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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.
| 1651. |
If we have `6% W//W` urea solution with density `1.060g//mL`, then calculate its strength in `g//L`? |
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Answer» `6 g ` urea is present in `100gm` solution. `6g "in" (100)/(1.060)mL` `(100)/(1.060)mL` `therefore1000mL=(6)/(100)xx1.060xx1000=10.6xx6=63.6` |
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| 1652. |
Calculate the amount of benzoic acid (C6HCOOH) required for preparing 250 mL of 0.15 M solution in methanol. |
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Answer» 0.15 M solution means that 0.15 moles of C6H5COOH is present in 1L = 1000 mL of the solution Molar mass of C6H6COOH = 72 + 5 + 12 + 32 + 1 = 122 g mol-1 Thus, 1000 mL of solution contains benzoic acid = 18.3 g 250 mL of solution will contain benzoic acid = \(\frac{18.3}{1000}\) x 250 = 4.575 g |
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| 1653. |
Calculate the amount of benzoic acid `(C_(6)H_(5)COOH)` required for preparing 250 ml of 0.15 M solution in methanol. |
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Answer» Given Molarity = 0.15 M Volume V = 250 ml Molecular weight of benzoic acid `(C_(6)H_(5)COOH)=122` Molarity (M) `=("Weight")/(GMw)xx(1000)/(V(ml))` `0.15=(W)/(122)xx(1000)/(250)` `W=(122xx0.15)/(4)=4.575" gms."` |
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| 1654. |
Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol. |
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Answer» 0.15 M solution means that 0.15 mole of benzoic acid is present in 1 L i.e. 1000 mL of the solution. Molar mass of benzoic acid (C6H5COOH) = 72 + 5 + 12 + 32 + 1 = 122 g mol-1 ∴ 0.15 mole of benzoic acid = 0.15 × 122g = 18.39 Thus, 1000 mL of the solution contain benzoic acid = 18.39 ∴ 250 ml of the solution will contain benzoic acid = \(\frac{18.3}{1000}\) × 250 = 4.575g |
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| 1655. |
Calculate the mass precentage of aspirin `(C_(9)H_(8)O_(4))` in acetonitrile `(CH_(3)CN)` when 6.5 gm of `C_(9)H_(8)O_(4)` is dissolved in 450g of `CH_(3)CN`. |
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Answer» Given The Mass of aspirin = 6.5 gms Mass of acetonitrile = 450 gms Mass of solution `= 6.5 + 450` = 456.5 gms `"Mass "%=(6.5)/(456.5)xx100=1.424%`. |
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| 1656. |
The molarity of a solution obtained by mixing 750 mL of 0.5dsM solution of HCI with 250 mL of 2M solution of HCI will be :A. 1.75 MB. 0.975 MC. 0.875 MD. 1.00 M |
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Answer» Correct Answer - c `M_(3)V_(3)=M_(1)V_(1)+M_(2)V_(2)` `M_(3)=((0.5M)xx(750mL)+(250 mL))/((750 mL+250 mL))` 0.875 M. |
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| 1657. |
The molarity of a solution obtained by mixing 750 mL of 0.5 M HCl with 250 mL of 2 M HCl will beA. `0.975 M`B. `0.875 M`C. `1.00 M`D. `1.75 M` |
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Answer» Correct Answer - B `M_(1)V_(1)+M_(2)V_(2)=M_(3)(V_(1)+V_(2))` `0.5xx750+2xx250=M_(3)(750+25)` `375+500=1000M_(3)` `therefore" "M_(3)=(875)/(1000)=0.875M.` |
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| 1658. |
