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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.
| 1201. |
Osmotic Pressure |
| Answer» Osomotic pressure is the extra pressure that is to be applied on the solution side when the solution and solvent are separated by a semi-permeable membrane to stop osmosis. | |
| 1202. |
How many ethyl alcohol must be added to `1.00 L` of water so that the solution will not freeze at `-4^(@)F` ? |
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Answer» Using relation `(DeltaT_(f))=K_(f)m` where `(DeltaT_(f))=-4^(@)F` or `-4^(@)F=-20^(@)C` `[.^(@)C=(5)/(9)(.^(@)F-32)]` `:. m=DeltaT_(f)/K_(f)=20/1.86=10.7` Therefore, amount of ethyl alcohol to be added `= m xx` molrcular weight `=10.7xx46.0` `=495 g` |
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| 1203. |
How much ethyl alcohol must be added to `1.00 L` of water so that the solution will not freeze at `-4^(@)F` ?A. `122`gB. `512`gC. `670`gD. `495`g |
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Answer» Correct Answer - C `-4^(@)f`is to be converted into `C^(@)` `(F-32)/(180)=(C)/(100)` ltbr `(-6-30)/(180)=C/(100)` `:. C=-27^(@)C` `:. DeltaT_(f)=-27^(@)C` 1 L `H_(2)O=1000g` water `DeltaT_(f)=(1000k_(f)w_(1))/(m_(1)w_(1))` `w_(1)("ethyl alcohol")=(DeltaT_(f)m_(1)w_(2))/(1000k_(f))=(27xx46xx1000)/(1000xx1.86)` `=(12466)/(186)=670g` |
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| 1204. |
The vapour pressure of a dilute solution of a solute is influeneced by:A. temperature of solutionB. melting point of soluteC. mole fraction of soluteD. degree of dissociation of solute |
| Answer» Correct Answer - B | |
| 1205. |
The vapour pressure of a dilute solution of a solute is influenced by :A. temperature of the solutionB. mole fraction of the soluteC. melting point of the soluteD. degree of dessociation of the solute |
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Answer» Correct Answer - a,b,d (a,b,d) are all correct. |
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| 1206. |
Mark out the incorrect combination in the depolar forceA. KCI in `H_(2)O` , ion depole forcesB. `C_(6)H_(6) " in"C CI_(4)`, dispersion forcesC. HCI in `CH_(3)CN`, ion dipole forcesD. HCI in water, hydrogen bonding |
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Answer» Correct Answer - c,d (c,d) are correct combination. |
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| 1207. |
Mark out the correct statement :A. Addition of mole of NaCI in 1L water increses boiling pointB. Addition of 1 mole of `CH_(3)OOH` to 1L water decreases boiling pointC. Addition of solid pointD. Addition of solid `CaCI_(2)" in ice "(0^(@)C)` decreases freezing point. |
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Answer» Correct Answer - a,b,c,d (a,b,c,d) are correct statements. |
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| 1208. |
In the depression of freezing point experment, it is found thatA. The vapour pressure of the solution is less than that of pure solventB. The vapour pressure of solution is more than the pure soventC. Only the solute molecules solidify at the freezing pointD. Only the solvent molecules solidify at the freezing point. |
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Answer» Correct Answer - a,d (a,d) are both statements. |
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| 1209. |
When 1 mL of 0.5 M `K_(2)SO_(4)` is added to 10 mL of 1 M `BaSO_(4)` precipitate of `BaSO_(4)` is formed. As a result,A. Freezing point is increasedB. Boiling point is increasedC. Freezing point is loweredD. Boilint point is lowerd |
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Answer» Correct Answer - b,c (b,c) are both correct statements. |
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| 1210. |
Which of the following is/are correct for a solution of a particular concerntration ?A. Molarity is always less than molalityB. Formality is equivalent to molarityC. Mole fraction is equivalent to mass fractionD. Normality of `H_(2)SO_(4)` solution is double than its molarity. |
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Answer» Correct Answer - b,d (b,d) are both correct statements. |
