArr 3Na^(+) + PO_(4)^(-3)` |
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| 239. |
The vapour pressure of water at `20^(@)` is `17.5 mmHg`. If `18 g` of glucose `(C_(6)H_(12)O_(6))` is added to `178.2 g` of water at `20^(@) C`, the vapour pressure of the resulting solution will beA. 15.750 mm HgB. 16.500 mm HgC. 17.325 mm HgD. 17.65 mm Hg |
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Answer» Correct Answer - C `(P^(@)-P_(s))/(P^(@))=(n)/(n+N)` `n=(18)/(180)=0.1, N=(178.2)/(18)=9.9` `(17.5-p_(s))/(17.5)=(0.1)/(0.1+9.9),(17.5-p_(s))/(17.5)=(1)/(100)` `1750-100 p_(s)=17.5 , 1732.5=100 p_(s)` `p_(s)=17.325` mm Hg. |
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| 240. |
1 Molar solution containsA. 1000g of soluteB. 1000g of solventC. 1 litre of solventD. 1 litre of solution |
| Answer» Correct Answer - D | |
| 241. |
10.6 gram of a substance of molecular weight 106 was dissolved in 100ml . 10 ml of this solution was pipetted out into a 1000ml flask and made up to the mark with distilled water. The molarity of the resulting solution isA. 1.0 MB. `10^(-2)M`C. `10^(-3)M`D. `10^(-4)M` |
| Answer» Correct Answer - B | |
| 242. |
The molarity of 0.006 mole of NaCl in 100ml solution isA. `0.6`B. `0.06`C. `0.066`D. None of these |
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Answer» Correct Answer - B `M=(n)/(V(l))=(0.006)/(0.1)=0.06` |
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| 243. |
Which of the following should be done in order to prepare 0.40 M NaCl starting with 100ml of 0.30M NaCl (mol. wt. of NACl - 58.5)A. Add 0.585 g NaClB. Add 20 ml waterC. Add 0.010ml NaClD. Evaporate 10ml water |
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Answer» Correct Answer - A 100 ml. of `0.30 M =(100xx0.3)/(1000)=0.03` mole of NaCl 100 ml of `0.40 M =(100xx0.4)/(1000)=0.04` mole of NaCl Moles of NaCl to be added = 0.04-0.03 =0.01 mole = 0.585 gm |
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| 244. |
What will be the value of molality for an aqueous solution of 10% w/W NaOH?A. `2.778`B. 5C. 10D. `2.5` |
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Answer» Correct Answer - A Let mass of solution = 100 g `therefore` Mass of NaOH = 10 g Mass of solvent = 90 g `therefore` Molality `=((10)/(40))/((90)/(1000))=(10)/(40)xx(1000)/(90)=2.778`. |
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| 245. |
Assertion :- On adding NACl to water its vapour pressure increases. Reason :- Addition of non-volatile solute increases the vapour pressure.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - D If a non-volatile solute is added to water its vapour pressure always decreases. Therefore, both assertion and reason are false. |
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| 246. |
Osmotic pressure of 0.1 M solution of NaCl and `Na_(2)SO_(4)` will beA. SameB. Osmotic pressure of NaCl solution will be more than `Na_(2)SO_(4)` solutionC. Osmotic pressure of `Na_(2)SO_(4)` solution will be more than NaClD. Osmotic pressure of `NaSO_(4)` will be less than that of NaCl solution |
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Answer» Correct Answer - C `Na_(2)SO_(4)` have more osmotic pressure than NaCl solution because `Na_(2)SO_(4)` gives 3 ions. |
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| 247. |
Some statements are given below for the same solution `Delta T_(b) =Delta T_(f)` 5% solution of urea will have more osmotic pressure than 10% solution of glucose elevation of B. pt. is due to increase in vapour pressure of solution on adding solute depression of F.pt. is due to decrease in vapoure pressure of solution on adding solute. Among the aboveA. B and D are trueB. A,B & D are falseC. B and C are falseD. only D is true |
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Answer» Correct Answer - D (A) False. As for a given solvent `K_(b) and K_(f)` are not equal. (B) True.as urea solution has higher molarity than glucose solution. `pi = MRT or `pi = CRT where` C = "molarity of solution" " As" R and T are constants `pi alpha M` (c) False. It is due to lowering of V.P. (D) True. |
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| 248. |
The osmotic pressure in atmospheres of 10% solution of canesugar at `69^(@)C` isA. 724B. 824C. `8.21`D. `7.21` |
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Answer» Correct Answer - C `pi V = nRT` `pi=(w)/(m)(RT)/(V)=(10)/(342)xx(0.821xx(273+69))/(0.1)=8.21` atm |
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| 249. |
The value of osmotic pressure of a 0.2 M aqueous solution at 293 K isA. 8.4 atmB. 0.48 atmC. 4.8 atmD. 4.0 atm |
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Answer» Correct Answer - C `pi=CRT=0.2xx0.0821xx293=4.81` atm. |
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| 250. |
At `25^(@)C` aqueous solution of glucose (molecular weight `= 180 g mol^(-1)`) is isotonic with a 2% aqueous solution containing an unknown solute. What is the molecular weight of the unknown soluteA. 60B. 80C. 72D. 63 |
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Answer» Correct Answer - C `pi_("glucose")=pi_("unknown solute")` `therefore (m_(1))/(M_(1))=(m_(2))/(M_(2))` or `(5)/(180)=(2)/(M_(2))rArr M_(2)=72`. |
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