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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.
| 1551. |
What is a vapour pressure of liquid? What is relative lowering of vapour pressure? |
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Answer» 1. The pressure of the vapour in equilibrium with its liquid is called vapour pressure of the liquid at the given temperature. 2. The relative lowering of vapour pressure is defined as the ratio of lowering of vapour. pressure to vapour pressure of pure solvent. Relative lowering of vapour pressure = (P°solvent - P°soluation) / P°solven |
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| 1552. |
Explain the effect of pressure on the solubility. |
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Answer» 1. The change in pressure does not have any significant effect in the solubility of solids and liquids as they are not compressible. However the solubility of gases generally increases with increase of pressure. 2. According to Le – chatlier’s principle, the increase in pressure will shift the equilibrium in the direction which will reduce the pressure. Therefore, more number of gaseous molecules dissolves in the solvent. 3. If pressure increases, solubility of gas also increases. |
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| 1553. |
A sample of 12 M Concentrated hydrochloric acid has a density 1.2g L-1 . Calculate the molality. |
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Answer» Given: Molarity = 12 M HCI Density of the solution = 1.2 g L-1 In 12 M HCl solution, there are 12 moles of HCl in 1 litre of the solution. Molality = (No. of moles of solute) / (Mass of solvent(in Kg.)) Calculate mass of water (solvent) Mass of 1 liter HCI solution = density x volume = 1.2g mL-1 x 1000 mL = 1200g Mass of HCI = (No. of moles of HCI) x (molar mass of HCI) = 12 mol x 36.5 g mol-1 = 438g Mass of waler = (mass of HCI solution) – (mass of HCI) = 1200 – 438 = 762 g Molalily = 12/0.762 = 15.75 m |
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| 1554. |
Define 1. Molality 2. Normality |
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Answer» 1. Molality (m): It is defined as the number of moles of the solute present in 1 kg of the solvent. m = (Number of moles of solute) / (Mass of the solvent (in kg) 2. Normality (N): It is defined as the number of gram equivalents of solute in 1 litre of the solution. N = (Number of gram equivalent of solute) / (Volume of soluation (in L) |
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| 1555. |
The estimated vantt Hoff factor for acetic acid solution in benzene is … |
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Answer» The estimated vantt Hoff factor for acetic acid solution in benzene is 0.5 |
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| 1556. |
Phenol dimerises in henzene having van’t Hoff factor 0.54. What is the degree of association? (a) 0.46 (b) 92 (c) 46 (d) 0.92 |
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Answer» (d) 0.92 \(\alpha=\frac{(1-i)n}{(n-1)}=\frac{(1-0.54)2}{(2-1)}\) = 0.46 x 2 = 0.92 |
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| 1557. |
What is the molality of a 10% w/w aqueous sodium hydroxide solution? (a) 2.778 (b) 2.5 (c) 10 (d) 0.4 |
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Answer» (b) 2.5 100% \(\frac{w}{w}\) aqueous NaOH solution means that 10 g of sodium hydroxide in 100g solution. Molality = (No. of moles of solute) / (Weight of solvent (in kg)) = (10/40) / 0.1 = 0.25 / 0.1 = 2.5 M |
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| 1558. |
The degree of association a is equal to …… |
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Answer» The degree of association a is equal to \(\frac{(i-1)n}{n-1}\) |
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| 1559. |
The correct equation for the degree of an associating solute, ‘n’ molecules of which undergoes association in solution, is ………\((a)\,\,\alpha=\frac{n(i-1)}{n-1}\)\((b)\,\,\alpha^2=\frac{n(1-i)}{n-1}\)\((c)\,\,\alpha=\frac{n(i-1)}{1-n}\)\((d)\,\,\alpha=\frac{n(1-i)}{n(1-i)}\) |