(a) A liquid mixture of benzene and toluene is composed of `1` mol of benzene and `1` mol of tolune if the pressure over the mixture at `300K` is reduced, at what pressure does the first bubble form? (b) What is the composition of the first bubble formed. (c) If the pressure is reduced further, at what pressure does the last trace of liquid disapper? (d) What is the composition of the last drop of liquid? (e) What will be the pressure, composition of the liquid and the composition of vapour, when `1` mol the mixture has been vaporized? Given `P_(T^(0))=40mmHg, P_(B^(0))=100mmHg` |
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Answer» Correct Answer - (a) `70 mmHg,` (b) `X_(b)=5//7` (c) `P=400//7=57.14 mm` (a) `P=X_(A)P_(A)+X_(B)P_(B)^(0)=0.5xx40+0.5xx100=70` (b) `Y_(A)=(0.5xx40)/(70)=(2)/(7) , Y_(B)=(5)/(7)` ltbRgt (c ) at last trace of liquid `Y_(A)=0.5 , Y_(B)=0.5` `(1)/(P)=(Y_(A))/(P_(A)^(0))+(Y_(B))/(P_(B)^(0))=(0.5)/(40)+(0.5)/(100)` `P=(400)/(7)` (d) `Y_(A)=(Y_(A)P_(A)^(0))/(P) rArr0.5=(X_(A)40)/(400//7) , X_(A)=(5)/(7) , X_(B)=(2)/(7)` (e) Let `X` mole of `B` be present in liquid phase. `{:(,"Mole ofA","Mole of B"),("Liquid",1-x,x),("Vapour",x,1-x):}` `P=40(1-x)+100 x` `(1)/(P)=(Y_(A))/(P_(A)^(0))+(Y_(A))/(P_(B))rArr(1)/(P)=(x)/(40)+(1+x)/(100)=(100x+(1-x)40)/(40xx100)` So `p^(2)=40xx100` `p=20sqrt(10)` `20sqrt(10)=40(1-x)+100x=40+60x` `x=(20sqrt(10)-10)/(60)=(sqrt(10)-2)/(3)` |
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| 1659. |
`29.2% (w//w) HCl` stock, solution has a density of `1.25 g mL^(-1)`. The molecular weight of `HCl` is `36.5 g mol^(-1)`. The volume `(mL)` of stock solution required to prepare a `200 mL` solution of `0.4 M HCl` is : |
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Answer» Correct Answer - 8 Molarity `(M_(1))`of stock concentrated solution can be calculated using following relation : `M_(1)(x xx d xx10)/(m_(B))` `=(29.2xx1.25xx10)/(36.5)=10` `M_(1)V_(1)` (Concentrated) `= M_(2)V_(2)` (Diluted) `10xxV_(1)=0.4xx200` `V_(1)=8mL`] |
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| 1660. |
A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of `290mm`at `300K` .The vapour pressure of propyl alcohol is `200mm` .if the mole fraction of ethyl alcohol is `0.6` ,its vapour pressure (in mm)at the same temperature will beA. `350`B. `300`C. `700`D. `360` |
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Answer» Correct Answer - A `P_(m)=P_(A)+P_(B)` `P_(A)^(@)X_(A)+P_(B)^(@)X_(B)` `:. P_(m)=290mm,P_(B)^(gamma)=20mm,X_(A)=0.6,X_(B)` `=1-0.6=0.4` `:. 290=P_(A)^(@)xx0.6+200xx0.4` `P_(A)^(gamma)=350mm` |
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| 1661. |
Concentrated aqueous sulphuric acid is `98% H_(2)SO_(4)` by mass and has a density of `1.80 g mL^(-1)`. Volume of acid required to make one litre of `0.1 MH_(2)SO_(4)` solution is:A. 16.65 mLB. 22.20 mLC. 5.55 mLD. 11.10 mL |
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Answer» Correct Answer - C `M=(x xx d xx 10)/(m_(B))=(98xx1.80xx10)/(98)=18` `M_(1)V_(1)=M_(2)V_(2)` `18xxV_(1)=0.1xx1000` `V_(1)=(100)/(18)=5.55mL`] |
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| 1662. |
Concentrated aqueous sulphuric acid is 98% acid required to make one litre of 1.80 g `mL^(-1)`. Volume of acid required to make one litre of 0.1 M `H_(2)SO_(4)` solution is :A. 16.65 mLB. 22.90 mLC. 11.10 mLD. 5.55 mL |
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Answer» Correct Answer - c Volume of 98% (W/W) solution of `H_(2)SO_(4)` `=("Mass")/("density")=((100g))/((1.80g mL^(-1))` =55.55 mL = 0.055L `"Molarity (M)"((98g))/((98g mol^(-1))xx(0.055L))` `=18.02 mol^(-1)L=18.02 M` `Vxx(18.02 M)=1Lxx(0.1M)` `V=((1L)xx(0.1M))/((18.02 M))=0.0055 L` =5.5 mL |
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| 1663. |
