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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.
| 151. |
The vapour pressure of water depends upon :A. surface area of containerB. volume of containerC. temperatureD. all |
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Answer» Correct Answer - C Vapour pressure is independent of surface area and volume of container. |
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| 152. |
The solubility of a gas in water depends uponA. Nautre of the gasB. TemperatureC. Pressure of the gasD. All of the above |
| Answer» Correct Answer - D | |
| 153. |
The cryoscopic constant value depends uponA. The number of particles of the solute in the solutionB. The number of particles of the solvent in the solutionC. The enthalpy of fusion of the solventD. The freezing point of the solvent |
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Answer» Correct Answer - C::D Explanation : The cryoscopic constant value is `K_(f)=(RT_(f)^(2)xxM_(o))/(Delta F_(f)xx1000)` `K_(f)` depends upon `Delta H_(f)` and `T_(f)`. Hence, choices (c ) and (d) are correct. It follows that (a) and (b) are not correct. |
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| 154. |
In the depression of freezing point experiment, it is found that theA. Vapour pressure of the solution is less than that of pure solventB. Vapour pressure of the solution is more than that of pure solventC. Only solute molecules solidify at the freezing pointD. Only solvent molecules solidify at the solution at room |
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Answer» Correct Answer - A::D The depression of freezing point is less than that of pure solvent and only solvent molecules solidify at the freezing point. |
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| 155. |
The molarity of an aqueous solution of `NaOH` containing 8 g in 2L of solution isA. `0.1M`B. `0.2m`C. `0.25M`D. `0.15M` |
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Answer» Correct Answer - A `M="Mass"/("Molar mass" xx V (dm^(3)))=8/(40 xx 2)=0.1M` |
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| 156. |
The normality of a solution of sodium hydroxide 100 ml of which contains 4 grams of NAOH isA. `0.1`B. 40C. `1.0`D. `0.4` |
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Answer» Correct Answer - C `N=(w xx 1000)/(eq.wt. xx "Volume in ml.") = (4xx1000)/(40xx100)=1.0 N`. |
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| 157. |
How many gram of HCl will be present in 150 ml of its 0.52 M solutionA. 2.84 gmB. 5.70 gmC. 8.50 gmD. 3.65 gm |
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Answer» Correct Answer - A `M = (w)/(mxx V(l)), 0.52 = (w)/(36.5xx0.15), w = 2.84 gm` |
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| 158. |
Which is behaviestA. 25 gm of mercuryB. 2 moles of waterC. 2 moles of carbon dioxideD. 4 gm atoms of oxygen |
| Answer» Correct Answer - C | |
| 159. |
Molarity is expressed asA. g/LB. L/molC. mol/LD. mol/1000g |
| Answer» Correct Answer - c | |
| 160. |
Molarity is expressed asA. Gram/litreB. Moles/litreC. Litre/moleD. Moles/1000 gms |
| Answer» Correct Answer - B | |
| 161. |
The normality of 10% (weight/volume) acetic acid isA. 1 NB. 10 NC. 1.66 ND. 0.83 N |
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Answer» Correct Answer - c Normality `= (w)/(E.V) = (10g)/(100 mL)xx(1)/(60)` `= (10g)/((100)/(1000)xx60) = 1.66 N` |
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| 162. |
What is the freezing point of a solution containing 8.1 g HBr in 100g water assuming the acid to be 90% ionised (`K_(f)` for water `= 1.86 "Kg mole"^(-1)`)A. `0.85^(@)C`B. `-3.53^(@)C`C. `0^(@)C`D. `-0.35^(@)C` |
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Answer» Correct Answer - B `{:(HBr,hArr,H^(+),+,Br^(-)),((1-alpha),,alpha,,alpha):}` Total `= 1+alpha therefore i=1+alpha=1+0.9=1.9` `Delta T_(f)= i K_(f)xx m=1.9xx1.86xx(8.1)/(81)xx(1000)/(100)=3.53^(@)C` `T_(f)=-3.53^(@)C`. |