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Answer» \((c)\,\,\alpha=\frac{n(i-1)}{1-n}\) \(\alpha=\frac{(i-1)n}{(1-n)}\,\,(or)\,\,\frac{n(i-1)}{(1-n)}\) |
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| 1560. |
Which of the following aqueous solutions has the highest boiling point? (a) 0.1 M KNO3 (b) 0.1 M Na3PO4 (c) 0.1 M BaCl2 (d) 0.1 M K2SO4 |
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Answer» (a) 0.1 M KNO3 Elevation of boiling point is more in the case of Na3PO4(no. of ions 4; 3 Na+ , \(PO_4^{3-}\)) |
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| 1561. |
A stream of air is bubbled slowly through liquid benzene in a flask at `20.0^(@)C` against an ambient pressure of `100.56kPa`. After the passage of `4.80L` of air, measured at `20.0^(@)C` and `100.56kPa` before it contains benzonb vapor, it is found that `1.705g` of benzene have been evaporated. Assuming that the air saturated benzene vapor when it leaves the flask. Calculate the equilibrium vapor pressure of the benzene at `20.0^(@)C`. |
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Answer» Correct Answer - `11.08kPa` Volume of Benzene `=4.80 lt` no of moles of Benzene `=(1.705)/(78)` Temperature `=20^(@)C=293K` `(nRT)/(V)=0.109` atm `=11.08 kPa`. |
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| 1562. |
A pressure cooker reduces cooking time becauseA. 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 - A It is fact that use of pressure cooker reduces cooking time because at higher pressure over the liquid due to cooker lid, the liquid boils at higher temperature and cooking occurs faster. |
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| 1563. |
A pressure cooker reduces cooking time becauseA. If both the assertion and reason are true and reason is a true explanation of the assertion.B. 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 - A It is fact that use of pressure cooker reduces cooking time because at higher pressure over the liquid due to cooker lid, the liquid boils at higher temperature and cooking occurs faster. |
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| 1564. |
Dry air is slowly passed through three solutions of different concentrations, `C_(1),C_(2)` and `C_(3)`, each containing (non-volatile) `NaCl` as solute and water as solvent, as shown in the Fig. If the vessel `2` gains weight and the vesel `3` loses weight then: A. `c_(1)gtc_(2)`B. `c_(1)ltc_(2)`C. `c_(1)ltc_(3)`D. `c_(2)gtc_(3)` |
| Answer» Correct Answer - B::D | |
| 1565. |
A pressure cooker reduces cooking time becauseA. Heat is more evenly distributedB. Boiling point of water inside the cooker is increasedC. The high pressure tenderizes the foodD. All of these |
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Answer» Correct Answer - B The temperature at which a liquid boils increases with increase in pressure. |
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| 1566. |
Pressure cooker reduces cooking time for food because (a) boiling point of water involved in cooking is increased (b) heat is more evenly distributed in the cooking space(c) the higher pressure inside the cooker crushes the food material (d) cooking involves chemical changes helped by a rise in temperature |
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Answer» (a) boiling point of water involved in cooking is increased Pressure cooker reduces cooking time for food because boiling point of water involved in cooking is increased. |
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| 1567. |
A pressure cooker reduces cooking time becauseA. the heat is more evenly distributed inside the cookerB. a large flame is usedC. boiling point of water is elevatedD. whole matter is converted into steam |
| Answer» Correct Answer - C | |
| 1568. |
The difference between the boiling point and freezing point of an aqueous solution containing sucrose (molecular mass `= 342 g mol^(-1))` in 100 g of water is `105.04`. If `K_(f)` and `K_(b)` of water are 1.86 and `0.51 Kg mol^(-1)` respectively, the weight of sucrose in the solution is aboutA. 34.2gB. 342gC. 7.2gD. 72g |
