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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.
| 1501. |
which of the following has been arranged in order of decreasing freezing point?A. `0.05MKNO_(3)gt0.04MCaCl_(2)gt0.140M` sugar `gt0.075M CuSO_(4)`B. `0.04M BaCl_(2)gt0.140M` sucrose `gt0.075M CuSO_(4)gt0.05M KNO_(3)`C. `0.075M CuSO_(4)gt0.140M` sucrose `gt0.04M BaCl_(2)gt0.05M KNO_(3)`D. `0.075M CuSO_(4)gt0.05M NaNO_(3)gt0.140M` sucrose `gt0.04M BaCl_(2)` |
| Answer» Correct Answer - A | |
| 1502. |
which of the following has been arranged in order of decreasing freezing point?A. `0.05M KNO_(3)gt0.04M CaCl_(2)gt0.140M sugargt0.075MCuSO_(4)`B. `0.04M BaCl_(2)gt0.140M sucrose gt0.075MCuSO_(4)gt0.05MKNO_(3)`C. `0.075M CuSO_(4)gt0.140M sucrose gt0.04MBaCl_(2)gt0.05MKNO_(3)`D. `0.075M CuSO_(4)gt0.05M NaNO_(3) gt0.140Msucrosegt0.04MBaCl_(2)` |
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Answer» Correct Answer - A Higher freezing point `rArr` lesser `DeltaT_(f) rArr` lesseer molarity `rArr` lesser number of particles |
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| 1503. |
represents the distillation of mixture of liquid `A` and liquid `B` which gives both of pure liquid `A` and `B` . Represents the azeotropic mixture of `HNO_(3)` and `H_(2)O` which distillation gives an azeotropic mixture and either of pure liquid. We cannot separate both the pure liquid, i.e., `H_(2)O` and `HNO_(3)`. a solution of `50%` of `B` on distillation results into A. Separation of an azeotropic mixture and pure `A`.B. Separation of an azeotropic mixture and pure `B`.C. Separation of both pure `A` and pure `B`.D. None of these |
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Answer» Correct Answer - C `A` and `B` do not form any azeotropic mixture, therefore they can be easily separable. |
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| 1504. |
Two components in the ratio of x:y form an azeotropic mixture. They are mixed in the ratio of `x:2y, how many moles one of the pure component y will be evaporated before getting azeotropic solution ? |
| Answer» More than y moles of a component will be evaporated. | |
| 1505. |
The solution of 2.5 g of a non-volatile substance in 100 g of benzene boiled at a temperature `0.42^(@)C` higher than the b.p. of pure benzene. Calculate mol. Wt. of the substance. (`K_(b)` of benzene is 2.67 K kg `"mole"^(-1)`) |
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Answer» Correct Answer - 158.9 g 158.9 g |
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| 1506. |
Which one of the following pairs of solution can we expect to be isotonic at the same temperature ?A. 0.1 M urea and 0.1 M NaClB. 0.1 M urea and `0.2 M MgCl_(2)`C. 0.1 M NaCl and `0.1 M Na_(2)SO_(4)`D. 0.1 M `Ca(NO_(3))_(2)` and `0.1 M Na_(2)SO_(4)` |
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Answer» Correct Answer - 4 |
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| 1507. |
The molal elevation/depression constant depends upon -A. `DeltaH_("solution")`B. Nature of solventC. Nature of soluteD. Freezing point of solution |
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Answer» Correct Answer - 2 |
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| 1508. |
0.1 M solution of A has osmotic pressure xNm-2 at 300 K. If 200 ml of A and 100 ml of 0.2 M solution of nonreactive solute B are mixed then the osmotic pressure will be :(a) 3x (b) 0.05 x (c) 1.33 x (d) 0.75 x |
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Answer» Option : (c) 1.33 x |
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| 1509. |
The osmotic pressure of 5% glucose solution at 300 K is :(a) 6.93 × 105 Nm-2 (b) 6.93 × 102 Nm-2(c) 6.83 Pa (d) 6.00 × 103 Pa |
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Answer» Option : (a) 6.93 × 105 Nm-2 |
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| 1510. |
Which of the following statements is not correct ?A. 5% aqueous solutions of NaCl and KCl are said to be isomolarB. 1 M sucrose solution and 1 M glucose solution are isotonicC. Molecular mass of acetic acid and benzoic acid is higher than normal mass in cryoscopic methods.D. For the same solution, `(DeltaT_b)/(DeltaT_f)=K_b/K_f` |