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| 1211. |
Which of following form/s an ideal solution ?A. Ethyl bromide + ethyhl iondideB. Ethyl alcohol + waterC. Chloroform + benzeneD. Benzen + toluene |
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Answer» Correct Answer - a,d (a,d) are both correct statemtnts. |
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| 1212. |
Which are the correct statements ?A. `CHCI_(3)+c CI_(4)`-Endothermic solution ?B. Acetoc acid + Pyridine - Hot solutionC. `HNO_(3)+"Water - Endothermic solution "`D. Water + HI - Minimum boiling point azetrope |
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Answer» Correct Answer - a,b (a,b) are both correct answer. |
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| 1213. |
The ebullioscopic constant for benzene is `2.52 K Kg mol^(-1)`. A solution of an organic substance in benzene boils at `0.125^(@)C` higher than benzene. Calculate the molality of solution? |
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Answer» Given values are: `K_(b) = 2.52 K m^(-1)` or `2.52 K kg m^(-1)` `DeltaT_(b) = 0.125^(@)C` Using the relation,`DeltaT_(b) =K_(b)m` `:. M = (Delta_(b)T) / K_(b)` `= (0.125)/(2.52)= 0.05` Hence, molality of the solution is `0.05 m`. |
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| 1214. |
Consider the following graphs Choose the correct statementsA. According to both graphs mole fraction of A `gt` mole fraction of B in condensateB. Graph I belongs to minimum boiling azenotropeC. Graph II belongs to maximum boiling azeotropeD. Graph II belongs to minimum boilings azeotrope whil graph I belongs to minimum boiling azeotrope |
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Answer» Correct Answer - 1,2,3 |
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| 1215. |
Dimer of acetic acid in benzene in benzen is equilibrium with acetic acid at a particular condition of temperature and pressure. If half of the dimer moleculaes are hypotheticaly separated out thanA. Osmotic pressure of the solution reducesB. Freezing point of the soluion reducesC. Boiling point of the solution reducesD. Vapour pressure of the soluteion reduces |
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Answer» Correct Answer - a,c (a,b) are both correct answer. |
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| 1216. |
Dimer acetic acid in benzene is in equilibrium with acetic acid at a particular condition of temperature and pressure. If half of the dimer molecules are hypothetically separated out thenA. Osmotic pressure of the solution reducesB. Freezing point of the solution reducesC. Boiling point of the solution reducesD. Vapour pressure of the solution reduces |
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Answer» Correct Answer - 1,3 |
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| 1217. |
What will be the concentration of sucrose solution which develops an osmotic pressure of 2 atm at `27^(@)C` ? |
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Answer» Osmotic pressure `rArr pi=CRT` `rArr 2=Cxx0.0821xx300` `rArr 0.081 M` |
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| 1218. |
What would be the osmotic pressure of 0.05 M solution of sucorse at `5^(@)C` ? Find out the concentration of a solution of glucose which would be isotonic with this solution of sucrose. (Molecular mass of sucrose = 342, Molecular mass of glucose = 180) |
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Answer» P=CRT `P=0.05xx0.082xx278 atm` Concentration of sucrose solution is 0.05 N or `0.05xx342 g=17.1 g//L` Isotonic solutions have same osmotic pressure `pi_("sugar")=pi_("glucose")` `C_("sugar")=C_("glucose")` `(17.1)/(342)=(x)/(180)` `x=9g//L` |
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| 1219. |
If `N_(2)` gas is bubbled throgh water at `20^(@)C`, how many millilitres (at stp) of `N_(2)` would dissolve in a litre of water ? Given that the partial pressure of `N_(2)` is equal to 742.5 torr and `K_(H)` for nitroge= `5.75xx10^(7)` torr. |