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Answer» Correct Answer - D `DeltaT_(b) = K_(b) xx m` and `DeltaT_(f) = K_(f) xx m` `:. DeltaT_(b) +DeltaT_(f) = (K_(b) +K_(f))m` Now, `T_(b) - T_(f) = (T_(b)^(@) +DeltaT_(b)) - (T_(f)^(@)-DeltaT_(f))` `105 = (DeltaT_(b) +DeltaT_(f)) + (T_(b)^(@) - T_(f)^(@))` `105 = (DeltaT_(b) +DeltaT_(f)) +100` `:. DeltaT_(b) + DeltaT_(f) = 5` `:. m = (DeltaT_(b) + DeltaT_(f))/(K_(b) +K_(f)) = (5)/(1.86 +0.51) = (5)/(2.37) = 2.11` Molalioty `= ("Moles of solute")/("Mass of solvent"(kg))` `:.` Moles of solute `= 2.11 xx 0.1 = 0.211` ? `:.` Mass of solute `= 0.211 xx 342 = 72.16g` |
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| 1569. |
Softening of hard water is done using sodium aluminium silicate (zeolite) . This causesA. adsorption of `Ca^(2+)` and `Mg^(2+)` ions of hard water relpacing `Na^(+)` ions.B. adsorption of `Ca^(2+)` and `Mg^(2+)` ions of hard water relpacing `Al^(3+)` ions.C. both (A) and (B)D. none of these |
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Answer» Correct Answer - A Softening of hard water: Ion exchange resins used for softening of hard water is based upon selective and competive adsorption of ions on resins. `Na_(2)Z+Ca^(+2) rarr CaZ+2 Na^(+)` The organic polymers containing groups like`-COOH, -SO_(3)H` and `-NH_(2)` etc. possess the property of selective adsorption of ions from solution. These are quite usefuk in the softening of water. |
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| 1570. |
A sample of charcoal weighing `6 g` was brought into contact with a gas contained in a vessel of one litre capacity at `27^(@)C`. The pressure of the gas was found to fall from `700` to `400` mm. Calculate the volume of the gas (reduced to STP) that is adsorbent under the condition of the experiment `("density of charcoal sample is" 1.5 g cm^(3))`. |
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Answer» The adsorption is taking place in a closed vessel hence if pressure falls is correspondingly increase in volume constant, excess of the volume of the gas would be adsorbed. `P_(1)V_(1)=P_(2)V_(2)` `V_(2)=P_(1) V_(1)/P_(2)=700xx1000/40=1750 mL` Actual volume of the flask `=1000- "volume of charcoal"=1000-6.00/1.50=996 mL` volume of the gas adsorbed `=1750-996=754 mL` Volume of the gas adsorbed per gram at STP `("U sing" (P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)))` `=(125.67xx400xx273)/(300xx760)=60.19 mL` |
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| 1571. |
A gas such as carbon monoxide would be most likely to obey the ideal gas law atA. high temperatures and low pressuresB. low temperatures and high pressuresC. high temperatures and high pressuresD. low temperatures and low pressures |
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Answer» Correct Answer - A Real gases show ideal gas behaviour at high temperatures and low pressures. |
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| 1572. |
After adding non-volatile solute, freezing point of water decreases to `-0.186^(@)C`. Calculate `Delta T_(b)` if : `K_(f) = 1.86 "K kg mol"^(-1) and K_(b) = 0.521 "K kg mol"^(-1)`A. `0.521`B. `0.0521`C. `1.86`D. `0.0186` |
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Answer» Correct Answer - B `Delta T_(f)=K_(f)xxm` `Delta T_(b)=K_(b)xxm` `(Delta T_(f))/(Delta T_(b))=(K_(f))/(K_(b))` `Delta T_(b) = (Delta T_(f)xxK_(b))/(K_(f))` `= (0.186xx0.521)/(1.86)=0.0521` |
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| 1573. |
When 2.0g of a non-volatile solute was dissolved in 90 gm of benzene, the boiling point of benzene is raised by 0.88K. Which of the following may be the solute? `K_(b)` for benzene `=2.53 K kg mol^(-1))`A. `CO(NH_(2))_(2)`B. `C_(6)H_(12)O_(6)`C. `NaCI`D. None of these |
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Answer» Correct Answer - A `Na^(+)Cl^(-)` an ionic solid is not soluble in benzene `therefore Delta T_(b)=K_(b)xxm` `m=(Delta T_(b))/(K_(b))=(0.88)/(2.53)=0.348` `m = (wxx1000)/(MxxW)` `0.348=(2xx1000)/(Mxx90)` `(["W = mass of solvent, w = mass of solute"])/(M = (2xx1000)/(90xx0.348)=63.86)` Molar mass of urea, `CO(NH_(2))_(2)` `=12+16+(14+2)xx2` `=28+32=60 gmol^(-1)` `therefore` Solute is urea (A) |
