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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.
| 1851. |
What is elevation of boiling point ? |
| Answer» The boiling point of a solution containing non-volatile solute is higher than the boiling point of pure solvent. The difference in boiling points between solution and pure solvent is called elevation of boiling point. | |
| 1852. |
Calculate the normality of 9.0 gram of oxalic acid, `(COOH)_(2)`, dissolved in 250 mL of solution. Strategy: Find the equivalent mass of oxalic acid by dividing its formula mass by its basicity, which is two, because oxalic acid is a dibasic acid. Use the numerical value of equivalent mass to get gram equivalent mass. Then convert grams of `(COOH)_(2)` to equivalent of `(COOH)_(2)`, which lets to calculate the nirmality |
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Answer» Each molecule of oxalic acid contains two acidic hydrogen atoms: `H-O-overset(O)overset(||)(C)-overset(O)overset(||)(C)-O-H` Formula mass of `(COOH)_(2)=90u` Basicity of `(COOH)_(2)=2` Thus, Equivalent mass of oxalic acid `=("Formula mass of" (COOH)_(2))/("Basicity of" (COOH)_(2))` `=(90u)/2=45 u` Hence, gram equivalent mass of oxalic acid is `45 g eq.^(-1)`. Number of equivalents of oxalic acid `=("Mass of"(COOH)_(2))/("Gram equivalent mass of" (COOH)_(2))` `=(9 g)/(45 g eq.^(-1))` `=0.2 eq`. `"Normality" (N)=(no . eq. (COOH)_(2))/V_(mL)xx(1000 mL)/L` `=(0.2 eq.)/(250 mL)xx(1000 mL)/L` `=0.8 eq. L^(-1)` or `0.8 N` |
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| 1853. |
20 mL of 0.5 M HCl is mixed with 30 mL of 0.3 M HCl, the molarity of the resulting solution is :A. 0.8 MB. 0.53 MC. 0.38 MD. 0.83 M |
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Answer» Correct Answer - C `M_(1)V_(1)+M_(2)V_(2)=M_(R)(V_(1)+V_(2))` `0.5xx20+0.3xx30=M_(R)xx50` `M_(R)=0.38M` where, `M_(R)=` resultant molarity of the mixture] |
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| 1854. |
Calculate the molarity, molality and mole fraction of ethyl alcohol in a solution of total volume 95 mL prepared by adding 50 mL of ethyl alcohol (density `= 0.789 mL^(-1)`) to 50 mL water (density `= 1.00 g mL^(-1)`). |
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Answer» No. of moles of ethyl alcohol `= ("Vol." xx"density")/("Mol. mass")` `= (50xx0.789)/(46)=0.8576` No. of moles of water `= ("Vol."xx"density")/("Mol. mass")=(50xx1)/(18)=2.7777` Molarity `= ("No. of moles")/("Vol. of sol. in mL")xx1000` `=(0.8576)/(95)xx1000=9.027M` Molarity `= ("No. of moles of solute")/("Mass of solvent in grams")xx1000` `=(0.8576)/(50)xx1000=17.152m` Mole fraction `= (0.8576)/(0.8576+2.7777)=(0.8576)/(3.6353)=0.236`. |
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| 1855. |
If `100 mL` of `1.00 M HCl` and `100mL` of `0.80 M NaOH` solution are mixed, the molarity of `Cl^(-)` ions in the resulting solutions will beA. `0.10 M`B. `0.50 M`C. `0.40 M`D. `0.30 M` |
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Answer» Correct Answer - 2 Millimoles of `HCl=V_(mL)xxM=(100 mL)(1.00 M)` `100` mmol Millimoles of `NaOH=(100 mL)(0.80 M)=80` mmol `" "HCl+NaOH rarr NaCl+H_(2)O` `{:("Rxn ratio:", 1 mmol,1 mmol,1 mmol,1 mmol),("Start:",100 mmol,80 mmol,0 mmol,),("Change:",-80 mmol,-80 mmol,+80 mmol,))/(After rxn: 20 mmol,0 mmol, 80mmol)` Molarity of unused `HCl=(20 mmol HCl)/(200 mL)` `=0.10 M HCl` Molarity of `NaCl` formed=`(80 mmol)/(200 mL) NaCl` `=0.40 M NaCl` Both `HCl` and `NaCl` are strong electrolytes, so the solution is `0.10 M` in `H^(+) (aq.), (0.10+0.40)M=0.50 M` in `Cl^(-)`, and `0.40 M` in `Na^(+)` ions. |
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| 1856. |
Give reason as to why when 30 mL of ethyl alcohol and 30 mL of water are mixed, the volume of the resulting solution is more than 60 mL. |
| Answer» Intermolecular hydrogen bonding is present in the molecules of both ethyl alcohol and water. Upon mixing. The extend of hydrogen bonding decreses. As a result, there is an incresese in volume of the resulting solution. | |
| 1857. |
What will happen to the freezing point of a solution when mercuric iodide is added to an aqueous solution of potassium iodide ? |