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| 163. |
Match the entries listed in Column I with appropriate listed in Column II. `{:(,"Column I",,"Column II"),((A),0.1 M BaCl_(2)" solution",(p),271 K),((B),0.1 M NaCl" solution",(q),270 K),((C ),0.1 M K_(3)[Fe(CN)_(6)]" solution",(r ),268 K),((D),0.1 M Al_(2)(SO_(4))_(3)" solution",(s),269 K):}` Given : Freezing point of 0.1 M sucrose solution = 272 K |
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Answer» Correct Answer - `A to q; B to p; C to s; D to r` `A to q, B to p, C to s, D to r` `Delta T_(f)` for 0.1 M sucrose sol. `=1^(@)(T^(@)=273 K)` (A) `Delta T_(f)` for `0.1 M BaCl_(2)`, i.e., 0.3 M particle conc. `= 3^(@)` (B) `Delta T_(f)` for 0.1 M NaCl, i.e., 0.2 M particle conc. `=2^(@)` (C ) `Delta T_(f)` for `0.1 M K_(3)[Fe(CN)_(6)]`, i.e., 0.4 M particle conc. `= 4^(@)C` (D) `Delta T_(f)` for `0.1 M Al_(2)(SO_(4))_(3)`, i.e., 0.5 M particle conc. `= 5^(@)C` |
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| 164. |
Dilution process of different aqueous solution , with water are given in List-I. The effects of dilution of the solutions on `[H^(+)]` are given in List-II. (Note : Degree of dissociation `(alpha)` of weak acid and weak base is ltlt 1, degree of hydrolysis of salt `lt lt 1 , [H^(+)]` represents the concentration of `H^(+)` ions) `|{:(," List-I",," List -II"),((P),"(10 mL of 0.1 M NaOH+20 mL of 0.1 acetic acid)dilutes to 60 mL",(1),"The value of "[H^(+)]" does not change on dilution"),((Q),"(20 mL of 0.1 M NaOH + 20 mL of 0.1 M acetic acid)diluted to 80 mL",(2),"The value of "[H^(+)]" change to half of its initial value on dilution"),((R ),"(20 mL of 0.1 M HCl + 20 mL of 0.1 M ammonia solution)diluted to 80 mL",(3),"The value of "[H^(+)]" change to two times of its initial value on dilution"),((S),"10 mL saturated solution of "Ni(OH)_(2)" in equilibrium with excess solid "Ni(OH)_(2)" is ",(4),"The value of "[H^(+)]" changes to "(1)/(sqrt(2))" times of its initial value on dilution"),(,"diluted to 20 mL (solid "Ni(OH)_(2)" is still present after dilution)",,),(,,(5),"The value of "[H^(+)]" changes to "sqrt(2)" times of its initial value on dilution"):}|` |
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Answer» Correct Answer - `P to 1; Q to 5; R to 4; S to 1` `P to 1, Q to 5, R to 4, S to 1` P. `underset("0.1M, 20ml")(CH_(3)COOH)+underset("0.1M, 10ml")(NaOH)to CH_(3)COONa+H_(2)O` `pH=pKa rArr [H^(+)]` will not change on dilution correct match : F - 1 Q. `{:(underset("0.1M, 20ml")(CH_(3)COOH)+underset("0.1M, 20ml")(NaOH)to CH_(3)COONa+H_(2)O),(" "- " " -" 0.05M"):}` `[OH^(-)]=sqrt(K_(H)C)=sqrt(((k_(w))/(k_(a))C))` `[H^(+)]_(1)=sqrt((k_(w)k_(a))/(C))` `([H^(+)]_(2))/([H^(+)]_(1))=sqrt((C_(1))/(C_(2)))=sqrt((0.05)/(0.025))=sqrt(2)` Correct match : Q-5 R. `underset("0.1M, 20ml")(NH_(4)OH)+underset("0.1M, 20ml")(HCl)to underset(0.05M)(NH_(4)Cl)` `[H^(+)]=sqrt(K_(H)C)` `([H^(+)]_(2))/([H^(+)]_(1))=sqrt((C_(2))/(C_(1)))=(1)/(sqrt(2))` Correct match : R-4 S. Because of dilution solubility does not change so `[H^(+)]` = constant. |
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| 165. |
The depression in f.pt. of `0.01m` aqueous solution of urea, soldium chloride and soldium sulphate is in the ration:A. `1:1:1`B. `1 : 2 : 3 `C. `1: 2: 3`D. `1: 2: 4` |
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Answer» Correct Answer - B 1 mole urea gives 1 mole 1 mole NaCl gives 2 mole 1 mole `Na_(2)SO_(4)` gives 3 mole ` Delta T_(f)` ratio 1: 2:3 |
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| 166. |
The osmotic pressure of decimolar solution of glucose at `30^(@)C` is :A. 24.88 atmB. 2.488 atmC. 248.8 atmD. 2488 atm |
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Answer» Correct Answer - B `C.RT = 0.1 xx 0.082 xx 303 = 2.488` |
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| 167. |