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| 1574. |
A very small amount of a non-volatile solute (that does not dissociate) is dissolved in `56.8 cm^(3)` of benzene (density `0.889 g cm^(3))`. At room temperature, vapour pressure of this solution is `98.88 mm Hg` while that of benzene is `100 mm Hg` . Find the molality of this solution. If the freezing temperature of this solution is `0.73` degree lower than that of benzene, what is the value of molal the freezing point depression constant of benzene? |
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Answer» `(Deltap)/(p_(0))=x_(B)` `(100-98.8)/(100)=x_(B)` `x_(B)=0.012` Molality `= (x_(B)xx1000)/((1-x_(B))m_(A))=(0.012xx1000)/(0.988xx78)=0.1557` `DeltaT=K_(f)xx` Molality `0.73=K_(f)xx0.1557` `K_(f)=4.688`. |
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| 1575. |
A system of greater disorder of molecules is more probable. The disorder of molecules is reflected by the entropy of the system. A liquid vapourizes to form a more disordered gas. When a solute is present, there is additional contribution to the entropy of the liquid due to increased randomness. As the entropy of solution is higher than that of pure liquid, there is weaker tendency to form the gas. Thus, a solute (non-volatile) lowers the vapour pressure of a liquid, and hence a higher boiling point of the solution. Similarly, the greater randomness of the solution opposes the tendercy to freeze. In consequence, a lower temperature must be reached for achieving the equilibrium between the solid (frozen solvent) and the solution. The elevation in boiling point `(DeltaT_(b))` and depression in freezing point `(DeltaT_(f))` of a solution are the colligative properties which depend only on the concentration of particles of the solute and not their identity. For dilute solutions, `(DeltaT_(b))` and `(DeltaT_(f))` are proportional to the molarity of the solute in the solution. A liquid possessing which of the following characteristics will be most suitable for determining the molecular mass of a compound by cryoscopic measurements?A. That having low freezing point and small enthhalpy of freezingB. That having high freezing point and small enthhalpy of freezingC. Greater than the normal boiling point of either of the liquid.D. Smaller than the normal boiling point of either of the liquid. |
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Answer» Correct Answer - B If `DeltaT_(f)` is high less error will be in measurement, also `K_(f)` should be high. Therefore, `T_(f)` should be high and `Delta_(fus)H` should be low. |
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| 1576. |
A system of greater disorder of molecules is more probable. The disorder of molecules is reflected by the entropy of the system. A liquid vapourizes to form a more disordered gas. When a solute is present, there is additional contribution to the entropy of the liquid due to increased randomness. As the entropy of solution is higher than that of pure liquid, there is weaker tendency to form the gas. Thus, a solute (non-volatile) lowers the vapour pressure of a liquid, and hence a higher boiling point of the solution. Similarly, the greater randomness of the solution opposes the tendercy to freeze. In consequence, a lower temperature must be reached for achieving the equilibrium between the solid (frozen solvent) and the solution. The elevation in boiling point `(DeltaT_(b))` and depression in freezing point `(DeltaT_(f))` of a solution are the colligative properties which depend only on the concentration of particles of the solute and not their identity. For dilute solutions, `(DeltaT_(b))` and `(DeltaT_(f))` are proportional to the molarity of the solute in the solution. To aqueous solution of `Nal`, increasing amounts of solid `Hgl_(2)` is added. The vapour pressure of the solutionA. decreases to a constant valueB. increases to a constant valueC. increases and then decreasesD. remains constant as `Hgl_(2)` is sparingly soluble in water |