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Answer» KI eill combine with mercuric iodide `(HgI_(2))` to from `K_(2)HgI_(4)`. The number of solute of solute perticles in solution will decrese. As a result, the depression ion freezing point temperature `(DeltaT_(f))` will decreas and the freezing point of the solution `(T_(f))` will increse. `2KI+HgI_(2)toK_(2)[HgI_(4)]` Potassium mercuro-iodide |
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| 1858. |
When mercuric iodide is added to the aqueous solution of potassium iodide, then:A. freezing point is raisedB. freezing point does not changeC. freezing point is loweredD. boiling poing does not change |
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Answer» Correct Answer - A When `HgI_(2)` is added to the aqueous solution of KI following reaction takes place. `2KI+HgI_(2)rarrK_(2)[HgI_(4)]hArr2K^(+)+[HgI_(4)]^(2-)` Thus there is a net decrease in number of ions present in solution. Two formula units of KI i.e., four ions decrease to three ions in `K_(2)[HgI_(4)]`. As number of ions decreases `DeltaT_(f)` decrease of f.p. is raised. |
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| 1859. |
If the density of methanol is `0.793 kg L^(-1)` what ia its volume needed for making 2.5 L of its `0.25 M` solution?A. `32.25 mL`B. `45.98 mL`C. `78.60 mLD. `25.22 mL` |
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Answer» Correct Answer - 4 It is basically a problem of dilution of solution, which can be solved by applying the molarity equation: `M_(i) V_(i) =M_(f)V_(f)` our objective is to get `V_(i)`. This is possible provided we know `M_(i)`. By definition, `M=n_(solute)/V_(L)` Since `n_(solute)=(mass_(solute))/(molar mass_(solute))` We can write `M=(mass_(solute))/(V_(L)xxmolar mass_(solute))` Since in this problem, we need to dilute `CH_(3)OH, V_(L)` is equal to volume of `CH_(3)OH` in litres and mass/volume is density. Thus `M=((density)_(CH_(3)OH)" in "Kg L^(-1))/((molar mass_(CH_(3)OH))" in "Kg mol^(-1))` `=(0.793 Kg L^(-1))/(0.032 Kg mol^(-1))` `=24.78 mol L^(-1)=M_(i)` Now `24.78 mol L^(-1) xxV_(i)=0.25 mol L^(-1)xx2.5 L` Thus, `V_(i)=((0.25 mol L^(-1))(2.5 L))/((23.78 mol L^(-1)))=0.02522 L` `=25.22 mL` |
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| 1860. |
In which mode of expression, the concentration of a solution remains independent of temperature?A. MolalityB. MolarityC. FormalityD. Normality |
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Answer» Correct Answer - 1 While expressing molarity, we take moles of solute per `1000 g` of solvent. Since the no. of moles and mass are independent of of temperature, the answer is (1). Each method of expressing concentration of the solutions has its own merits and demerits. Mass%, ppm, mole fraction and molarity are independed of temperature, whereas molarity is a function of temperature. This is because volume depends on temperature and the mass does not. Similarly formally and noemally involving depend on temperature. |
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| 1861. |
‘Henry’s Law is special case of Raoult’s Law. Explain how? |
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Answer» According to Raoult’s Law- P1 = P1 ◦ χ 1 According to Henry’s Law- P = KH χ g As vapour pressure of liquid increases with increase in mole fraction of it in solution as mole fraction of gas in solution increases with increase in pressure of the gas. In this condition KH can be taken equivalent to P1◦ . |
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| 1862. |
Why is the vapour pressure of a liquid remains constant at constant temperature? |
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Answer» At equilibrium, the rate of evaporation = rate of condensation. Hence the vapour pressure of a liquid is constant at constant temperature. |
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| 1863. |
A solution is prepared by mixing 50ml of chloroform and 50ml of acetone. The volume of the resulting solution will be 100 ml or less than 100ml or more than 100 ml? |
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Answer» When chloroform and acetone are mixed they exhibit negative deviation from ideal behavior because intermolecular hydrogen bonds are formed between molecules of acetone and chloroform. Since the hydrogen bonds are comparatively stronger forces of attraction, the volume of the solution decreases. Therefore, the volume of the resulting solution will be less than 100ml. |