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Answer» Correct Answer - A Both NaCl and KCl have different molecular masses. |
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| 1511. |
A plant cell shrinks when it is kept inA. Hypotonic solutionB. Hypertonic solutionC. Isotonic solutionD. Water |
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Answer» Correct Answer - 2 |
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| 1512. |
A mango kept in a salt solution shrinks. Hence the liquid content in mango with respect to the salt solution is : (a) isotonic (b) hypertonic (c) hypotonic (d) equimolar |
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Answer» Option : (c) hypotonic |
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| 1513. |
A plant cell shrinks when it is kept inA. hypotonic solutionB. hypertonic solutionC. isotonic solutionD. pure water |
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Answer» Correct Answer - B Hypertonic solution has high osmotic pressure . When a plant cell is placed in hypertonic solution water will diffuse out of the cell resulting in shrinking of the cell. |
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| 1514. |
Grapes placed in three beakers X ,Y and Z containing different type of solutions are shown in figures. If beaker X contains water, Y and Z containA. Y-hypotonic solution , Z-hypertonic solutionB. Y-hypertonic solution, Z-hypotonic solutionC. Y and Z -isotonic solutionsD. Y and Z - hypotonic solutions |
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Answer» Correct Answer - A In hypotonic solution, the water is drawn in and the grape swells while in hypertonic solution the water is drawn out and the grape shrinks |
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| 1515. |
By dissolving `13.6 g` of a substance in `20 g` of water, the freezing point decreased by `3.7^(@)C`. Calculate the molecular mass of the substance. (Molal depression constant for water `= 1.863K kg mol^(-1))` |
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Answer» `m=(1000K_(f)xxw)/(WxxDeltaT)` Given, `K_(f)=1.863"K kg mol"^(-1)` `w = 13.6g,W=20g,Delta T=3.7^(@)C` `m=(1000xx1.863xx13.6)/(20xx3.7)=342.39`. |
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| 1516. |
`10 g` of glucose (molar mass 180) and `20 g` of sucrose (molar mass 342) are dissolved in `100 g` of water. What will be the vapour pressure of the resultant solution if the vapour pressure of water is `30 mm Hg`? |
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Answer» Mass of glucose = 10g No. of moles of glucose = 0.0556 Mass of sucrose = 20g No. of moles of sucrose = 0.0585 Mass of water = 100g No. of moles of water = 5.556 Total number of moles `= 5.556+0.0585+0.0556` `=5.67` Mole fraction of water `= (5.556)/(5.67)` Vapour pressure of solution = Mole fraction of water `xx p_(0)` `=(5.556)/(5.67) xx 35 = 34.3` mm Hg. |
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| 1517. |
Insulin is dissolved in suitable solvent and the osmotic pressure `(pi)` of solution of various concentration `(g//cm^(3))` C is measured at `27^(@)C`. The slope of plot of `pi` against C is found to be `4.1 xx 10^(-3)`. The molecular mass of inulin is:A. `6 xx 10^(6)`B. `3xx 10^(6)`C. `6 xx 10^(3)`D. `3 xx 10^(3)` |
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Answer» Correct Answer - A `Delta T = I xx K_(f) xx m` `(273 - 269.28) = I xx 1.86 xx 1` `3.72 = I xx 1.86` `I = 2` `alpha = (i-1)/(n - 1)` `1 = (2 - 1)/(n - 1) " or " n = 2` Thus, the complex should give two ions in the solution, i.e., the complex will be `[Pd(H_(2)O)_(3) Cl_(3)] Cl. 3H_(2)O]` |
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| 1518. |
Insulin `(C_(2)H_(10)O_(5))_(n)` is dissolved in a suitable solvent and the osmotic pressure `(pi)` of solutions of various concentrations `(g//cm^(3))C` is measured at `20^(@)C`. The slope of a plot of `pi` against `C` is found to be `4.65 xx 10^(-3)`. The molecular weight of insulin is:A. `4.8xx10^(5)`B. `9xx10^(5)`C. `3xx10^(5)`D. `5.17xx10^(6)` |