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Answer» `"Mole fraction of "N_(2)(x_(N_2))=("Partial pressure of" N_(2))/(K_(H)"for"N_(2))` `=((0.987 bar))/((76.48 k bar))=((0.987 bar))/((76480 bar))=1.29xx10^(-5)` `x_(N_2)=n_(N_2)/(n_(N_2)+n_(H_2)o)=n_(N_2)/(n_(H_2)o)` `(N_(2))" in the denominator has been negleeted as the gas is very little soubel in water").` `x_(N_2)=((n))/((55.5 mol))=n/(55.5),` `n=x_(N_2)xx(55.5 mol)=7.16xx10^(4) mol` ` =(7.16xx10^(-4))xx(1000mL)=0.716 milli mol .` `n_(H_2)=x_(N_(2))xxn_(H_2)o=(1.29xx10^(-5)55.55 mol)=7.16xx10^(-4) mol.` Calculation of valume of `N_(2)` at S.T.P. `"1 mole of" N_(2)" at S.T.P. occupy"=22400 mL` `7.16xx10^(-4) "mole of" N_(2) "at S.T.P. occupy"=((2240mL))/((1 mol))xx(7.16xx10^(-4)mol)=16.04 mL` |
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| 1220. |
Partial pressure of `N_(2)` gas at 298 K is 0.987 bargt If is bubbled through water at 298 K, how many millimoles of `N_(2)` gas would be dissolved in 1 litre of water ? (Given : `K_(H)` for `N_(2)` at 298 K =76.48 bar). |
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Answer» `p=K_(H).x rArr x=(p)/(K_(H))` `rArr x=(0.987)/(76.48xx10^(13))=1.29xx10^(-5)` 1 litre of water =55.5 mole of water `rArrx=(n_(N_(2)))/(n_(N_(2))+55.5)~~(n_(N_(2)))/(55.5)=1.29xx10^(-5)` `rArr n_(N_(2))=7.16xx10^(-4)"mol"` =0.716 millimole |
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| 1221. |
If at a particular temperature, the density of `18MH_(2)SO_(4)` is `1.8g cm^(-3)`, calculate (a) molality, (b) `%` concentrating by weight of solute and solvent (c) mole fraction of water and `H_(2)SO_(4)`. |
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Answer» Correct Answer - (a) `500` molal (b) `2%` (c) `0.1` (a) Molarity of the solution `M=18` mol `L^(-1)` solution hence, moles of solute `(H_(2)SO_(4))n_(1)=18` mol, mass of solute `w_(1)=18xx98g` mass of solution `(w_(1)+w_(2))=1000xx1.8=1800g` mass of water (solvent `w_(2)`) `=(1800-18xx98)=36g` We know molarity (conc. in mol `kg^(-1)` solvent) `=(1000w_(1))/(m_(1)w_(2))=(1000xx18xx98)/(98xx36)=500` molal (b) `%` concentration by weight of solute `=(18xx98)/(18xx98+36)xx100=(18xx98)/(1800)xx100=98%` `%` concentration by weight of solvent `(36)/(1800)xx100=2%` (c) `X_(H_(2)SO_(4))=(18)/(18+2)=(18)/(20)=0.9` `X_(H_(2)SO_(4))=(2)/(0.20)=0.1` |
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| 1222. |
A molal solution is one that contains one mole of a solute in:A. `1000g` of the solutionB. `1000 c.c` of the solutionC. `1000 c.c` of the solventD. `1000g` of the solvent |
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Answer» Correct Answer - D Definitation of molality |
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| 1223. |
If `O_(2)` gas in bubbled through water at 293 K, how many millimoles of `O_(2)` gas would dissove in 1 littre of water ? Assume that `O_(2)` exerts a partial pressure of 0.987 bar and `k_(H)` for `O_(2)=34.86` k bar. |
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Answer» Correct Answer - 1.57 milli that `O_(2)` Step I. Calculation of `x_(O_(2))` `x_(O_(2))=rho_(O_(2))/k_(H)=((0.98"bar"))/((43860"bar"))=2.83xx10^(-5)` Step II. Calculation os milli-moles of `O_(2)` If n moles os `O_(2)` are present in 1 L of water (or 1000 g of water) `X_(N_(2))=n/(n+55.5)=n/(55.5)` `n=X_(N_(2))xx55.5=2.83xx10^(-5)xx55.5` `=157.06xx10^(-5)mol=157.06xx10^(-2)`milli-moles =1.57 milli-moles |
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| 1224. |
The Vapour pressure of water at 293 K is 0.0231 bar and the vapour pressure of the solution of 108.24 g of a compound in 1000 g of water at the same temperature. Calculate the molar mass of the solute. |
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Answer» Vapour pressure of solvent pA0 = 0.0231 bar Vapour pressure of solution pA0 = 0.0228 bar Lowering in vapour pressure pA0 - pA = 0.0231 - 0.0228 = 0.0003 bar Weight of solvent WA = 1000 g Weight of solute WB = 108.24 g Molar mass of the solvent MA = 18 Molar mass of the solute MB = ? We know that pA˚ - pA / p0A = WB MA/WA MB MB = WB MA PA˚/ WA (PA˚ - PA) = 108.24 X 0.0231 X 18/ 0.0003 X 1000 = 150 g mol-1 |
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| 1225. |