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| 1864. |
At `80^(@)C`, the vapour p[ressure of pure liquid A is 520 mm of Hg and that of pure liquid B is 1000 mm of Hg. If a mixture solution of A and B boils at `80^(@)C` and 1 atomoshere pressure, the amount of A in the mixture is (1 atm = 760 mm of Hg)A. 60 mol precentB. 52 mol precentC. 34 mol presentD. 48 mol precent. |
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Answer» Correct Answer - a According to available information, `P_(A)^(@)=520 mm Hg, P_(B)^(@)=1000mm Hg` `P_(A)^(@)X_(A)+P_(B)^(@)X_(B)=760 mm Hg`. `P_(A)^(@)X_(A)+P_(B)^(@)(1-X_(A))760` `520 X_(A)+1000(1-X_(A))=760` `520X_(A)+1000-1000X_(A)=760` `-480X_(A)=-240` or `X_(A)=240/480=1/2 "or 50 mol precent". |
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| 1865. |
4 L of 0.02 M aqueous solution of NaCl was diluted by adding 1 L of water. The molality of the resultant solution is……..A. 0.004B. 0.008C. 0.012D. 0.016 |
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Answer» Correct Answer - D `M_1V_1=M_2V_2` 0.02 M x 4 L = `M_2 ` x 5 L `M_2=(0.02xx4)/5`=0.016 M Here, molality will be equal to molarity as for NaCl, molecular weight is equal to its equivalent weight. |
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| 1866. |
Which of the following solutions will exhibit highest boiling point?A. `0.01 M Na_(2)SO_(4)`B. `0.01M KNO_(3)`C. `0.015 M` ureaD. `0.015 M` glucose |
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Answer» Correct Answer - A `DeltaT = ixx K_(b) xx m` `ixx m` of `Na_(2)SO_(4)` is highest, hence its boiling point will also be highest. `{:(Na_(2)SO_(4),i xx m = 3 xx0.001 = 0.03),(KNO_(3),i xxm = 2 xx 0.01 =0.02),(Urea,i xx m =1 xx 0.015=0.015),(Glucose,ixxm = 1xx 0.015 =0.015):}` |
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| 1867. |
Which of the following solutions will exhibit highest boiling point?A. `0.01 M Na_(2)SO_(4)`B. `0.01 M KNO_(3)`C. `0.015M` ureaD. `0.015M`glucose |
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Answer» Correct Answer - A Elevation is boiling point is a colligative property which depends upon the number of solute particles. Greater the number of solute particles in a solution, higher the extent of elevation in boiling point. `Na_(2)SO_(4)rarr2Na^(+)+SO_(4)^(2-)` |
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| 1868. |
Which of the following solutions will exhibit highest boiling point?A. 0.01 M `Na_(2)SO_(4)`B. 0.01 M `KNO_(3)`C. 0.015 M ureaD. 0.015 M glucose |
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Answer» Correct Answer - A `DeltaT=ixxk_(b)xxm` `ixxm` of `Na_(2)SO_(4)` is highest, hence its boiling point will also be highest `Na_(2)SO_(4)" "ixxm=3xx0.01=0.03` `KNO_(3)" "ixxm=2xx0.01=0.02` `"Urea "ixxm=1xx0.015=0.015` `"Glucose "ixxm=1xx0.015=0.015`] |
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| 1869. |
`1.355 g` of a substance dissolved in `55 g`of `CH_(3)COOH` produced a depression in the freezing point of `0.618^(@)C`. Calculate the molecular weight of the substance `(K_(f)=3.85)` |
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Answer» `DeltaT=K_(f)xx(w_(B)xx1000)/(m_(B)xxw_(A))` where, `w_(B)=` mass of solute, `m_(B)=` molar mass of solute, `w_(A) =` mass of solvent `0.618=3.85xx(1.355xx1000)/(m_(B)xx55)` `m_(B)=153.47`. |
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| 1870. |
An aqueous solution of a non-volatile solute boils at `100.17^(@)C`. At what temperature will the solution freeze? (Given: `K_(b) = 0.512` and `K_(f) = 1.86`) |
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Answer» We know that, `Delta T_(b) =` molality `xx K_(b)` 0.17 = molality `xx 0.512` Molality of the solution `=(0.17)/(0.512)m` Let depression in freezing point be `Delta T_(f)` `Delta T_(f)`= molality `xx K_(f)` `=(0.17)/(0.512)xx1.86 = 0.62^(@)C` Thus, the freezing point of the solution `= 0.00-0.62 = -0.62^(@)C`. |
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| 1871. |