The vapour pressure of water at room temperature is 23.8 mm Hg. The vapour pressure of an aqueous solution of sucrose with mole fraction 0.1 is equal toA. `2.39 mm Hg`B. `2.42 mm Hg`C. `21.42 mm Hg`D. `21.44 mm Hg` |
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Answer» Correct Answer - C `DeltaP = p^(@) xx x_(B) = 23.8 xx 0.1 = 2.38` `:.` Vapour pressure of sol `= 23.8 - 2.38` `= 21.42 mm Hg` |
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| 1664. |
if solution of sucrose in water is labed as 20%w/W.Caliculate mole fraction of each componrnt in the solution. |
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Answer» 20% w/w solution of sucrose in water means that 20 g of sucrose in present in 100 g of the solution i.e., 20 g of sucrose is present in (100 - 20) g = 80 g of water. Mole fraction of sucrose (xs) = 0.2 / ( 0.2+0.8) = 0.2 And, mole fraction of water (xw) = 1 - xs =1-0.2 =0.8 |
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| 1665. |
Assertion:Osmosis is one sided movement of solvent particles. Reason :In Osmosis , the net movement of solvent particles from dil.to conc.solution and from conc.to dil. Solution takes place through semipermeable membrance,showing finally the direction of dil.to conc.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both the assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If assertion is false but reason is true. |
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Answer» Correct Answer - D Osmosis is net movement of solvent particles from dil.to conc. And conc.to dil.i.e., a bilateral process. The more movement is form dil.to conc.thus net flow from dil.to conc.is noticed. |
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| 1666. |
Two liquids X and Y from an ideal solution at`300K`, Vapour pressure of the Solution containing `1`mol of X and `3`mol of Y is `550`mmHg . At the same temperature, if `1` mol of Y is further added to this solution ,vapour pressur of the solutions increases by `100`mmHg Vapour pressure (in mmHg) of X and Y in their pure states will be,respectivelyA. 400 and 600B. 500 and 600C. 200 and 300D. 300 and 400 |
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Answer» Correct Answer - A `550 = P_(A)^(@) xx (1)/(4) +P_(B)^(@) xx (3)/(4)` `560 = P_(A)^(@) xx (1)/(5) +P_(B)^(@) xx (4)/(5)` `P_(A)^(@) = 400 P_(B)^(@) = 600` tor |
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| 1667. |
Two liquids X and Y from an ideal solution at`300K`, Vapour pressure of the Solution containing `1`mol of X and `3`mol of Y is `550`mmHg . At the same temperature, if `1` mol of Y is further added to this solution ,vapour pressur of the solutions increases by `100`mmHg Vapour pressure (in mmHg) of X and Y in their pure states will be,respectivelyA. 200 and 300B. 300 and 400C. 400 and 600D. 500 and 600 |
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Answer» Correct Answer - C `550=P_(x)^(0).(1)/(4)+P_(y)^(0).(3)/(4)` ………(i) after adding 1 mole of y further `560=P_(x)^(0).(1)/(5)+P_(y)^(0).(4)/(5)` …………(ii) solving (i) & (ii) `P_(x)^(0)=400 MM` `P_(y)^(0)=600 MM` |
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| 1668. |
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. 17.675 mm of HgB. 15.750 mm of HgC. 16.500 mm of HgD. 17.325 mm of Hg |
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Answer» Correct Answer - D `(P^(0)-P_(s))/(P^(0))underline~(n)/(N)implies(17.5-P_(s))/(17.5)=((18)/(180))/((178.2)/(18))impliesP_(S)=17.325mm` |
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| 1669. |
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 Hg`B. `16.500 mm Hg`C. `17.325mm Hg`D. `17.675mm Hg` |