The number of moles present in 2 litre of `0.5 M NaOH` is:A. 2B. 1C. `0.1`D. `0.5` |
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Answer» Correct Answer - B `0.5 =(n)/(2),n=1` |
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| 1226. |
The density of `NH_(4)OH` solution is `0.6g//mL`. It contains `34%` by weight of `NH_(4)OH`. Calculate the normality of the solution:A. `4.8N`B. `10N`C. `0.5N`D. `5.8N` |
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Answer» Correct Answer - D wt. of `1000ml` of solution `600gm`. & wt.of `NH_(4)OH = 600 xx (34)/(100) = 204` `n_(NH_(4)OH) = (204)/(35) = 5.8` `M = N = (5.8)/(1) = 5.8` |
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| 1227. |
What volume of `98%` sulphuric acid should be mixed with water to obtain `200mL` of `15%` solution of sulphuric acid by weight ? Given density of `H_(2)O=1.00 g cm^(-3)`, sulphuric acid `(98%)=1.88g cm^(-3)` and sulphuric acied `(15%)=1.12g cm^(-3)`. |
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Answer» Correct Answer - `V_(1)=18.2 mL` It is a case of dilution, simplest way is to determine normally of the `98%` and `15% H_(2)SO_(4)`. (i) For `98% H_(2)SO_(4) : 100g H_(2)SO_(4)` solution `=(100)/(1.88)cm^(3)` solution has `H_(2)SO_(4)=98g` `N_(1)(98%)=(1000w_(1))/(E_(1)V)=(100xx98)/(49xx(100)/(1.88))=37.6N` where `w_(1)` is the weight of solute of equivalent weight `E,` in `V mL` solution (ii) For `15% H_(2)SO_(4) : 100g H_(2)SO_(4)` solution `=(100)/(1.12)cm^(3)` has `H_(2)SO_(4)=15g` `N_(2)(15%)=(1000w_(1))/(E_(1)V)=(1000xx15)/(49xx(100)/(1.12))=3.43` Using `N_(1)V_(1)=N_(2)V_(2)` `37.6xxV_(1)=3.43xx200` `V_(1)=18.2 mL` |
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| 1228. |
For an ideal solution of two components A and B, which of the following is true?A. `DeltaH_("mixing") lt 0` (zero)B. `A - A, B-B` and `A-B` interations ae identicalC. `A -B` interaction is stronger than `A-A` and `B-B` interactionsD. `DeltaH_("mixing") gt 0` (zero) |
| Answer» Correct Answer - B | |
| 1229. |
The normally of `4%` (wt/vol).`NaOH` is:A. `0.1`B. `1.0`C. `0.05`D. `0.01` |
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Answer» Correct Answer - B `100ml` of solution contains 4 gm of `NaOH` `M = (0.1)/(100) xx 100 = 1` `M = N = 1` |
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| 1230. |
The solubility of `N_(2)` in water at `300K` and `500` torr partial pressure is `0.01 g L^(-1)`. The solubility (in `g L^(-1))` at 750 torr partial pressure isA. `0.02`B. `0.005`C. `0.015`D. `0.0075` |
| Answer» Correct Answer - C | |
| 1231. |
The weight of solute present in 200mL of `0.1M H_(2)SO_(4)`:A. `2.45g`B. `4.9g`C. `1.96g`D. `3.92g` |
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Answer» Correct Answer - C `n =(0.1 xx 98)/(5) =1.96gm` |
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| 1232. |
The normally of `0.3 M` phosphorus acid `(H_(3) PO_(3))` isA. `0.1`B. `0.9`C. `0.3`D. `0.6` |
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Answer» Correct Answer - D `N = 2 xx 0.3 = 0.6` |
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| 1233. |
The nature of mixture obtained by mixing 50mL of `0.1M H_(2)SO_(4)` and `50mL` of `0.1M NaOh` is:A. AcidicB. BasicC. NeutralD. Amphoteric |
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Answer» Correct Answer - A `[H^(+)]` is more than `[OH^(-)] :.` Acidic |
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| 1234. |
At what partial pressure, nitrogen will have a solubility of 0.05 g `L^(-1)` in water at 293 K ? Given that `k_(H) for N_(2)` at 293 K is 76.48 k bar. Assume that the density of the solution is the same as that of the pure solvent. |
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Answer» Correct Answer - 246.26 bar Step I. Calculation of `x_(N_(2))` `"Mass of 1 L of solution"=1000mLxx1" g mL"^(-1)=1000 g` `"Mass of solution (water)=(1000-0.05)~~1000 g` `"No. of moles of water "=((1000g))/((18g mol^(-1)))=55.5 mol` `"No. of moles of nitrogen"=((0.05g))/((28" g mol"^(-1)))~~1.79xx10^(-3)mol` `X_(N_(2))=(n_(N_(2)))/(n_(N_(2))+n_(H_(2)O))=((1.79xx10^(-3)mol))/((1.79xx10^(-3)+55.5 mol))=3.22xx10^(-5)` Step II. Calculation os partial pressure of `N_(2)` `rho_(N_(2))=K_(H)xxX_(N_(2))=(76.48xx10^(3)"bar")xx(3.22xx10^(-5))=246.26"bar"` |