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Answer» Correct Answer - D `2NaI_(2) + HgI_(2) rarr Na_(2)Hgl_(4)` As `HgI_(2)` is added, the number of molecules will decrease due to the formation of `Na_(2)Hgl_(4)`. Therefore, pressure will increase. As more `HgI_(2)` is added, pressure will remain constant (and`Nal` is consumed) after that as `HgI_(2)` is sparingly soluble. |
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| 1577. |
`1 Lit`. Of aq. Solution of urea having density`=1.060gm//mL` is found to have `DeltaT_(b)=0.5^(@)C`, if temp. of this solution increase to `101.5^(@)C` then calculate amount of water which must have gone is vapour state upto this pt. given `K_(b)=0.5K kg mol^(-1)` for water |
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Answer» `0.5=0.5xx(1060)/(1060)(M)/(60) rArr 1060xx60=M` `1` Lit solution`=1060gm` (Mass of water)`|=1000gm` urea`=60gm` `DeltaT_(b)=("molality")=K_(b)` (Molality) `rArr 3=(60)/(60)xx(1000)/(W_("water")` `W_("water")=(1000)/(3)gm`. mass of water vaporised`=1000-(2000)/(3)gm=666.67gm` |
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| 1578. |
An M/10 solution of potassium ferrocyanide is `46%` dissociated at `300 K`. What will be its osmotic pressure? |
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Answer» Normal osmotic pressure`=W/(MwxxV)xxRxxT` When no dissociation has taken place `W/(Mw)=0.1`,`V=1 L`,`R=0.821`,`T=300 K` Normal osmotic pressure =`0.1/1xx0.0821xx300` `=2.463 atm` Potassium ferrocyanide is an electrolyte. It dissociates as `K_(4)[Fe(CN)_(6) hArr 4 K^(o+) + [Fe(CN)_(6)]^(4-)` Total numberof particles =`1-alpha+4alpha=1+4alpha` `alpha=0.46`,so `1+4alpha=1+4xx0.46=2.84` `"Observed osomotic pressure"/"Normal osomotic pressure"=2.84/1` Observed osmotic pressure =`2.84xx2.463=6.995` |
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| 1579. |
A `0.1M` solution of potassiu ferrocyanide is `46%` dissociated at `18^(@)C`. What will be its osmotic pressure? |
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Answer» Correct Answer - `p = 6.785 atm` `{:(pi=icSt,alpha =(i-1)/(n-1)),(0.46=(i-1)/(5-1),i=2.84):}` `pi = 2.84 xx 0.1 xx 0.082 xx 291` `pi = 6.785 atm` |
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| 1580. |
According to Freundlich adsorption isotherm, which of the following is correct?A. `x/m prop p^(0)`B. `x/m prop p^(1)`C. `x/m prop p^(1//n)`D. All the above are correct for different ranges of pressure. |
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Answer» Correct Answer - D `x/m prop P^(1//n)` " " where `n ge 1` |
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| 1581. |
The coagulating power of electrolytes having inos `Na^(o+),Al^(3+)` and `Ba^(2+)` for arsenic sulphide sol increases in the orderA. `Al^(3+) lt Ba^(2+) lt Na^(+)`B. `Na^(+) lt Ba^(2+) lt Al^(3+)`C. `Ba^(2+) lt Na^(+) lt Al^(3+)`D. `Al^(3+) lt Na^(+) lt Ba^(2+)` |
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Answer» Correct Answer - B According to Hardy Schulze rule, greater the charge on cation, greater is its coagulating power for negatively charged solution. So, order of coagulating power : `Na^(+) lt Ba^(2+) lt Al^(3+)`. |
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| 1582. |
What is the molarity of `H_(2)SO_(4)` solution that has a density 1.84 g/c c at `35^(@)`C and contains 98% by weight?A. `4.18M`B. `8.14M`C. `18.4`D. `18M` |
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Answer» Correct Answer - C `98%H_(2)SO_(4)`means`100g`solution contain`98gH_(2)SO_(4)` `(100)/(1.84)c c`contain `98gH_(2)SO_(4)` `1000c c`solution contains `=(98)/(100)xx1.84xx1000gH_(2)SO_(4)` `=(98)/(100)xx(1.84xx1000)/(98) "moles of" H_(2)SO_(4)` `=18.4M` |
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| 1583. |
The vapour pressure of a solvent decreased by 10 mm of Hg when a non-volatile solute was added to the solvent. The mole fraction of solute in solution is 0.2, what would be the mole fraction of solvent if the decrease in vapour pressure is 20 mm of Hg?A. 0.8B. 0.6C. 0.4D. 0.2 |