The volume of water of be added to `50 cm^(3)` of a decimolar solution to convert it to a centimolar solution will beA. `500 cm^(3)`B. `450 cm^(3)`C. `400 cm^(3)`D. `100 cm^(3)` |
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Answer» Correct Answer - B Before dilution equiv After dilution `M_(1) V_(1) = M_(2) V_(2)` `0.1 xx 50 = 0.01 xx V_(2)` `:. V_(2)=500 cm^(3)` :. Volume of water added `=500-50=450 cm^(3)` |
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| 168. |
How many moles of water are present in 180 g of waterA. 1 moleB. 18 moleC. 10 moleD. 100 mole |
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Answer» Correct Answer - C Moles of water `=(180)/(18)=10` mole. |
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| 169. |
A `5.2` molal aqueous of methyl alcohol, `CH_(3)OH`, is supplied. What is the molefraction of methyl alcohol in the solution ?A. `0.100`B. `0.190`C. `0.086`D. `0.050` |
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Answer» Correct Answer - C `X_("methyl alcohol")=(5.2)/(5.2+(1000)/(18))=0.086` |
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| 170. |
2 mole of ethanol is dissolved in 8 mole of water. The mole fraction of water in the solution isA. 0.2B. 0.8C. 0.4D. 0.1 |
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Answer» Correct Answer - B Total number of moles in the solution =10 `:. "Mole fraction of water"=("Moles of water")/("Total moles")` `=8/10=0.8` |
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| 171. |
200 ml of water is added to 500 ml of 0.2 M solution. What is the molarity of this diluted solutionA. 0.5010 MB. 0.2897 MC. 0.7093 MD. 0.1428 M |
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Answer» Correct Answer - D No. of millimoles `= 500xx0.2=100` Thus, molarity of diluted solution `=(100)/(700)=0.1428 M` |
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| 172. |
3.65grams of HCI dissolved in 16.2g of water. The mole fraction of HCI in the resulting solution isA. `0.4`B. `0.3`C. `0.2`D. `0.1` |
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Answer» Correct Answer - D `X=(n)/(n + N)` `n=(w)/(m)=(3.65)/(36.5)=0.1, N=(W)/(M)=(16.2)/(18)=0.9` `X=(0.1)/(0.1+0.9)=0.1`. |
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| 173. |
What are the units of mole fraction ?A. Moles/litreB. Moles/`"litre"^(2)`C. Moles-litreD. Dimensionless |
| Answer» Correct Answer - D | |
| 174. |
The system that forms maximum boiling azeotrope is `CS_(2)`, acetoneA. Carbondisulphide-acetoneB. Benzene - tolueneC. Acetone - chloroformD. n-hexane - n-heptane |
| Answer» Correct Answer - C | |
| 175. |
Which values can be obtained from the information represented by the vapour pressure curve of a liquidlt/brgt A. Normal boiling pointlt/brgtB. Normal freezing point lt/brgtC. Enthalpy of vaporizationA. A onlyB. A and B onlyC. A and C onlyD. A, B and C |
| Answer» Correct Answer - C | |
| 176. |
The weight of sodium carbonate required to prepare 500 ml of a semi-normal solution isA. 13.25 gB. 26.5 gC. 53 gD. 6.125 g |
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Answer» Correct Answer - A `N=(w xx1000)/(eq.wt. xx " volume in ml.")eq.wt.=(106)/(2)=53` `w=(0.5xx53xx500)/(1000)=13.25`. |
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| 177. |
Volume of water needed to mix with 10 mL 10N `HNO_(3)` to get 0.1 N `HNO_(3)` is :A. 1000 mlB. 990 mlC. 1010 mlD. 10 ml |
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Answer» Correct Answer - B `N_(1)V_(1)=N_(2)V_(2)` `10xx10=0.1(10+V)` `V=(10xx10)/(0.1)-10=1000-10=990 ml`. |
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| 178. |
How much `K_(2)Cr_(2)O_(7) (M.W. = 294.19)` is required to prepare one litre of `0.1N` solution?A. 2.9424 gB. 0.4904 gC. 1.4712 gD. 0.2452 g |
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Answer» Correct Answer - D `W=(N xx eq.wt. xx V(ml))/(1000)=(0.05xx49.04xx100)/(1000)=0.2452`. |