The mole fraction of `CH_(3)OH` in an aqueous solution is 0.02 and its density is `0.994 g cm^(-3)`. Determine its molarity and molality. |
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Answer» Let `x` mole of `CH_(3)OH` and y moles of water be present in solution Mole fraction of `CH_(3)OH=(x)/(x+y)=0.02` So, `(y)/(x)=49or(1)/(y)=(1)/(49)` Molality `= (x)/(18xxy)xx1000=(1000)/(18xx49)=1.13m` Volume of solution `= ("Total mass")/("density")=(32x+18y)/(0.994)mL` `=(32x+18y)/(0.994xx1000)"litre"=(32x+18y)/(994)` litre Molarity `= (x)/(32x+18y)xx994` `=(994)/(32+18xxy//x)=(994)/(32+18xx49)=1.0875M`. |
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| 1872. |
If `100mL` of `1N` sulphuric acid were mixed with 100mL of 1N sodium hydroxide, the solution will beA. AcidicB. BasicC. NeutralD. Slightly acidic |
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Answer» Correct Answer - C `100 mL` of `1N H_(2)SO_(4)` will completely neutralise `100 mL` of `1M NaOH` and therefore the resulting solution will be neutral. |
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| 1873. |
Calculate normality of the mixure obtained by mixing `100ml` of `0.1N HCL` and `50 ml` of `0.25N NaOH` solutionA. ` 0.0467 N `B. ` 0.0367 N `C. ` 0.0267N `D. ` 0.0167N ` |
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Answer» Correct Answer - D Meq.of HCl=`100xx0.1=10` Meq. of NaOH=`50xx0.25=1.25` `:.`HCl and NaOH neturalized each other with equal. Eq. Meq. Of NaOH left=`12.5-10=2.5` Volume of new solution =`100+50=150ml.` `N_(NAOH)`left=`(2.5)/(150)=0.0167N` Hence normality of the mixture obtained is `0.0167N` |
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| 1874. |
When `1.80g` glucose dissolved in `90g of H_(2)O`,the mole fraction of glucose isA. `0.00399`B. `0.00199`C. `0.0199`D. `0.998` |
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Answer» Correct Answer - B Mole fraction of glucose `(n)/(n+N)=(0.01)/(0.01+5)` `0.00199` |
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| 1875. |
`0.5M` of `H_(2)SO_(4)` is diluted from 1 litre to 10 litre, normality of resulting solution isA. 1NB. 0.1NC. 10ND. 11N |
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Answer» Correct Answer - B `0.5M H_(2)SO_(4)=1N H_(2)SO_(4)` `underset(("conc"))(N_(1)V_(1)) = underset(("dil"))(N_(2)V_(2))` ` 1Nxx1L = N_(2)xx10L` `N_(2)=1//10N = 0.1N` |
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| 1876. |
A`5` molar solution of `H_(2)SO_(4)`is diluted from `1`litre to `10`litres. What is the normality of the solution?A. `0.25N`B. `1N`C. `2N`D. `7N` |
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Answer» Correct Answer - B Molarity of `H_(2)SO_(4)=5M` Normality of `H_(2)SO_(4)(NI)=2xx5=10N` `N_(1)V_(1)=N_(2)V_(2)` `10xx1=N_(2)xx10` `N_(2)=1N` |
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| 1877. |
0.5 M of `H_(2)SO_(4)` is diluted from `1` litre to 10 litre, normality of resulting solution isA. `1N`B. `0.1N`C. `10N`D. `11N` |
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Answer» Correct Answer - B Molarity of `H_(2)SO_(4)=0.5` Normality of `H_(2)SO_(4)(NI)=0.5xx2=1` `N_(1)V^(1)=N_(2)V_(2)` `1xx1=N_(2)xx10 or N_(2)=(1)/(10)=0.1N` |
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| 1878. |
0.5 M of `H_(2)SO_(4)` is diluted from `1` litre to 10 litre, normality of resulting solution isA. 1 NB. 0.1 NC. 10 ND. 11 N |
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Answer» Correct Answer - B `M_(1)V_(1)=M_(2)V_(2)` `0.5xx1=M_(2)xx10` `M_(2)=0.05` `N=Mxx` basicity of acid `=0.05xx2=0.1N`] |
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| 1879. |
Calculate the Quantity of substance in a Titrated solution: A flask contains 20 mL of a solution with an unknown amount of HCl. This solution is titrated with 0.207 M NaOH. It takes 4.47 mL NaOH to complete the rection. What is mass and the molority of the HCl? Strategy: Convert the volume of NaOH to moles of NaOH (using the molar of NaOH). Then convert moles NaOH to moles HCl, using the chemical equation. Finally, convert moles HCl to gram HCl and molaruty HCl. |