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Answer» Correct Answer - C `(Deltap)/(p^(@)) = X` or `(Deltap_(1))/(Deltap_(2)) = (X_(1))/(X_(2))` `:. X_(2) = (X_(1) xx Deltap_(2))/(Deltap_(1)) = 0.2 xx (20)/(10) = 0.4` |
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| 1584. |
A `5%` solution of cane sugar (molecular weight =`342`) is isotonic with a `1%` solution of substance `X`. The molecular weight of `X` isA. 342B. 171.12C. 65.6D. 136.8 |
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Answer» Correct Answer - C Isotonic solutions: Same osmotic pressure (i.e., same `C_("effective")` and assume `c~~m`) `rArr (5//342)/(95//1000) = (1//Mw_(2))/(99//1000) rArr Mw_(2)=65.64 g "mol"^(-1)` |
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| 1585. |
A `5%` solution of cane sugar (molecular weight=342) is isotonic with `1%` solution of substance `X`.The molecular weight of `X` isA. 171.2B. 68.4C. 34.2D. 136.2 |
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Answer» Correct Answer - B For non-electrolytic solutions, if isotonic, `C_(1)=C_(2)` =`(5 xx 1000)/(342 xx 100)=(1 xx 1000)/(Mw_(2) xx 100)` `:. Mw_(2) =68.4` |
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| 1586. |
A `5%` solution of cane sugar (molecular weight =`342`) is isotonic with a `1%` solution of substance `X`. The molecular weight of `X` isA. 171.2B. 68.4C. 34.2D. 136.2 |
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Answer» Correct Answer - B `(% "age"(1))/("M. Mass" (1)) = (%"age"(2))/("M.Mass"(2))` or `(5)/(342) = (1)/(X)` or `X = (342)/(5) = 68.4` |
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| 1587. |
The density of a solution containing `13%` by mass of sulphuric acid is `1.09g//mL`. Calculate the molarity and normality of the solutionA. ` 1.445M `B. ` 14.45M `C. ` 144.5M `D. ` 0.1445M ` |
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Answer» Correct Answer - A Volume of ` 100g `of the solution =` (100)/(d)` `=(100)/(1.09)mL = (100)/(1.09xx1000)` litre `(1)/(1.09xx10)` litre Numbers of moles of `H_(2)SO_(4)` in `100 g` of the solution `=(13)/(98)` Molarity `= ("No. of moles of water")/("Volume in litre")` `(13)/(98)xx(1.09xx10)/(1)=10445 M` |
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| 1588. |
Density of `2.05 M` solution of acetic acid in water is `1.02g//mL`. The molality of same solution is:A. `"1.14 mol kg"^(-1)`B. `"3.28 mol kg"^(-1)`C. `"2.28 mol kg"^(-1)`D. `"0.44 mol kg"^(-1)` |
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Answer» Correct Answer - C `m=(Mxx1000)/(1000d-Mm_(B))=(2.05xx1000)/(1000xx1.02-2.05xx60)` `=2.28"mol kg"^(-1)`] |
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| 1589. |
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogenous solution. These are called colligative properties. Application of colligative properties are very useful in day-to-day life. One of its example is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is `0.9`. Given : Freezing point depression constant of water `(K_(f)^("water")) = 1.86 K "mol"^(-1)` Freezing point depression constant of ethanol `(K_(f)^("ethonal")) = 2.0 K kg "mol"^(-1)` Boiling point elevation constant of water `(K_(b)^("water")) = 0.52 K kg "mol"^(-1)` Boiling point elevation constant of ethanol `(K_(b)^("ethonal")) = 1.2 K kg mol^(-1)` Standard freezing point of water `= 273 K` Standard freezing point of ethonal `= 155.7 K` Standard boiling point of water `= 373 K` Standard boiling point of ethanol `= 351.5 K` Vapour pressure of pure water `= 32.8 mm Hg` Vapour pressure of pure ethonal `= 40 mm Hg` Molecular weight of water `= 18 g "mol"^(-1)` Molecular weight of ethonal `= 45 g"mol"^(-1)` In answering the following questions, consider the solution to be ideal ideal solutions and solutes to be non-volatile and non-dissociative. Water is added to the solution M such lthat the molecules fraction of water in t he solution becomes `0.9`. The boiling point of this solution is :A. 380.4 KB. 376.2 KC. 375.5 KD. 354.7 K |