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| 179. |
The molal cryoscopic constant for water isA. `1.86 K "molality"^(-1)`B. 5.26 K `molality_(-1)`C. `55.5 K "molality"^(-1)`D. `0.52 K "molality"^(1)` |
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Answer» Correct Answer - A It is the numerical value of molal cryoscopic constant for water. It changes with change of solvent. |
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| 180. |
When mercuric iodide is added to the aqueous solution of potassium iodide, then:A. f.pt. is raisedB. f. pt. is loweredC. f.pt. does not changeD. b.pt. does not change |
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Answer» Correct Answer - A `2 KI + HgI_(2) rarr K_(2) HgI_(4)`. As a result of this reaction, no. of ions decrease. So the lowering in F pt. is less or the actual F. pt is more. |
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| 181. |
Consider separate solutions of `0.500 M C_(2)H_(5)OH(aq)`,`0.100 M Mg_(3)(PO_(4))(aq)`,`0.250 M KBr(aq)`, and `0.125 M Na_(3)PO_(4)(aq)` at `25^(@)C`. Which statement is true about these solutions, assuming all salts to be strong electrolytes?A. They all have the same osmotic pressureB. `0.100 M Mg_(3)(PO_(4))_(2)(aq)` has the highest osmotic pressureC. `0.125 M Na_(3)PO_(4)(aq)` has the highest osmotic pressureD. `0.500 M C_(2)H_(5)OH(aq)` has the highest osmotic pressure |
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Answer» Correct Answer - A `{:(0.5 M C_(2)H_(5)OH(aq),0.1M Mg_(3)(PO_(4))_(2)(aq)),(i=1,i=5),("effective molarity "=0.5,"effective molarity "=0.5 m),(0.25 M KBr (aq),0.125 M Na_(3)PO_(4)(aq)),(i=2,i=4),("effective molarity "=0.5 M,"effective molarity "=0.5 M):}` Hence all colligative properties are same. Note : This equation is solved by assuming that the examinar has taken `Mg_(3)(PO_(4))_(2)` to be completely soluble. However the fact is that it is insoluble (sparingly soluble). |
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| 182. |
A 0.0020 m aqueous solution of an ionic compound `Co(NH_(3))_(5) (NO_(2))Cl` freezes at `-0.00744^(@)C`. The number of moles of ions which 1 mole of ionic compound produces on being dissolve in water is `(K_(f)=-1.86^(@)C//m)` |
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Answer» Correct Answer - 2 No. of moles of ions produce from 1 mole of ionic compound = i `because T_(f)=K_(f)i.m.` `therefore i=(T_(f))/(K_(f)xxm)=(0.00744)/(1.86xx0.0020)=2`. |
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| 183. |
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. `0.549`B. `0.200`C. `0.786`D. `0.478` |
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Answer» Correct Answer - D `P_(T)=p_(p)^(@)x_(p)+p_(h)^(@)x_(h)=440xx(1)/(5)+120xx(4)/(5)` `=88+96=184, p_(p)^(@)x_(p)=Y_(p)P_(T), (88)/(184)=y_(p)` |
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| 184. |
An aqueous solution is `1.00` molal in `KI`. Which change will cause the vapor pressure of the solution to increase?A. Addition of waterB. Addition of NaClC. Addition of `Na_(2)SO_(4)`D. Addition of 1.00 molal KI |
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Answer» Correct Answer - A According to raoults low `P_(s)=PX_(A)` (`X_(A)` = mole fraction of solvent) and on addition of water the mole fraction of water in the solution increases therefore vapour pressure increases. |
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| 185. |
The vapour pressure of chloroform `(CHCl)_(3)` and dichlorocethene `(CH_(2)Cl_(2))` at `298 K` is `200 mm Hg ` and `415 mm Hg`, respectively. Calculate a. The vapour pressure of the solution prepared by mixing `25.5 g` of `CHCl_(3)` and `40 g` of `CH_(2)_Cl(2)` at `298 K`. b. Mole fractions of each components in vapour phase . |