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Answer» Correct Answer - C `P_(A)=P_(A)^(@)x_(A)=17.5xx`(178.2//18)/(178.2/(18)+(18)/(180))=17.325` |
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| 1670. |
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. 17.325 mm HgB. 17.675 mm HgC. 15.75 mm HgD. 16.5 mm Hg |
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Answer» Correct Answer - A `p=p^(0)x_(A)` `=17.5xx(178.2//18)/(178.2//18+(18)/(180))` `=17.325`mm Hg |
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| 1671. |
A solution has `1:4` mole ratio of pentane to hexane . The vapour pressure of pure hydrocarbons at `20^@C`are `440` mmHgfor pentane and `120`mmHg for hexane .The moleA. `0.786`B. `0.478`C. `0.549`D. `0.200` |
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Answer» Correct Answer - B `P=P_(C_(5)H_(12))^(O)xxX_(C_(5)H_(12))+P_(C_(6)H_(14))^(O)xxX_(C_(6)H_(14)),` `p=440xx(1)/(5)+120xx(4)/(5)=(920)/(5)=184 y_(C_(5)H_(12))=(P_(C_(5)H_(12)))/(P_(t))` Now,`Y_(C_(5)H_(12))=(88)/(184)=0.478` |
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| 1672. |
Vapour pressure of chloriform `(CHCI_(3))` and dichloromethane `(CH_(2)CI_(2))` at `25^(@)C` are 200 mm Hg and 41.5 mm Hg respectively. Vapour pressure of solution obtained bu mixing 25.5 g of `CHCI_(3)` and 40.0g of `CH_(2)CI_(2)` at the same temperature will be (Mlecular mass of `CHCI_(3)=1199.5` u and molecular mass of `CH_(2)CI_(2)=85u`A. 173.9 mm HgB. 615.0 mm HgC. 347.9 mm HgD. 90.63 mm of Hg |
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Answer» Correct Answer - d `P_(CHCI_(3))^(@)=200 mm Hg. P_(CH_(2)CI_(2))^(@)=41.5 mm Hg` Moles of `CHCI_(3)` `("Mass")/("Molecular weight")=((25.5g))/((119.5g mol^(-1)))=0.213 mol` Moles of `CH_(2)CI_(2)=((40g))/((85gmol^(-1)))=0.470 mol` `X_(CHCI_(3))=((0.213))/((0.213+0.470))=0.31` `X_(CH_(2)CI_(2))=(0.470)/(0.213+0.470)=0.69` `P_(T)=P_(CHCI_(3))^(@)X_(CHCI_(3))+P_(CH_(2)CI_(2))^(@)X_(CH_(2)CI_(2))` =`200xx0.31+41.5xx0.69 =62+28.63=90.63 mm of Hg |
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| 1673. |
Statement-I: If decimolal solution of sodium chloride boils at `101.2^(@)C`, then decimolal solution of calcium chloride will also boil at the same temperature. Because Statement-II: For same molal concentration of aqueous solutions of electrolytes, the elevation of boiling point may not be same. |
| Answer» Correct Answer - D | |
| 1674. |
`25.3 g` sodium carbonate, `Na_(2)CO_(3)`, was dissolved in enough water to make `250 mL` of solution. If sodium carbonate dissociates completely, molar concentration of `Na^(+)` and carbonate ions are respectively:A. 0.955 M and 1.910 MB. 1.910 M and 0.955 MC. 1.90 M and 1.910 MD. 0.477 M and 0.477 M |
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Answer» Correct Answer - B Molarity `= ("Number of moles of solute")/("Volume of solution (in mL)")xx1000` `=(25.3xx1000)/(106xx250)=0.9547 ~~ 0.955 M` `Na_(2)CO_(3)` in aqueous solution remains dissociated as `{:(Na_(2)CO_(3),hArr,2Na^(+),+,CO_(3)^(2-)),(" x",," 2x",," x"):}` Since, the molarity of `NA_(2)CO_(3)` is 0.955 M, the molarity of `CO_(3)^(2-)` is also 0.955 M and that of `Na^(+)` is `1xx0.955=1.910 M` |
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| 1675. |
`25.3 g` sodium carbonate, `Na_(2)CO_(3)`, was dissolved in enough water to make `250 mL` of solution. If sodium carbonate dissociates completely, molar concentration of `Na^(+)` and carbonate ions are respectively:A. 0.477 M and 0.477 MB. 0.955 M and 1.910 MC. 1.910 M and 0.955 MD. 1.90 M and 1.910 M |