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Answer» Correct Answer - B 1. Molality `m = (x_(B)xx1000)/((1-x_(B))m_(A))=(0.1xx1000)/(0.9xx46)` `=2.415` `DeltaT=K_(f)xxm` `=2xx2.415=4.83` Freezing point of solution `= 155.7 - 4.83 = 150.9 K` 2. `p=p^(0)x_(A)` `=40xx0.9=36` mm Hg 3. When water becomes solvent, the molality of solution will be : `m=(x_(B)xx100)/(x_(A)xxm_(A))=(0.1xx1000)/(0.9xx18)=6.172` `DeltaT=K_(b)xxm=0.52xx6.172=3.209` Boiling point of solution `= 373 + 3.209` `=376.2 K`] |
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| 1590. |
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogenous solution. These are called colligative properties. Application of colligative properties are very useful in day-to-day life. One of its example is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is `0.9`. Given : Freezing point depression constant of water `(K_(f)^("water")) = 1.86 K "mol"^(-1)` Freezing point depression constant of ethanol `(K_(f)^("ethonal")) = 2.0 K kg "mol"^(-1)` Boiling point elevation constant of water `(K_(b)^("water")) = 0.52 K kg "mol"^(-1)` Boiling point elevation constant of ethanol `(K_(b)^("ethonal")) = 1.2 K kg mol^(-1)` Standard freezing point of water `= 273 K` Standard freezing point of ethonal `= 155.7 K` Standard boiling point of water `= 373 K` Standard boiling point of ethanol `= 351.5 K` Vapour pressure of pure water `= 32.8 mm Hg` Vapour pressure of pure ethonal `= 40 mm Hg` Molecular weight of water `= 18 g "mol"^(-1)` Molecular weight of ethonal `= 45 g"mol"^(-1)` In answering the following questions, consider the solution to be ideal ideal solutions and solutes to be non-volatile and non-dissociative. Water is added to the solution M such lthat the molecules fraction of water in t he solution becomes `0.9`. The boiling point of this solution is :A. 380.4 KB. 376.2KC. 375.5 KD. 354.7K |
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Answer» Correct Answer - B `therefore` Water is solvent `DeltaT_(b)=K_(b)("water")xx(n_("ethanol")xx1000)/(w_("water"))=0.52xx(0.1xx1000)/(0.9xx18)=3.2` `DeltaT_(b)("solution")=373+3.2=376.3K` |
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| 1591. |
A 5% solution of cane sugar (molar mass = 342) is isotonic with 1% solution os a substance X. The molar mass of X is:A. 171.2B. 68.4C. 34.2D. 136.2 |
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Answer» Correct Answer - b `pi=(W_(B)xxRxxT)/(M_(B)xxV)` R, T and V are constants for isotonic solutions `pi_(x)=pi_(sugar)1/M_(x)=5/342` `or M_(X)=342/5=68.4` |
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| 1592. |
The relative lowering of vapour pressure of an aqueous solution containing a non-volatile solute, is 0.0125. The molality of the solution is |
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Answer» c. As we know `(P^(@) - P)/P^(@) = chi_(2)` = mole fraction of solute The ration `(P^(@)-P)//P^(@)` is the relative lowering of vapour pressure, which is equal to `0.0125` here. `:. chi_(2) = 0.0125` The relation between `m` and `chi` is: `m=(chi_(2) xx 1000)/(chi_(1) xx Mw_(1)) = (0.0125 xx 1000)/(( 1- 0.0125 xx 18))` `[{:(,Mw_((H_(2)O))),(=,18g mol^(-1)):}]` ` = (0.0125 xx 1000)/(0.9875 xx 18)` =`0.70` ` :. m =0.70` |
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| 1593. |
What is the molarityk and molality of a 13% solution (by weight) of sulphric acid with a density of `1.02 mL^(-1)`? To what volume should `100 mL` of this acid be diluted in order to preapre a `1.5 N` solution? |
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Answer» Correct Answer - Molality=`1.52 m`;`normality=2.70 V=180 L` Molarity =`(% "by weight" xx "Density" xx 10)/("Molecular weight")` =`(13 xx 1.02 xx 10)/98` =`132.6/98` =`1.35 M` `13%` solution by weight means that `13 g` of solute is dissolved in `87 g` of solvent. Thus, molality =(`("Weight of solute")/("Molecular weight of solute"))/("Weight of solvent")xx1000` =`(13/98)/87xx1000` =`(13 xx 1000)/(98 xx 87)` =`1.52 m` Normality = Molarity `xx` Ew factor `:. N=1.35 xx 2=2.70N` For dilution `N_(1)V_(1)=N_(2)V_(2)` `100 xx 2.70N =1.5 N xx V_(2)` `V_(2)=180 mL` So, the acid should be diluted upto `180 mL` to prepare `1.5 N` solution. |