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Answer» Correct Answer - D `piV=(w_(B))/(m_(B))xxRT` `pi=(w_(B)/(V))(RT)/(m_(B))` `pi=cxx(RT)/(m_(B))xx1000`…..(i) where C = concentrated in g/cc, Comparing eqs.(i)and(ii),…(iii) Slope `=(RT)/(m_(R))xx1000` `m_(B)=(RT)/(Slope)xx1000` `(0.0821xx293xx1000)/(4.65xx10^(-3))=5.17xx10^(6)J` |
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| 1519. |
Insulin `(C_(2)H_(10)O_(5))_(n)` is dissolved in a suitable solvent and the osmotic pressure `(pi)` of solutions of various concentrations `(g//cm^(3))C` is measured at `20^(@)C`. The slope of a plot of `pi` against `C` is found to be `4.65 xx 10^(-3)`. The molecular weight of insulin is:A. `4.8xx10^(5)`B. `9xx10^(5)`C. `293xx10^(3)`D. `8.314xx10^(5)` |
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Answer» Correct Answer - C `pi=CRT` `pi=(c)/(M)RT C=` moles//liter, `c =kg//m^(3)` `(pi)/(c)=(RT)/(M)` M=(RT)/(pi//c) [pi//c=8.314xx10^(-3)]` `[T=298K]` `M=(8.314xx293)/(8.314xx10^(-3))=293xx10^(3)` |
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| 1520. |
What is solubility of a solute? |
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Answer» Solubility : It is defined as amount of a solute present per unit volume in its saturated solution at a specific temperature. It is expressed in mol L-1 or mol dm-3. |
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| 1521. |
The osmotic pressure of a solution isA. directly proportional to the concentration at constant temperatureB. inversely proportional to the molecular mass of the solute at constant temperatureC. indirectly proportional to the concentration at constant temperatureD. directly proportional to the temperature |
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Answer» Correct Answer - A::B::D Osmotic pressure is directly proportional to concentraction and temperature and inversely proportional to molecular mass. |
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| 1522. |
Molarity of `HCI(aq)` if its density is 1.17 g/ccA. 36.5B. 18.25C. 32.05D. 42.1 |
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Answer» Correct Answer - C Molarity `= (m_(B) xx 1000)/(M_(B) xx V)` `= (1.17 xx 1000)/(36.5 xx 1) = 32.05M` |
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| 1523. |
Which of the following statements regarding adsorption is/are correct?A. Extent of adsorption of gases on charcoal increases with increase in pressure of the gasB. Extent of adsorption independent of temperatureC. Extent of chemisorption by a given mass of adsorbent is limtedD. Extent of adsorption is dependent on the nature of adsorbent |
| Answer» Correct Answer - A::C::D | |
| 1524. |
Adsorption is the phenomenon in which a substance :A. accumulates on the surface of the other substanceB. goes into the body of the other substanceC. remains close to the other substanceD. none of these |
| Answer» Correct Answer - A | |
| 1525. |
If `P^(@)` and `P_(5)` are the vapour pressure of the solvent and its solution respectively and `x_(1)` and `x_(2)` are the mole fraction of the solvent and solute respectively, thenA. `P_(s) = P^(@) 2x_(2)`B. `P^(@) - P_(s) = P^(@)x_(2)`C. `P_(s) = P^(@)x_(1)`D. `(P^(@)-P_(s))//P_(s)=x_(1)//(x_(1)+x_(2))` |
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Answer» Correct Answer - B::C `P_(s)=P^(@)x_(1)` `P_(s)=P^(@)(1-x_(2))` `P_(s)=P^(@)-P^(@)x_(2)` `P^(@)-P_(s)=P^(@)x_(2)` |
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| 1526. |
The depression of freezing point is directly proportional toA. mole fraction of the solutionB. molarity of the solutionC. molality of the solutionD. molarity of the solvent |
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Answer» Correct Answer - C `DeltaT_(f) = K_(f)m :. DeltaT_(f) prop m` (molality of the solution). Hence the lowering in vapour pressure will be maximum. |
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| 1527. |
Which of the following has minimum gold number ?A. Potato starchB. Gum arabicC. GelatinD. Albumin |
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Answer» Correct Answer - C Gelatin has minimum gold number. Therefore, (C) is correct option. |
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| 1528. |
Explain the adsorption of nitrogen on iron. |