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Answer» Correct Answer - 3 `M(Na_(2)CO_(3))=n_(Na_(2)CO_(3))/V_(mL)xx(1000 mL)/L` `=(25.3 g Na_(2)CO_(3)//106 g Na_(2)CO_(3) mol^(-1))/(250 mL)xx(1000 mL)/L` `=0.955 mol L^(-1) Na_(2)CO_(3)(aq) rarr 2Na^(+)(aq)+CO_(3)^(2-)(aq)` `1 mol 2 mol 1 mol` Thus `M(Na^(+))=2M(Na_(2)CO_(3))=2(0.955 M)=1.910 M` `M(CO_(3)^(2-))=M(Na_(2)CO_(3))` `=0.955 M` |
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| 1676. |
214.2 gram of sugar syrup contains 34.2 gram of sugar. Caluculate (i) molality of the solution and (ii) mole fraction of the sugar in the syrup-A. 0.555m, 0.0099B. 0.4565m,0.0110C. 0.355m, 0.0199D. None of these |
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Answer» Correct Answer - A Mass of sugar=34.2 gram. Number of moles of sugar`=34.2/("mol. mass")=34.2/34.2=0.1` Mass of water `=(214.2-34.2)=180" gram or "180/1000kg` Number of moles of water `=180/18=10` `"Molality"=("No. of moles of sugar")/("Mass of water in kg")=0.1/180xx1000=0.555m` (ii) Total no. of moles`=10.0+0.1=10.1` Mole fraction of sugar `=("No. of moles of sugar")/("Total number of moles")= 0.1/10.1=0.0099` |
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| 1677. |
214.2 gram of sugar syrup contains 34.2 gram of sugar. Caluculate (i) molality of the solution and (ii) mole fraction of the sugar in the syrup- |
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Answer» (i) Mass of sugar = 34.2g No. of moles of sugar `= (34.2)/("Mol.mass")=(34.2)/(342)=0.1` Mass of water `=(214.2-34.2)` `= 180g = (180)/(1000)kg` No. of moles of water `= (180)/(18)=10` Molality `= ("No. of moles of sugar")/("Mass of water in kg")=(0.1)/(180)xx1000` `=0.555m` (ii) Total number of moles `=10.0+0.1=10.1` Mole fraction of sugar `= ("No.of moles of sugar")/("Total number of moles")` `=(0.1)/(10.1)=0.0099`. |
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| 1678. |
`12.2gm` benzoic acid `(M =122)` in `100g H_(2)O` has elevation of boiling point of `0.27^(@)C, K_(b) = 0.54 K` kg/mole. If there is `100%` dimerization the no.of molecules of benzoic acid in associated sate is:A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B `0.27 =ixx(0.1)/(100) xx 1000 xx 0.54` `i=(1)/(2)` |
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| 1679. |
The osmotic pressure of a `5% (weight// volume)` solution of cane sugar at `150^(@)C` isA. `4 atm`B. `3.4 atm`C. `5.07 atm`D. `2.45 atm` |
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Answer» Correct Answer - C `piV=nRT`,`pi xx 100/1000 =5/342 xx0.0821 xx 423` `:. pi =5.078` atm |
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| 1680. |
The osmotic pressure of a `5% (weight// volume)` solution of cane sugar at `150^(@)C` isA. 4 atmB. 5.07 atmC. 3.55 atmD. 2.45 atm |
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Answer» Correct Answer - B `pi = (5 xx 0.0821 xx 1000 xx 423)/(342 xx 1000) = 5.07` atm. |
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| 1681. |
Calculate osmotic pressure of 5% solution of cane sugar (sucrose) at `15^(@)C`. |
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Answer» m = mol. Mass of sucrose `(C_(12)H_(22)O_(11))=342` w = 5g," " V = 100 mL = 0.1 litre `S = 0.082" "T=(15+273)=288K` Applying the equation `PV=(w)/(m)ST`, `P=(5)/(342)xx(1)/(0.1)xx0.082xx288` `=3.453` atm |
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| 1682. |