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Answer» Step 1. Converting volume NaOH to moles NaOH `n_(NaOH)=V_(L)xxM` `=4.47xx10^(-3) L NaOH "soln"xx(0.207 mol NaOH)/(1 L NaOH "soln")` `=0.000925 mol NaOH` `=9.25xx10^(-4) mol NaOH` Step 2. According to the balanced equation `(NaOH+HCl rarr NaCl+H_(2)O)`, the number of moles of HCl is the same as that os NaOH: `n_(HCl)=9.25xx10^(-4) mol NaOHxx(1 mol HCl)/(1 mol NaOH)` `=9.25xx10^(-4) mol HCl` Step 3. Dividing the number of moles of HCl by the volume gives the molarty of HCl: HCl molarity`=("Moles" HCl)/V_(L)` `=(9.25xx10^(-4) mol HCl)/(20xx10^(-3) L HCl)` `=0.0462 M HCl` Step 4. Multiplying the number of moles of HCl by the molar mass of HCl gives the mass of HCl: `mass_(HCl)=n_(HCl)xx Molar mass_(HCl)` `=(9.25xx10^(-4) mol HCl)((36.5 g HCl)/(1 mol HCl))` `=0.0338 g HCl` |
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| 1880. |
1 litre solution containing 490 g of sulphuric acid is diluted to 10 litre with water. What is the normality of the resulting solution ?A. 0.5 NB. 1.0 NC. 5.0 ND. 10.0 N |
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Answer» Correct Answer - B `N=(w_(B)xx1000)/(E_(B)xxV)=(490xx1000)/(49xx1000)=10` `N_(1)V_(1)=N_(2)V_(2)` `10xx1=N_(2)xx10` `N_(2)=1`] |
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| 1881. |
The density (in g `mL^(-1)`) of 3.60 M sulphuric acid solution this is 29% `H_(2)SO_(4)` (molar mass =98 g `mol^(-1)`) by mass will be :A. 1.45B. 1.64C. 1.88D. 1.22 |
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Answer» Correct Answer - d By definition, `3.6 M (3xx98=352.8 g) of H_(2)SO_(4)` are present in 1000 mL of solution. 29.0 g of acid are present in solution=100 g 325.8 of acid are present in soluion `((100g))/((29.0 g))xx(352.8 g)` =1216 g `"Density of solution"=("Mass of solution")/("Volume of solution")` `((1216g))/((1000mL))=`1.22 g `mol^(-1)` |
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| 1882. |
The density `("in" g mL^(-1))` of a `3.60 M` sulphuric acid solution that is `29% H_(2)SO_(4)` (Molar mass `=98 g mol^(-1)`) by mass will be:A. `1.22`B. `1.45`C. `1.64`D. `1.88` |
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Answer» Correct Answer - 1 `Density _("soln")=(mass_("soln"))/(volume_("soln"))` `3.60 M` sulphuric acid solution means that 3.6 mole `H_(2)SO_(4)` is dissolved in every `1 L (1000 mL)` of solution. Thus `mass_(H_(2)SO_(4))=(n_(H_(2)SO_(4)))("moles mass"_(H_(2)SO_(4)))` `=(3.6 mol)(98 g mol^(-1))` `=352.8 g` By definition `mass % H_(2)SO_(4)=(mass_(H_(2)SO_(4)))/(mass_(solution))xx100%` Thus `mass_(soln)=(mass H_(2)SO_(4))/(mass % H_(2)SO_(4))xx100%` `=(352.8 g)/(29%)xx100%` `=1216 g` Substituting this result, we get `"density"_(soln)=((1216 g)/(1000 mL))/("Density of soln")` `=1.216 g mL^(-1)` `=1.22 g mL^(-1)` |
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| 1883. |
The density `("in" g mL^(-1))` of a `3.60 M` sulphuric acid solution that is `29% H_(2)SO_(4)` (Molar mass `=98 g mol^(-1)`) by mass will be:A. 1.45B. 1.64C. 1.88D. 1.22 |
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Answer» Correct Answer - D `M=(x xxd xx10)/(m_(B))` `d=(Mxxm_(B))/(x xx10)=(3.6xx98)/(29xx10)~~1.22"g mL"^(-1)`] |
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| 1884. |
Which of the following aqueous solution shas osmotic presseure nearest to that of an equimolar solution of `K_(4)[Fe(CN)_(6)]`A. `Na_(2)SO_(4)`B. `BaCl_(2)`C. `Al_(2)(SO_4)_(3)`D. `C_(12)H_(22)O_(11)` |
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Answer» Correct Answer - C We know that the value of collgative property is directly proportional to no. of particles of solute .Since`K_(4)[Fe(CN)_(6)]`and `Al_(2)(SO_(4))_(3)`both from `5` ion on dissociation .Hence osmotic pressure of `Al_(2)(SO_(4))_(3)`solution will be nearest to that of equimolar solution of `K_(4)[Fe(CN)_(6)]` |
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| 1885. |
20 mL of a solution os sulphuric acid neuitralise 21.2 mL of 30% soution of sodium carbonete. How much water should be added to 100 mL of this solutio to bring down its down its strength to decinoral? |