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Answer» (i). Molar mass of `CH_(2)Cl=12xx1+1xx2+35.5xx2=85" g "mol^(-1)` Molar mass of `CHCl_(3)=12xx1+1xx1+35.5xx3=119.5" g "mol^(-1)` Moles of `CH_(2)Cl_(2)=(40g)/(85g" "mol^(-1))=0.47` mol Moles of `CHCl_(3)=(25.5g)/(119.5" g "mol^(-1))=0.213`mol Total number of moles `=0.47+0213=0.683`mol `x_(CH_(2)Cl_(2))=(0.47mol)/(0.683" mol")=0.688` `x_(CHCl_(3))=1.00-0.688=0.312` Using equation `P_("total")=p_(1)^(0)+(p_(2)^(0)-p_(1)^(0))x_(2)=200+(415-200)xx0.688` `=200+147.9=347.9" mm Hg"` (ii) Using the relation `y_(1)=P_(i)//p_("total")` we can calcualte the mole fraction of the components in gas phase `(y_(i))` `Pp_(CH_(2)Cl_(2))=0.688xx415mm" Hg"=285.5" mm Hg"` `p_(CHCl_(3))=0.312xx200" mm Hg"=62.4" mm Hg"` `y_(CH_(2)Cl_(2))=285.5" mm Hg"//347.9mm" "Hg=0.82` `y_(CHCl_(3))=62.4" mm Hg"//347.9" mm Hg"=0.18` |
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| 186. |
The vapour pressure of chloroform `(CHCl)_(3)` and dichlorocethene `(CH_(2)Cl_(2))` at `298 K` is `200 mm Hg ` and `415 mm Hg`, respectively. Calculate a. The vapour pressure of the solution prepared by mixing `25.5 g` of `CHCl_(3)` and `40 g` of `CH_(2)_Cl(2)` at `298 K`. b. Mole fractions of each components in vapour phase .A. 90.92 mm HgB. 615.0 mm HgC. 347.9 mm HgD. 285.5 mm Hg |
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Answer» Correct Answer - A `n_(CHCl_(3))=(25.5)/(119.5)=0.213` `n_(CH_(2)Cl_(2))=(40)/(85)=0.47` `P_(T)=P_(A)^(@)X_(A)+P_(B)^(@)X_(B)` `=200xx(0.213)/(0.683)+41.5xx(0.47)/(0.683)` `=62.37+28.55=90.92` |
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| 187. |
The number of moles present in 2 litre of `0.5 M NaOH` is:A. `0.5`B. `0.1`C. 1D. 2 |
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Answer» Correct Answer - C `M=(n)/(V(l)), 0.5 =(n)/(2)= n = 1` |
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| 188. |
If 0.50 mol of `CaCl_(2)` is mixed with 0.20 mol of `Na_(3)PO_(4)`, the maximum number of moles of `Ca_(3)(PO_(4))_(2)` which can be formed, isA. `0.70`B. `0.50`C. `0.20`D. `0.10` |
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Answer» Correct Answer - D `3 CaCl_(2)+2 Na_(3)PO_(4)to Ca_(3)(PO_(4))_(2)+6NaCl` `therefore` 2Moles of `Na_(3)PO_(4)=3` mole of `CaCl_(2) = 1` mole `Ca_(3)(PO_(4))_(2)` `therefore 0.2` mole of `Na_(3)PO_(4)=0.3` mole of `CaCl_(2)=0.1` mole of `Ca_(3)(PO_(4))_(2)`. |
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| 189. |
A solution contains `25% H_(2)O, 25% C_(2)H_(5)OH` and `50%CH_(3)COOH` by mass. The mole fraction of `H_(2)O` would beA. `0.25`B. `2.5`C. `0.503`D. `5.03` |
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Answer» Correct Answer - C `X_(H_(2)O)=(n_(H_(2)O))/(n_(H_(2)O)+n_(C_(2)H_(5)OH)+n_(CH_(3)COOH))` |
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| 190. |
The concentration of an aqueous solution of `0.01 M CH_(3)OH` solution is very nearly equal to which of the followingA. `0.01% CH_(3)OH`B. `0.01m CH_(3)OH`C. `x_(CH_(3)OH)=0.01`D. `0.01N CH_(3)OH` |
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Answer» Correct Answer - D For methyl alcohol N = M. |
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| 191. |
10 grams of a solute is dissolved in 90 grams of a solvent. Its mass percent in solution isA. `0.01`B. `11.1`C. 10D. 9 |
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Answer» Correct Answer - C % by w.t `=("w.t. of the solute (g)")/("wt. of the solution g")xx100` `=(10)/(90+10)xx100=10` |
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| 192. |
The freezing point of a 0.01 M aqueous glucose solution at 1 atmosphere is `-0.18^(@)C`. To it, an addition of equal volume of 0.002 M glucose solution will , produced a solution with freezing point of nearlyA. `-0.036^(@)C`B. `-0.108^(@)C`C. `-0.216^(@)C`D. `-0.422^(@)C` |
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Answer» Correct Answer - B `Delta T_(f)=(K_(f)xx1000xx w)/(m xx W)=-0.108^(@)C` |