| Answer» When nitrogen gas is brought in contact with iron at `83 K`, it is physisorbed on iron surface as nitrogen molecules `N_(2)`. As the temperature is increased the amount of introgen adsorbed decreases rapidly and at room temperature, partically there is no adsorption of nitrogen on iron. At `773 K` and above, nitrogen is chemisorbed on the surface as nitrogen atoms. | |
| 1529. |
Explain the factors on which the solubility of a substance (solute) depends. |
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Answer» The extent of dissolution of a substance (solute) depends upon the following factors : (1) Nature of a solute : A solute may be crystalline, amorphous, ionic or covalent. Hence accordingly its tendency to dissolve changes. The substances having similar intermolecular forces tend to dissolve in each other. (2) Nature of solvents : Solvents are classified as polar and nonpolar. Polar solutes dissolve in polar solvents. For example, Ionic compounds dissolve in polar solvent like water. A solvent other than water is called a nonaqueous solvent. For example, C6H6,CCl4, etc. Solutions in these solvents are called nonaqueous solutions. The solvent may be a gas, a liquid or a solid. (3) Amount of a solvent : More amount of a solvent, will dissolve more quantity of the solute. (4) Temperature : Depending on the nature of a solvent and a solute the solubility changes with termperature. The effect depends on the heat of solution, hydration energy, etc. Generally as the temperature increases, solubility of solid increases and that of gases decreases. (5) Pressure :
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| 1530. |
The solubility of gases depends upon some factors.1. Can you suggest the factors? 2. Which is the law behind it? 3. What are the limitations of this law? |
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Answer» 1. Nature of the gas, Nature of the solvent, Pressure, Temperature. 2. Henry’s law. It states that the mass of a gas dissolved per unit volume of the solvent at a given temperature is proportional to the pressure applied on it. 3. Henry’s law is valid only under the following conditions :
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| 1531. |
The vapour pressure of a solution (P) and the vapour pressure of the solvent `(P^(@))` are related to each others as `(x_(1)` is the mole fraction of solvent)A. `P = P^(@) x_(2)`B. `P = P^(@) x_(1)`C. `P^(@) = P x_(1)`D. `P^(@) = P x_(2)` |
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Answer» Correct Answer - B Vapour pressure of solution (P) = Vapour pressure of the solvent in solution = Mole fraction of the solvent `xx V.P` of pure solvent `= x_(1) xx P^(@) = P^(@) x_(1)`. |
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| 1532. |
Which of the following are correctly matched ?A. Butter-gelB. Milk-emulsionC. Fog-aerosolD. Dust-solid sol |
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Answer» Correct Answer - A::B::C (A, B, C) are correct matches. |
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| 1533. |
The solubility of a gas in water depends uponA. nature of the gasB. temperatureC. pressure of the gasD. All of the above |
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Answer» Correct Answer - D Solubility of a gas in water depends upon all the given factors. |
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| 1534. |
Which of the following is correctly matched?A. `{:("Solvent","Ebullioscopic constant"),("Acetone",0.51):}`B. `{:("Solvent","Ebullioscopic constant"),("Benzene",2.53):}`C. `{:("Solvent","Ebullioscopic constant"),("Water",1.86):}`D. `{:("Solvent","Ebullioscopic constant"),("Chloroform",1.22):}` |
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Answer» Correct Answer - B `K_(b)` for benzene `= 2.53 K kg mol^(-1)`. |
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| 1535. |
In a mixture A and B compounds show negative deviation asA. `DeltaV_("mix") gt 0`B. `DeltaH_("mix") lt 0`C. A-B interaction is weaker than A-A and B-B interactionD. None of the above reason is correct |
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Answer» Correct Answer - B For a solution showing -ve deviation `DeltaH_("mix") lt 0` |