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| 1594. |
The density of 2.0 M solution of a solute is 1.2 g `mL^(-1)`. If the molecular mass of the solute is 100 g `mol^(-1)`, then the molality of the solution is :A. 2.0 mB. 1.2 mC. 0.6 mD. 2.4 m. |
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Answer» Correct Answer - a Volume of solution containing 2 moles of solute =1L =1000 mL mass of solution =`Vxxd` =`1000mL xx1.2 g mL^(-1)=1200 g` Molality of solvent =1000 g=1.0 kg. `"Molality of solution (m)"=("No. of moles of colute")/("Mass of solvent in kg")` `=((2 moles))/(1 kg)=2 "mol kg"^(-1)=2.0 m` |
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| 1595. |
Which has highest boiling point under point under 1 atm pressue ?A. 0.1 M NaCIB. 0.1 M ScuroseC. 0.1 M `BaCI_(2)`D. 0.1 M Glucose. |
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Answer» Correct Answer - c 0.1 M `BaCI_(2)` because it will form maximum ions (3) in solution `Deltat_(b)=iKbm` |
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| 1596. |
An aqueous solution freezes at-`0.186^(@)C(K_(f)=1.86,K_(b)=0.512)`. What is the elecation in boiling point ?A. 0.186B. 0.512C. 0.86D. 0.0512. |
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Answer» Correct Answer - d For the same solution `(DeltaT_(b))/K_(b)=(DeltaT_(b))/K_(f)` of `DeltaT_(b)=(DeltaT_(f))/(T_(f))xxK_(b)` `=((0.186^(@)C))/1.86xx0.512.` =`0.0512^(@)C` |
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| 1597. |
An aqueous solution freezes at `-0.186^(@)C (K_(f)=1.86^(@)` ,`K_(b)=0.512^(@)`. What is the elevation in boiling point?A. 0.186B. 0.512C. `(0.512)/(1.86)`D. 0.0512 |
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Answer» Correct Answer - D `DeltaT=K_(f)xxm` `0.186=1.86xxm` `m=0.1` `DeltaT=K_(b)xxm=0.512xx0.1 =0.0512`] |
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| 1598. |
0.5 molar aqueous solution of a weak acid (HX) is 20% ionised. If `K_(f)` for water is `1.86 K kg mol^(-1)`, the lowering in freezing point of the solution isA. `-0.56K`B. `-1.12K`C. `0.56K`D. `1.12K` |
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Answer» Correct Answer - D `{:(,"HX",hArr,H^(+)+,X^(-)),("Initial","1 mole",,,),("After ionization",1-a,,a,a):}` Total no. of moles at equilibrium `= 1+ alpha` `:. i=1 +a = 1+ 0.20 = 1.20` `Delta T_(f) = iK_(f)m` `=1.20 xx 1.80 xx 0.5 = 1.20K`. |
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| 1599. |
In a `0.2` molal aqueous solution of weak acid `HX` (the degree of dissociation `0.3`) the freezing point is (given `K_(f) = 1.85 K molality^(-1)`):A. `+0.480^(@)C`B. `-0.480^(@)C`C. `-0.360^(@)`D. `+0.360^(@)` |
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Answer» Correct Answer - 2 We have `(DeltaT_(f))_("observed")=(K_(f))_("observed")` molality `" "HX(aq)hArrH^(+)(aq)+X^(-)(aq)` `{:("Moles before ionization",+,1 mol,0 mol,0 mol),("Moles after ionization",-,1-alpha mol,alpha mol,alpha mol):}` `i=("Total moles after ionization")/("Total moles before ionozation")` `=(1-alpha+alpha+alpha)/1=1+alpha` `=1+0.3` `=1.3` we also have `i=(K_(f) ("observed"))/(K_(f)("calculated"))` or `K_(f) ("observed")=i K_(f)("calculate")` `=(1.3)(1.85)` `=2.405` Thus `DeltaT_(f)=(2.405 K kg mol^(-1))(0.28 mol kg^(-1))` `=0.481 K` or `0.481^(@)C` Hence Freezing point of solution =Freezing point of water `-DeltaT_(f)` `=(0.000^(@)C)- (0.481^(@)C)` `= -0.481^(@)C` |
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| 1600. |
Oxygen gas is slightly soluble in water but it is highly soluble in blood. Explain. |
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Answer» The vital constituent of blood, namely haemoglobin reacts with oxygen increasing the solubility of oxygen. Haemoglobin + 4O2(g) → Haemoglobin O8 This oxygenated blood is circulated to the various parts of body, for the supply of oxygen. |
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