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| 1235. |
The volume strength of `H_(2)O_(2)` solution is 10. what does it mean:A. at S.T.P. 10g solution of `H_(2)O_(2)` gives 10mL of `O_(2)`B. at S.T.P. 1g equivalent of `H_(2)O_(2)` gives 10mL of `O_(2)`C. at S.T.P. 10 litre solution of `H_(2)O_(2)` gives 10mL of `O_(2)`D. at S.T.P. 1mL solution of `H_(2)O_(2)` gives 10mL of `O_(2)` |
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Answer» Correct Answer - D Volume strength 10 means, 1 litre solution of `H_(2)O_(2)` gives 10L oxygen at S.T.P. so 1 ml solution will give 10 ml of oxygen at S.T.P. |
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| 1236. |
Calculate the number of molecules of exalic acid `(H_(2)C_(2)O_(4)2H_(2)O)` in 100 mL of 0.2 N oxalic acid solution. |
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Answer» Calculation of mass of oxlic acid `" Normality of solution"=("Mass of oxalic acid"//"Equivalent mass")/("Volume of solution in litres")` `"Normality mass of acid"= ("Molar mass")/("Basicity")=(2xx1+2xx12+4xx16+2xx18)/2=126/2` `63 g equiv^(-1)` Volume of solution =100mL=100/1000=0.1L `(0.2"equiv"L^(-1))=("Mass of oxalic acid")/(63"gequiv"^(-1)xx(0.2L))` `"Mass of oxalic acid"=(0.2equiv.L^(-1))xx(63 g equiv.^(-1))xx(0.1L)=1.26 g` Calculation of no. of molecules of oxalic acid. Molar mass of xalic acid =`126 g mol^(-1)` Now, 126 gof exalic acid has molecules=`(N_(0)=6.022xx10^(23))` 1.26 g of oxalic acid has molecules=`(6.022xx10^(23))/((126g))xx(1.26g)=6.022xx10^(21)` |
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| 1237. |
Find out the weight of `H_(2)SO_(4)` in `150 mL, (N)/(7)H_(2)SO_(4)`. |
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Answer» `N = ("Weight in gram")/("equivalent weight" xx volume("in"))` wt. in gram = eq. wt `xx N xx` volume `= 49 xx (1)/(7) xx (150)/(1000) = (21)/(20) = 1.05 g` |
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| 1238. |
If `250 mL` of a solution contains `24.5 g H_(2)SO_(4)` the molarity and normality respectively are:A. `1M,2N`B. `1M,0.5M`C. `0.5M,1N`D. `2M,1N` |
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Answer» Correct Answer - A `M = ((24.5)/(98))/(250) xx 100 = (24.5)/(98) xx 4 = 1` `N = n` factor `xx M = 2 xx 1 = 2` |
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| 1239. |
How much volume of `10 M HCI` should be diluted with water to prepare `2.00L` of `5M HCI`? |
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Answer» In dilution the following equation is applicable: `{:(M_(1)V_(1),=,M_(2)V_(2)),(10 xx V_(1),=,5 xx2.00),(V_(1),=,(5xx2.00)/(10)=1.00L):}` |
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| 1240. |
If `0.4gm` of `NaOH` is present in 60 ml of solution. What is the molarity and normality `[M. wt`. Of `NaOH = 40]` |
| Answer» Molarity `=("wt. of solute" xx 1000)/(M. Wt. "of solute" xx "volume of solution" (mL)) = (0.4)/(40 xx 80) xx 1000 = 0.125 M = 0.125 N` | |
| 1241. |
The normality of `1.5M H_(3)PO_(4)` is- |
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Answer» Basicity of `H_(3)PO_(4)` is 3 `N = M xx n = 4.5 N` |
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| 1242. |
What is the normality of 0.3 M `H_(3)PO_(4)` solution ? |
| Answer» Normality of solution= `"Molarity" xx "Basicity" =0.3xx3=0.9 N` | |
| 1243. |
Defing the term mole fration and ideal solution. |
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Answer» For the definition of mole fraction, consult section 3. For the definition of mole fraction, consult section 6. |
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| 1244. |
Mole fration of a solute in 2.0 molal aqueous solutions is :A. 1.77B. 1.87C. 0.0347D. 0.347 |
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Answer» Correct Answer - c `"Mole fraction of solute"(X_(B))=n_(B)/(n_(B)+n_(A))` `"No, of moles of solute"(n_(B))=2 mol` `"No. of moels of water"(n_(A))=((1000g))/((18g mol^(-1)))` =55.55 mol `X_(B)=((2 mol))/((57.55 mol))=0.0347` |