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Answer» Calculation of normality of suplhric acid solution strength of `Na_(2)CO_(3)` solution=`130g//100mL =300gL^(-1)` ` "Normality of" Na_(2)CO_(3) "solution"=("Strength")/("Equivalent mass")= (300gL^(-1))/(53g equiv^(-1))` `=5.66"equiv." L^(-1)=5.66 N`. `N_(1)xxV_(1)=N_(2)xxV_(2)or N_(1)=(N_(2)xxV_(2))/V_(1)` `(H_(2)SO_(4))=(Na_(2)CO_(3))` `N_(1)=((5.66N)xx(21.2mL))/((20.0mL))=6.0N`. Calculation of volume of water to be added Normality of `H_(2)SO_(4)` solution=6.0N volume of `H_(2)SO_(4)` solution=100mL Normality of diluted `H_(2)SO_(4)` solution=0.1N Volume of diluted `H_(2)SO_(4)` solution may be calculated as: `"underset"(("diluted" H_(2)SO_(4)))(N_(1)xxV_(1))=underset(("Concentreated"H_(2)SO_(4)))(N_(2) xx V_(2))` `V_(1)=(N_(2)xxV_(2))/N_(1)=((6.0N)xx(100mL))/((0.1N))=6000mL` Volume of water to be added =(6000-100)=5900mL |
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| 1886. |
The freezing point of a `0.05` molal solution of a non-electrolyte in water is: (`K_(f) = 1.86 "molality"^(-1)`)A. `-1.86^(@)C`B. `-0.93^(@)C`C. `-0.093^(@)C`D. `0.93^(@)C` |
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Answer» Correct Answer - C As we know ,the depression in freezing point is `DeltaT_(f)=K_(f)xx i xx "molarity"` From question,molal concentration ` = C _(m)=0.05m` for, non-electrolytes,`i=1 K_(f)=1.86` `:. DeltaT_(f)=1.86xx1xx0.05=0.0930^(@)C` hence, F.P.of solution `=0-0.093^(@)C=-0.093^(@)C` |
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| 1887. |
The vapour pressure of two pure liquids, A and B that form an ideal solution are 300 and 800 torr respectively, at temperature T. A mixture of the vapour of A and B for which the mole fraction of A is 0.25 is slowly compressed at temperature T, Calculate (a) the composition of the first drop of the condesate, (b) the total pressure when this drop is formed, (c) the composition of the solution whose normal boiling point is T, (d) the pressure when only the last bubble of vapour remains, and (e) the composition of the last bubble. |
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Answer» Correct Answer - (a)0.47, (b) 565 torr, (c)`"X"_("A")=0.08," X"_("B")=0.92,` (d)`"X"_("A")^(1)=0.11," X"_("B")^(1)=0.89` (a) `"P"_("A")="X"_("A")xx"P"_("A")^(0)` `"P"_("T")="X"_("A")("P"_("A")^(0)-"P"_("B")^(0))+"P"_("B")^(0)` `564.7="X"_("A")(300-800)+800` `"X"_("A")=0.47` (b) `("Y"_("A"))/("P"_("A")^(0))+("Y"_("B"))/("P"_("B")^(0))=1/"P"_("T")` `0.25/300+0.75/800=1/"P"_("T")` `"P"_("T")=564.7 " torr"` (c) At boiling temp vapour pressure=760 torr `300"X"_("A")+800"X"_("B")=760` `"X"_("A")(300-800)+800=760` `"X"_("A")=0.08` `"X"_("B")=0.92` (d) When only the last bubble of vapour remains, we can assme that the composition of vapour is now the composition of the codensation `therefore " X"_("A")=0.25, " X"_("B")=0.75` ` "P"_("T")= "X"_("A")"P"_("A")^(0)+"X"_("B")"P"_("B")^(0)` `=0.25xx300+0.75xx800` `=675 "torr"` (e) Composition of last bubble`="X"_("A")` `"X"_("A")="Y"_("A")` `"X"_("A")="P"_("A")/"P"_("T")` `"X"_("A")=75/675=0.11," X"_("B")=1-0.11=0.89` |
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| 1888. |
The vapour pressures of two pure liquids `A` and `B` that form an ideal solution are `300` and 800 torr, respectively, at tempertature `T`. Calculate a. The composition of the first drop of the condensate. b.The total pressure when this drop is formed. c. The composition of the solution whose normal boiling point is `T`. d. The pressure when only the last bubble of vapour remains. e. Composition of the last bubble. |
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Answer» Given `p_(A)@ = 300` torr, `chi_(A)^(l) = 0.25, chi_(B)^(l) = 1 - 0.25 = 0.75` a. By the condensation of only one drop, we can assume that the composition of the vapour remanns the same. `chi_(A)^(V) = (p_(A)@chi_(A)^(l))/p_(T)` and `chi_(B)^(V) = (p_(B)@chi_(B)^(l))/p_(T)` or `chi_(A)^(V)/ chi_(B)^(V) = p_(A)@chi_(A)/p_(B)@(1-chi_(A))` Putting various known values, we get `chi_(A)^(V) = 0.111` and `chi_(B)^(V) = 0.888` b. `p=p_(A)@chi_(A)^(V) + p=p_(B)@chi_(B)^(V)` = `300 xx 0.11+ 800 xx 0.888` `= 743.7` c. `760 = 300chi_(A) + 800chi_(B)` `chi_(A) = 0.08` and `chi_(B) = 0.92` d. When only the last bubble of vapour remains, we can assume the composition of vapour is now the composition of the condensate.n Hence, `P= (0.25 xx 300) + 0.75 xx 800 = 675` torr e. Composition of last bubble `(chi_(A) = p_(A)@chi(A))/P = (0.25 xx 300 )/675 = 0.11` `chi_(B) = 0.89` |
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| 1889. |