A `10%"(w/w)"` solution of cane sugar has undergone partial inversion according to the reaction: `"sucrose+water"rightarrow"glucose+ fructose"`. If the boiling point of solution is `100.27^(@)"C".` (a) What is the average mass of the dissolved materials? (b) What fraction of the sugar has inverted? `"K"_("b")("H"_(2)"O")=0.512" K mol"^(-1)" kg"` |
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Answer» Correct Answer - (a)210.65, (b)62.35% `Delta"T"_("b")=[1+("n"-1)alpha]"K"_("b")("W"_("solute")xx1000)/("m"_("solute")xx"W"_("solvent"))` `0.27=[1+(2-1)alpha]xx(0.512xx10xx1000)/(342xx90)` `alpha=0.623` `"m"_("avg")=("m"_("sucrose"))/(1+alpha)=342/(1.623)` Average mass of dissolved material `=209.55"gm mol"^(-1)` |
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| 1683. |
2g mixture of glucose and sucrose is dissolved in 1 litre water at 298K to develop osmotic pressure of 0.207 atm. Calculate percentage composition of glucose and sucrose by mole as well as by mass. |
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Answer» Correct Answer - Percentage by mole : glucose = 65.194, sucrose = 34.806 ; Percentage by mass : glucose = 50, sucrose = 50] |
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| 1684. |
What do you understand by the term that `K_(f)` for water is `1.86 K kg mol^(-1)`? |
| Answer» It means that the freezing point of water is lowered by `1.86 K` when `1 mol` of a non-volatile solute is dissolved in `1000 g` of water. | |
| 1685. |
Calculate the osmotic pressure of 5% solution of cane sugar(sucrose) at `15^(@)C`. |
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Answer» Osmotic pessure `(pi)=(W_(B)xxRxxT)/(M_(B)xxV)` Mass of sucrose `(W_(B))=5g` Molar msaa of sucrose `(C_(12)H_(22)O_(11))M_(B)=342 g mol^(-1)` Volume of solution in litres (V)=100mL=100/1000=0.1L Soution constant `(T)=15^(@)C=15+273=288 K` `pi=((5g)xx(0.821L atm K^(-1)mol^(-1))xx(288K))/((342 gmol^(-1))xx(0.1L))=3.457 atm` |
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| 1686. |
Calculate the osmotic pressure at `25^(@)C` of a solution containing 1g of glucose and 1g of sucrose in 1 litre of solution. If it were not known that the solute was a mixture of glucose and sucrose, what would be the molecular weight of solute corresponding to the calculated osmotic pressure ? |
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Answer» Correct Answer - 0.207 atm, 236.384 Use ,`piV=(n_(1)+n_(2))RT` to calculate osmotic pressure `pixx1=((1)/(180)+(1)/(342))xx0.0821xx298` `=0.207` atm Use , `piV=(w)/(m)RT` to calculate molar mass of solute `0.207xx1=(2)/(m)xx0.0821xx298` m`=236.384`] |
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| 1687. |
Calculate the molal depression constant of a solvent which has a. Freezing point `16.6^(@)C` and latent heat of fusion `180.75 J g^(-1)` . b. Freezing point `20.0^(@)C` and latent heat of fusion `200.00 J g^(-1)`. |
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Answer» a. `K_(f)=(RT_(f)^(2))/(1000xxl_(f))` `R = 8.314 JK^(-1)mol^(-1)` `T_(f) = 16.6^(@)C = 273+16.6 = 289.6 K` `l_(f) = 180.75 Jg^(-1)` Substituting values in Eq. (i), we get `K_(f) = (8.314 xx (289.6))^(2))/(1000 xx 180.75) = 3.86` b. `K_(f) = (RT_(f)^(2))/(1000 xx l_(f))` `T_(f) = 273 + 20.0 = 293.0` `l_(f) = 200.0 J g_(-1) = ((8.314 xx (293.0)^(2)) /(1000 xx 200.0) = 3.56` |
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| 1688. |
Calculate the molal depression constant of a solvent which has freezing point `16.6^(@)C` and latent heat of fusion `180.75 "J g"^(-1)` :A. 2.68B. 3.86C. 4.68D. 2.86 |
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Answer» Correct Answer - B `K_(f)=(RT_(0)^(2))/(1000L)" "T_(0)=273+16.6=289.6K` `=(8.314xx(289.6)^(2))/(1000xx180.75)=3.86`] |