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| 1536. |
A solution contains non-volatile solute of molecular mass`M_(2)`which of the following can be used to calculate the molecular mass of solute in terms of osmotic pressure? (`m_(2)` =mass of solute,V=volume of solution,`pi` =osmotic pressure)A. `M_(2)=(m_(2)/V)RT`B. `M_=(m_/V)(RT)/pi`C. `M_(2)=(m_(2)/V)piRT`D. `M_(2)=(m_(2)/V)pi/(RT)` |
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Answer» Correct Answer - 2 For dilute solutions, it has been found that the osmotic pressure is proportional to the molarity `(M)` of the solution at a given temperature `T`. Thus `pi=MRT` Here `pi` is the osmotic pressure and `R` is the gas constant. or `pi=n_(2)/V RT` where V is the volume of a solution in liters containing `n_(2)` moles of the solute. If `m_(2)` grams of solute, of molar mass, `M_(2)` is present in the solution, then `n_(2)=m_(2)//M_(2)` and we can write `pi=m_(2)/(M_(2)V)RT` or `M_(2)=(m_(2)/V) (RT)/pi` |
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| 1537. |
A solution contains non-volatile solute of molecular mass `M_(2)`. Which of the following can be used to calculate the molecular mass of solute in terms of osmotic pressure? Here `n_(2)=` mass of solute, V = volume of solution, `pi =` osmotic pressure.A. `M_(2) = [(m_(2))/(pi)]VRT`B. `M_(2) = [(m_(2))/(V)] (RT)/(pi)`C. `M_(2) = [(m_(2))/(V)] nRT`D. `M_(2) = [(m_(2))/(V)] (pi)/(RT)` |
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Answer» Correct Answer - B `M_(2) = (m_(2)RT)/(piV) = ((m_(2))/(V)) (RT)/(pi)`. |
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| 1538. |
Which of the following will have the highest freezing point at one atmosphere ?A. 0.1 M NaCl solutionB. 0.1 M sugar solutionC. 0.1 M `BaCl_2` solutionD. 0.1 M `FeCl_3` solution |
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Answer» Correct Answer - B For the same concentration of different solvents any colligative property `prop` I For NaCl, i=2 Sugar solution , i=1 `BaCl_2`, i=3 , `FeCl_3` , i=4 Thus, for sugar solution depression in freezing point is minimum i.e., highest freezing point. |
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| 1539. |
Arrange the following aqueous solutions in the order of their increasing boiling points (i)`10^(-4)` M NaCl , (ii)`10^(-4)` M Urea , (iii)`10^(-3)` M `MgCl_2` , (iv)`10^(-2)` M NaClA. (i) lt (ii) lt (iv) lt (iii)B. (ii) lt (i) = (iii) lt (iv)C. (ii) lt (i) lt (iii) lt (iv)D. (iv) lt (iii) lt (i) = (ii) |
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Answer» Correct Answer - C For `{:(10^(-4) "M NaCl", "i=2"),(10^(-4) "M Urea", "i=1"), (10^(-3) M MgCl_2, "i=3"),(10^(-2) "M NaCl", "i=2"):}` More the value of I,C, more will be the elevation in boiling point hence increasing order of boiling point is `10^(-4)` M Urea lt `10^(-4)` M NaCl lt `10^(-3)` M `MgCl_2` lt `10^(-2)` M NaCl. |
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| 1540. |
Which of the following has the highest freezing point ?A. 1 m NaCl solutionB. 1 m KCl solutionC. 1 m `AlCl_3` solutionD. 1 m `C_6H_12O_6` solution |
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Answer» Correct Answer - D `C_6H_12O_6` is a non-electrolyte hence furnishes minimum number of particles and will have maximum freezing point . `DeltaT_f=iK_fm` or `DeltaT_f prop i` and `DeltaT_f=T_f^@ -T_f` |
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| 1541. |
Which of the following conentration factor is affected by change in temperature?A. MolarityB. MolalityC. Mol fractionD. Weight fraction |
| Answer» Correct Answer - A | |
| 1542. |
Which one of the following is incorrect for ideal solution? (a) ∆Hmix = 0(b) ∆Umix = 0 (c) ∆P = PObserved – PCalculated by raoults law = 0 (d) ∆Gmix = 0 |
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Answer» (d) ∆Gmix = 0 For an ideal solution, ∆Smix ≠ 0; Hence ∆Gmix ≠ 0 ∴ Incorrect is ∆Gmix = 0 |
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| 1543. |
When partial pressure of solvent in solution of nonvolatile solute is plotted against its mole fraction, nature of graph is :(a) a straight line passing through origin.(b) a straight line parallel to mole fraction of solvent.(c) a straight line parallel to vapour pressure of solvent.(d) a straight line intersecting vapour pressure axis. |