A mixture of two liquids A and B have vapour pressures 3.4 × 104 Nm-2 and 5.2 × 10 Nm-2. If the mole fractions of A is 0.85, find the vapour pressure of the solution. |
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Answer» Given : Vapour pressure of pure liquid A = \(P_A^0\) = 3.4 × 104 Nm-2 Vapour pressure of pure liquid B Mole fraction of A = xA = 0.85 Mole fraction of B = xB = 1 – xA = 1 – 0.85 = 0.15 The vapour solution is given by, Psoln = XA\(P_A^0\) + XB\(P_B^0\) = 0.85 × 3.4 × 10 + 0.15 × 5.2 × 104 = 2.89 × 10 + 0.78 × 104 = (2.89 + 0.78) × 104 Psoln = 3.67 × 104 Nm-2 Vapour pressure of a solution = 3.67 × 104 Nm-2 |
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| 1890. |
The freezing point of a `1.00 m` aqueous solution of `HF` is found to be `-1.91^(@)C`. The freezing point constant of water, `K_(f)`, is `1.86 K kg mol^(-1)`. The percentage dissociation of `HF` at this concentration is:-A. `2.7%`B. `30%`C. `10%`D. `5.2%` |
| Answer» Correct Answer - A | |
| 1891. |
The vapour pressures of pure liquids A and B are 450 mm and 700 mm of Hg respectivly at 360 K. Calculate the composition of the composition of the liqid mixture if the total vpour pressure is 600 mm of Hg. Also find the compositon of the mixture in the capour phase. |
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Answer» Calculation of the composition of liquid mixture For the two liquids A and B `P_(A)=P_(A)^(@)x_(A),P_(B)=P_(B)^(@)x_(B)=P_(B)^(@)(1-x_(A)),P=P_(A)+P_(B)` `600=450x_(A)+700(1-x_(A))` `600=450x_(A)+700-700x_(A)` `250x_(A)=100 or x_(A)=100/250=0.4.` `x_(B)=1-x_(A)=1-0.4=0.6` Calculation of the mixture in the capour phase `P_(A)=0.4xx450mm=180mm,P_(B)=0.6xx700mm=420 mm`. Mole fraction in the vapour phase may be calculated as : `x_(a)=((180mm))/((600mm))=0.3,x_(B)=((420mm))/((600mm))=0.7` |
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| 1892. |
The vapour pressures of pure liquids A and B are 400 mm Hg and 650 mm Hg respectively at 330 K. Find the composition of liquid and vapour if total vapour pressure of solution is 600 mm Hg. |
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Answer» Given : \(P_A^0\)= 400 mm Hg; \(P_B^0\) = 650 mm Hg, PT = 600 mm Hg, T = 330 K; xA = ? xB = ?, y1 = ? y2 = ? (x is mole fraction in liquid phase while y is mole fraction in vapour phase.) PT = (\(P_A^0\) - \(P_B^0\))xB + \(P_A^0\) 600 = (650 – 400)xB + 400 = 250xB + 400 ∴ xB = \(\frac{600-400}{250}\) = 0.8 ∵ xA + xB = 1 ∴ xA = 1 – xB = 1 – 0.8 = 0.2 The composition of A and B in liquid mixture is, xA = 0.2 and xB = 0.8. If P1 and P2 are vapour pressures (or partial pressures) of A and B in vapour phase then by Raoult’s law, P1 = xA × \(P_A^0\) = 0.2 × 400 = 80 mm Hg P2 = xA ×\(P_B^0\) = 0.8 × 650 = 520 mm Hg If y1 and y2 are mole fractions of A and B respectively in vapour phase then, By Dalton’s law, P1 = y1PT ∴ y1 = \(\frac{P_1}{P_T}\) = \(\frac{80}{600}\) = 0.1333 P2 = y2PT ∴ y2 = \(\frac{P_2}{P_T}\) = \(\frac{520}{600}\) = 0.8667 (or y2 = 1 – y1 = 1 – 0.1333 = 0.8667) ∴ Composition of liquid : xA = 0.2 and xB = 0.8 ∴ Composition of vapour : yA = 0.1333 and yB = 0.8667 |
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| 1893. |
On mixing , heptate and octane form an ideal solution. At 373 K. The vapour pressure of the two liquid components (heptaane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solution obtainde by mixing 25.0 g of heptane and 35 g of octane will be (molar mass of heptane=100 g `mol^(-1)` )A. 0.0186 KB. 0.0372 KC. 0.0558 KD. 0.0744 K |
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Answer» Correct Answer - c Fro `NaSO4, i=3` Since `alpha` is eaual to 1 `"Molality (m)"=((0.01mol))/(1kg)` `DeltaT_(f)=ixxK_(f)xxm` `=3xx(1.86 " K kg"mol^(-1)) xx (0.01 mol kg^(-1))` =0.0558 K. |
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| 1894. |
Two liquids `A` and `B` form an ideal solution. At `300 K`, the vapour pressure of a solution containing `1 mol` of `A` and `3 mol` fo `B` is `550 mm Hg`. At the same temperature, if `1 mol`more of `B` is added to this solution, the vapour pressure of the solution increases by `10 mm Hg`. Determine the vapour pressure of `A` and `B` in their pure states.A. `450, 150 mm Hg`B. `250, 300 mm Hg`C. `125, 150 mm Hg`D. `150, 450 mm Hg` |