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| 1689. |
The osmotic pressure in atmosphere of `10%`solution of cane sugar at `69^(@)C`isA. `724`B. `824`C. `8.21q`D. `7.21` |
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Answer» Correct Answer - C `piV=nRt` `pi=(wRT)/(mV)=(10)/(342)xx(0.821xx(273+69))/(0.1)=8.21` atm |
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| 1690. |
What is the volume of solution containing 1g mole of sugar that will give rise to an osmotic pressure of 1atmosphere at `0^(@)C` ? |
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Answer» Applying the equation `PV = n.ST`, `V=(n)/(P)xxSxxT` Given, n = 1,P = 1 atm, S = 0.0821 and T = 273 K `V=(1)/(1)xx0.0821xx273=22.4`litre. |
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| 1691. |
If latent heat of fusion of ice is 80 cals per g at `0^(@),` calculate molal depression constant for water.A. 18.36B. 186.3C. 1.863D. 0.1863 |
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Answer» Correct Answer - C `"K"_("f")=("RT"_("f")^(2))/(1000l_("f"))" Here R"=2"cals","T"_("f")=0+273=273"K, "l_("f")=80"cals"` `"K"_("f")=(2xx273xx273)/(1000xx80)=1.863` |
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| 1692. |
What is the volume of a solution containing 1 g mole of sugar that will give rise to osmotic pressure of 1 atmosphere at `0^(@)C` ? |
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Answer» Correct Answer - 22.4 L `pi=1 atm, n=1"g mol",R=0.0821"L atm K"^(-1)mol^(-1),T=0^(@)C=0+273=273 K` `pi=n/VRTor V=n/piRT` `V=((1"gmol")xx(0.0821"L atm K"^(-1)mol^(-1))xx(273 K))/((1"atm"))=22.4 L`. |
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| 1693. |
The osmotic pressure of a sugar solution at `24^(@)C` is `2.5 atm`. The concentration of the solution in mole per litre isA. 10.25B. 1.025C. 1025D. 0.1025 |
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Answer» Correct Answer - D `pi=CRT rArr 2.5=C xx 0.0821 xx 297` `:. C=0.1025 "mol" L^(-1)` |
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| 1694. |
Osmotic pressure of a sugar solution at `24^(@)C` is 2.5 atmosphere .Determine the concentration of the solution in gram mole per litre.A. `0.0821"moles/litre"`B. `1.082"moles/litre"`C. `0.1025"moles/litre"`D. `0.0827"moles/litre"` |
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Answer» Correct Answer - C Here it is given that `pi=205"atm,"" T"=24+273=297"A"^(@),"S"=0.0821"lit.atm. " "deg"^(-1)"mol"^(-1),"C"=?` We know that `pi="CST "or "C"=pi/"ST"=2.5/(0.0821xx297)=0.1025" mole/litre"` |
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| 1695. |
Osmotic pressure of a sugar solution at `24^(@)C` is 2.5 atmosphere .Determine the concentration of the solution in gram mole per litre.A. `0.0821 "moles/litre"`B. `1.082 "moles/litre"`C. `0.1025 "moles/litre"`D. `0.0827 "moles/litre"` |
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Answer» Correct Answer - C Here it is given that `pi=2.5 "atm",T=24+273=297K` `S=0.0821 "lit, atm.deg"^(-1)mol^(-1),C=?` We know that `pi=CST` or `C=(!=)/(ST)=(2.5)/(0.0821xx297)=0.0120 "moles/litre"` |
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| 1696. |
5mL of `N HCI, 20 mL` of `N//2 H_(2)SO_(4)` and 30 mL of `N//3 HNO_(3)` are mixed together and volume made to one litre. The normality of the resulting solution isA. `(N)/(5)`B. `(N)/(10)`C. `(N)/(20)`D. `(N)/(40)` |