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Answer» Option : (a) a straight line passing through origin. |
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| 1544. |
A graph plotted between `(P)/(d) vs d` (where `p` is osmotic pressure of solution of a solute of mol. Wt. `m` and `d` is its density temperature `T`. Pick out the correct statements about the plots :A. `[(P)/(d)]_(drarr0)=(ST)/(m)`B. The intercept of the plot `(ST)/(m)`C. The intercept of the plot `=[(P)/(d)]_(drarr0)`D. `[(P)/(d)]_(drarr0)` is independent of temperature |
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Answer» Correct Answer - A::B::C `P=MST` `P=(n)/(V)xxST` `P=(w)/(VxxM)xxST` `(P)/(d)=(ST)/(M)` |
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| 1545. |
In the graph plotted between vapour pressure (V.P.) and temperature (T), A. PQ is the curve for solvent , XY is the curve of solution and `DeltaT` is depression in freezing pointB. PQ is the curve for solution. XY is the curve for solvent and `DeltaT` is elevation is boiling pointC. PQ is the curve for solvent, XY is the curve for solution and `DeltaT` is molal elevation in boiling pointD. PQ is the curve for solvent, XY is the curve for solution and `DeltaT` is elevation in boiling point |
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Answer» Correct Answer - D PQ corresponds to increase in vapour pressure of solvent with temperature and XY corresponds to increase in vapour pressure of solution with temperature . `DeltaT` is the elevation in boiling point of a solution. |
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| 1546. |
A solution containing 12.5 g of non-electrolyte substance in 185 g of water shows boiling point elevation of 0.80 K. Calculate the molar mass of the substance. (`K_b=0.52 K kg mol^(-1)`)A. `"53.06 g mol"^(-1)`B. `"25.3 g mol"^(-1)`C. `"16.08 g mol"^(-1)`D. `"43.92 g mol"^(-1)` |
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Answer» Correct Answer - D `M_B=(K_bxxW_B)/(DeltaT_bxxW_A)=(0.52xx12.5)/(0.80 xx 0.185)=43.92 g mol^(-1) (because 185 g =0.185 kg ) ` |
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| 1547. |
The elevation in boiling point of a solution of 9.43 g of `MgCl_2` in 1 kg of water is ( `K_b` = 0.52 K kg `mol^(-1)` , Molar mass of `MgCl_2 = 94.3 g mol^(-1)`)A. 0.156B. 0.52C. 0.17D. 0.94 |
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Answer» Correct Answer - A `MgCl_2 hArr Mg^(2+) + 2Cl^(-) , i=3` `DeltaT_b=iK_bm=3xx0.52xx9.43/(94.3xx1)`=0.156 |
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| 1548. |
On dissolving `3.24 g` of sulphur in `40 g` of benzene, the boiling point of the solution was higher than sulphur? (`K_(b)` for benzene = `2.53 K kg mol^(-1)` , atomic mass of sulphur `= 32 g mol^(-1)`). |
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Answer» The given values are: `W_(B) = 3.24 g`, `W_(A) = 40 g` `DeltaT_(b) = 0.81 K`, `K_(b) = 2.53 Kg mol^(-1)` Using formula, `Mw_(B) = (K_(b) xx 1000 xx W_(B)) /(DeltaT_(b) xx W_(A))` On substituting all the values, we get `:. Mw_(B) = (2.53 xx 1000xx 3.24) / (0.81 xx 40) = 253` Let the molecular formula of sulphur = `S_(x)` Atomic mass of sulphur = `32` Molecular mass = `32 xx x` `:. 32x = 253` `x= (253)/(32) = 7.91 ~~ 8` `:.` Molecular formula of sulphur `= S_(8)` |
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| 1549. |
Define molaritv. |
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Answer» Molarity is defined as the number of moles of solute present in 1 litre of the solution. Molarity (M) = (Number of moles of solute) / (Volume of the soluation (in L)) |
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| 1550. |
18 g of glucose is dissolved in 90 g of water. The relative lowering of vapour pressure is equal to …. |
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Answer» Answer: 0.1 \(\frac{P°-P}{P°}\) = x2 x2 = No. of moles of glucose \(\frac{18}{180}\) = 0.1 \(\frac{P°-P}{P°}\) = 0.1 |
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