| Answer» Correct Answer - A | |
| 1895. |
Calculate the biling point of a solution containing 0.456 g of comphor (molar mass=152 g mol^(-1)) dossoved in 31.4 g of acetone(b.p.= 329.45 K). Given that the molecular elevation constant per 100 g of solvent is 17.2 K. |
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Answer» Calculation of elevaltion in boiling point `(T_(b))`. `DeltaT_(b)=(K_(b)xxW_(B))/(M_(B)xxW_(A))` Mass of camphor `(W_(B))=0456 g` Mass of acetone `(W_(A))`=31.4 g= 0.0314 kg `" Molal elevation contant "(K_(b))=(17.2K per 100 g)/1000=1.72 "K kg mol"^(-1)` `" Molar mass of camphor "(M_(B))=152 g mol^(-1))` `DeltaT_(b)=((1.72 K kg mol^(-1))xx(0.456 g))/((152 g mol^(-1))xx(0.0314 kg))=0.164 K` Calculation of boiling point of solution. Boiling point of acetone= 329.45 K Elevation in b.p tempertaure=0.164 K Boiling point of solution=329.45K+0.164 K= 329.614 K |
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| 1896. |
Two liquids X and Y form an ideal solution at 300 K. Vapour pressure of the solution containing 3 moles of X and 1 mole of Y is 550 mm Hg. At the same temperature, if 1 mole of Y is further added, to this solution, the vapour 1 mole of Y is further added, to this solution, the vapour pressure of solution increases by 10 mm Hg. Vapour pressure (in mm g) of X and Y in their pure states will be repectivelyA. 200and 300B. 300 and 400C. 600 and 400D. 500 and 600 |
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Answer» Correct Answer - c For detailed solution, consult solved example 2.37. |
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| 1897. |
Two liquids A and B form ideal solution. At `300 K`, the vapour pressure of a solution containing 1 mole of A and 3 moles of B is `550 mm` of Hg. At the same temperature, if one more mole of B is added to this solution, the vapour pressure of the solution increases by `10 mm` of Hg. Determine the vapour pressure of a and B in their pure states.A. `p_A^@`=600 mm Hg and `p_B^@`=400 mm HgB. `p_A^@`=550 mm Hg and `p_B^@`=560 mm HgC. `p_A^@`=450 mm Hg and `p_B^@`=650 mm HgD. `p_A^@`=400 mm Hg and `p_B^@`=600 mm Hg |
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Answer» Correct Answer - D Vapour pressure of solution containing 1 mole of A + 3 moles of B =550 mm Hg Vapour pressure of solution containing 1 mole of A + 4 moles of B = (550+10) = 560 mm Hg `P_"Total"=p_A^(@) xx x_A + p_B^(@) xx x_B` or 550 =`p_A^(@) xx x_A + p_B^(@) xx x_B` `=p_A^@ xx 1/4 + p_B^@xx3/4[because x_A=1/(1+3)=1/4, x_B=3/(1+3)=3/4]` `550=p_A^@/4+3/4xxp_B^@` or 2200=`p_A^@+3p_B^@`...(i) Again , we have 560=`p_A^@xx1/5+p_B^@xx4/5 (because x_A=1/(1+4)=1/5, x_B=4/(1+4)=4/5)` 2800=`p_A^@+4p_B^@`...(ii) Solving equations (i) and (ii) , we get `p_B^@`=600 mm Hg , `p_A^@`=400 mm Hg |
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| 1898. |
Mole fraction of ` C_(3)H_(5)(OH)_(3) `in a solution of ` 36 g ` of water and ` 46 g `of glycerine is :A. ` 0.46 `B. ` 0.36 `C. ` 0.20 `D. ` 0.40 ` |
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Answer» Correct Answer - C Mole of ` H_(2)O = (36)/(18)=2` Mol e of glycerine =` (46)/(92) =0.5 ` total Mo le =` 2=0.5 =2.5 ` Mole fraction of glycerine = ` (n_(1))/(n_(1)+n_(2)) =(0.5)/(2.5)X_(0)=0.2 ` |
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| 1899. |
Relative decrease in vapour pressure of an aqueous `NaCl` is `0.167`. Number of moles of `NaCl` present in `180g` of `H_(2)O` is:A. 2 molB. 1 molC. 3 molD. 4 mol |
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Answer» Correct Answer - B RLVP=i(n)/(n+V)` so `0.167=(2xxn)/(n+(180)/(18))` so `n=1` |
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| 1900. |
A solution containing 7g of a solute (molar mass `210 "g mol"^(-1)`) in 350g of acetone raised the boiling point of acetone from `56^(@)C` to `56.3^(@)C`. The value of ebullioscope contant of acetone in `"K kg bol"^(-1)` is :A. 2.66B. 3.15C. 4.12D. 2.86 |
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Answer» Correct Answer - B `DeltaT=K_(b)xx(w_(B)xx1000)/(m_(B)xxw_(A))` `0.3=K_(b)xx(7xx1000)/(210xx350)` `K_(b)=(0.3xx210xx350)/(7xx1000)=3.15" K kg mol"^(-1)`] |
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