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Answer» Correct Answer - D For `HCI`, `N_(1)V_(1) = N_(2)V_(2)` `N xx 5 = N_(2) xx 1000` `N_(2) = 5 xx 10^(-3) N =0.5 xx 10^(-2)N` For `H_(2)SO_(4)`, `N_(1)V_(1) = N_(2)V_(2)` `(N)/(2) xx 20 = N_(2) xx 1000` `N_(2) = 10^(-2)N` For `HNO_(3)`, `N_(1)V_(1) = N_(2)V_(2)` `(N)/(3) xx 30 = N_(2) xx 1000` `N_(2) = 10^(-2) N` Total normality `= N(HCI) + N(H_(2)SO_(4)) + N (HNO_(3))` `= (0.5 xx 10^(-2) +106(-2) +10^(-2)) N` `= 2.5 xx 10^(-2)N` `(25)/(1000) N = (N)/(40)` |
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| 1697. |
The substance A when dissolved in solvent B shows the molecular mass corresponding to `A_(3)`. The vant Hoffs factor will be:A. 1B. 2C. 3D. `1//3` |
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Answer» Correct Answer - D `{:(3A ,rarr ,A_(3)),(1-alpha,,alpha//3):}` `i = 1 - alpha +alpha//3 alpha 100%` `= 1 - 1 +(1)/(3) =(1)/(3)` |
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| 1698. |
From the colligative properties of solution which one is the best method for the determination of mol.wt of proteins & polymers:A. osmotic pressureB. lowering in vapour pressureC. loweing in freezing pointD. elevation in B.Pt |
| Answer» Correct Answer - A | |
| 1699. |
Addition of non-volatile solute to a solvent always inceases the colligative properties such as osmotic pressure, `DeltaP, DeltaT_(b)` and `DeltaT_(f)`. All these colligative properties are direactly propertional to molality if solutions are dilute. The increase in colligative properties on addition of non-volatile solute is due to incease in number of solute particles. For different aqueous solutions of `0.1 N NaCl, 0.1 N` urea, `0.1 N Na_(2)SO_(4)` and `0.1 N Na_(3)PO_(4)` solution at `27^(@)C`, the correct statement are : `1g` mixture of glucose and urea present in `250 mL` aqueous solution shows an osmotic pressure of `0.74` atm at `27^(@) C`. Assuming solution to be dilture, which are correct ? (P) Percentage of urea in solute mixture is `17.6` (Q) Relative lowering in vapour pressure of this solutions is `5.41 xx 10^(-4)`. (R) The solution will boil at `100.015^(@)C`, if `K_(b)` of water is `0.5 K "molality"^(-1)` (S) If glucose is replaced by same amount of sucrose, the solution will show higher osmotic pressure at `27^(@)C` (T) If glucose is repalced by same amount of `NaCl`, the solution will show lower osmotic pressure at `27^(@)C`.A. 1,2,3B. 1,2,3,5C. 2,4,5D. 1,4,5 |
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Answer» Correct Answer - A (1) `0.74=(0.0821)/(250xx10^(-3))xx300((x)/(60)+(1-x)/(180))` `x=0.176` `%` urea `=(0.176)/(1)xx100=17.6` (2) `(dP)/(P)=X_("solute")=((0.176)/(60)+(0.824)/(180))/(0.176/(60)+(0.824)/(180)+(250)/(18))=5.4xx10^(-4)` (3) `DeltaT_(b)=mK_(b)=((40.176)/(60)+(0.824)/(180))xx(1000)/(250)xx0.5=0.015` `DeltaT_(b)=100.015`. |
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| 1700. |
Addition of non-volatile solute to a solvent always inceases the colligative properties such as osmotic pressure, `DeltaP, DeltaT_(b)` and `DeltaT_(f)`. All these colligative properties are direactly propertional to molality if solutions are dilute. The increase in colligative properties on addition of non-volatile solute is due to incease in number of solute particles. For different aqueous solutions of `0.1 N NaCl, 0.1 N` urea, `0.1 N Na_(2)SO_(4)` and `0.1 N Na_(3)PO_(4)` solution at `27^(@)C`, the correct statement are : (P) The order of osmotic pressure is, `NaCl gt Na_(2)SO_(4) gt Na_(3)PO_(4) gt ` urea (Q) `pi = (DeltaT_(b))/(K_(b)) xx ST` for urea solution (R) Addition of salt on ice increases its melting point (S) Addition of salt on ice brings in melting earlierA. 2,3,4B. 2,4C. 1,2,3D. 3,4 |
| Answer» Correct Answer - B | |