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| 1245. |
5.85 g of NaCI is dissoloved in water and the solution is made upto 0.5 litre. The molarity of solution is:A. 1 MB. 2 MC. 0.1 M `BaCI_(2)`D. 0.2 M |
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Answer» Correct Answer - d `"Molarity of solution (M)"=("Mass of NaCI/Molar Mass")/("Volume of solutin in liters")` ` =((5.85g)//(58.5g mol^(-1)))/((0.5 L))` `=((0.1 mol))/((0.5 L))=0.2 mol L^(-1)=0.2 M` |
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| 1246. |
The boiling point 0.2 mol `kg^(-1)` solution of X in water is greater than equyimolal solution of Y in water. Which one of the folloing statements in true is this case ?A. Molecular mass of X is less than that of YB. Y is undergoing dissociation in water wheres X dose not undergoing dissociation in water.C. X is undergoing dissociation in water.D. Molecular mass X is greater than that of Y. |
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Answer» Correct Answer - c Scince bot solution are equimolal, the solution undergoing dissociation has higher boiling point, Therefore, solution X is undergoing dissociation in waster. |
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| 1247. |
Osmotic pressure can be increased byA. increasing temperatureB. decreasing temperatureC. increasing volumeD. None of these |
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Answer» Correct Answer - A `pi = CRT` `pi prop` Temperature |
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| 1248. |
The osmotic pressure of a solution can be increased byA. increasing the volumeB. increasing the number of solute moleculesC. decreasing the temperatureD. removing semipermeable membrane |
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Answer» Correct Answer - B `pi=n/V RT` Hence, `pi prop n ` |
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| 1249. |
People taking lot of salt experience puffiness or swelling of the body due toA. water retention in tissue cells and intercellular spaces because of osmosisB. water loss from the cells through skin tissuesC. capillary action of water through skin poresD. excessive thirst and drinking more water . |
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Answer» Correct Answer - A Due to osmosis water moves into the tissues and intercellular spaces causing retention of water |
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| 1250. |
The main ingredient of automobile antifreeze mixtures isA. `CH_(2)(OH)CH(OH)CH_(2)(OH)`B. `C(CH_OH)_(4)`C. `CH_(2)(OH)CH(OH)CHO`D. `CH_(2)(OH)CH_(2)(OH)` |
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Answer» Correct Answer - 4 A familiar application of `DeltaT_(f)`, is the addition of "permanent" antifreeze, mostly ethylene glycol, `(CH_(2)OH)_(2)` to the water in automobile radiator. Because the boiling point of the solution is elevated, addition of a solute as a winter antifreeze also helps protect against loss of the coolant by summer "boil-over" The amount by which the freezing anf boiling point change depend on the concentration of the ethylene glycol solution. However, the addition of too much EG is counter productive. The freezing point of pure glycol is about `-12^(@)C`. A solution that is mostly glycol would have a somewhat lower freezingpoint due to the persence of water as a solute. Suppose we graph the `DeltaT_(f)` of `H_(2)O` below `0^(@)C` as glycol is added, and also graph the DeltaT_(f) of glycol below `-12^(@)C` as `H_(2)O` is added. These two curves would intersect at same temperature, indicating the limit of lowering that can occur. At thses high concentrations, the solutions do not behave ideally, so the temperature could not be accurately predicted by the equations we have interoduced, but the main ideas still apply. Most antifreeze labels recommened a `50:50` mixture by volume `(f_(p)-34^(@)F, bp 265^(@)F` with a 15 -psi pressure cap on the radiator), and cite the limit of possible protection with a 70:30 mixture by volume of antifreeze: water `(f_(p)-84^(@)F, bp 276^(@)F` with a 15 - psi pressure cap). |
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