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| 193. |
Molal depression constant for water is `1.86^(@)C` kg/mole. The freezing point of a 0.05 molal solution of a non-electrolyte in water isA. `-1.86^(@)C`B. `-0.93^(@)C`C. `-0.093^(@)C`D. `0.93^(@)C` |
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Answer» Correct Answer - C `Delta T_(f)=K_(f)xx` molality `=1.86xx0.05=0.093^(@)C` Thus freezing point `=0-0.093=-0.093^(@)C`. |
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| 194. |
Pressure cooker reduces cooking time for food becauseA. Heat is more evenly distributed in the cooking spaceB. Boiling point of water involved in cooking in increased materialC. The higher pressure inside the cooker crushes the food materialD. Cooking involves chemical changes helped by a rise in temperature |
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Answer» Correct Answer - B Due to higher pressure inside the boiling point is elevated. |
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| 195. |
During the evaporation of liquidA. The temperature of the liquidB. The temperature of the liquid will fallC. May rise or fall depending on the natureD. The temperature remains unaffected |
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Answer» Correct Answer - B In the process of evaporation, high energy molecules leave the surface of liquid, hence average kinetic energy and consequently the temperature of liquid falls. |
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| 196. |
The boiling point of a solution of 0.11 gm of a substance in 15 gm of ether was found to be `0.1^(@)C` higher than that of the pure ether. The molecular weight of the substance will be `(K_(b)=2.16)`A. 148B. 158C. 168D. 178 |
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Answer» Correct Answer - B `m=(K_(b)xx w xx1000)/(Delta T_(b)xx W)` `K_(b)=2.16, w=0.11, W=15 g, Delta T_(b) =0.1` `m=(2.16xx0.11xx1000)/(0.1xx15)=158.40 ~= 158`. |
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| 197. |
The boiling point of a solution of `0.0150g` of a substance in `15.84g` of ether was found to be `100^(@)C`higher than that of pure ether.What is the molecular weight of the substance [Molecular elevation constant of ether per`100`A. 144.5B. 143.18C. 140.28D. 146.66 |
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Answer» Correct Answer - B `m=(K_(b)xx wxx 1000)/(Delta T_(b)xx W)=143.18` |
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| 198. |
A solution containing 30 gms of non-volatile solute in exactly 90 gm water has a vapour pressure of 21.85 mm Hg at `25^(@)`C. Further 18 gms of water is then added to the solution. The resulting solution has a vapour pressure of 22.15 mm Hg at `25^(@)`C. Calculate the molecular weight of the soluteA. 74.2B. 75.6C. 67.83D. 78.7 |
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Answer» Correct Answer - C We have, `(p^(@)-21.85)/(21.85)=(30xx18)/(90xx m)`, for 1 case …….(i) Wt. of solvent `= 90+18=108 gm` `(p^(@)-22.15)/(22.15)=(30xx18)/(108xx m)`, for II case …..(ii) By eq. (1) `p^(@)m-21.85m=21.85xx6=131.1` By eq. (2) `p^(@)m-22.15m =22.15xx5=110.75` `0.30m=20.35` `m=(20.35)/(0.30)=67.83` |
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| 199. |
Weight of Urea required 200ml of 2 M solution will beA. 12 gmB. 24 gmC. 20 gmD. 60 gm |
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Answer» Correct Answer - B Molarity `= (W_(B))/(M_(B))xx(1000)/(V_((ml)))` V = 200 ml Molarity = 2 M and `M_(B) = 60 g` `W_(B) = (M_(B)xx V_((ml))xx" Molarity")/(1000)=(60xx200xx2)/(1000)=24 gm`. |
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| 200. |
A solution made by dissolving 40 g NaOH 1000 g of `H_(2)O` isA. 1 molarB. 1 normalC. 1 molalD. None of the above |
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Answer» Correct Answer - C `M=(40)/(40xx1)=1` m sol |
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