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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.
| 1701. |
What is the expected Van’t Hoff’s factor ‘i’ value for K3[Fe(CN)6] ? |
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Answer» Therefore, i = 4 |
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| 1702. |
What are colligative properties? Why are they also called democratic properties? Is osmotic pressure a colligative property? Prove it. |
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Answer» These are the properties of solution that depend on the number of solute particles in the solution not at all on the nature of solute particle. Above mentioned is also the reason as why are they sometime called as democratic properties. Yes, osmotic pressure is a colligative property because osmotic pressure of a solution is proportional to its molarity at a given temperature. |
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| 1703. |
Assertion : Out of various colligative properties, only osmotic pressure is used to determine the molecular masses of polymers. Reason : Ploymer solutions donot possiss a contant boiling point of freezing point.A. If both assertion are reason are correct and reason is correct explanation for assertion.B. If both assertion and reason are corrct but reason is not correct explanation for assertion.C. If assertion is corrct but reason is incorrect.D. If assertion and reason both are incrrect. |
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Answer» Correct Answer - c Correct Reason. Ploymer solutions have very small elevation in boiling point and depression in freezing point. |
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| 1704. |
Which of the following is not a colligative property? (a) optical activity (b) osmotic pressure (c) elevation boiling point (d) depression in freezing point |
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Answer» (a) optical activity |
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| 1705. |
The molar mass of bio molecules is determined by osmotic pressure and not by other colligative properties. Why? |
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Answer» Osmotic pressure is measured around room temperature whereas bio molecules are generally unstable at higher temperatures. |
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| 1706. |
Assertion: Osmotic pressure is a colligative property. Reason: Osmotic pressure of a solution at any temperature T depends on the molar concentration.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - A | |
| 1707. |
Assertion: Isotonic solutions do not show phenomenon of osmosis. Reason: Isotonic solutions have equal osmotic pressure.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - B | |
| 1708. |
A solution which has lower osmotic pressure compared to that of other solution is called …….. . |
| Answer» Correct Answer - C | |
| 1709. |
A solution of benzene and toluene is an example of ………….solution. |
| Answer» Correct Answer - A::D | |
| 1710. |
In cold countries …………is used as an anti-freezing agent in the radiator of car. |
| Answer» Correct Answer - C | |
| 1711. |
Mole fraction of the solute in a 1 molal aqueous solution is :A. 1.77B. 0.177C. 0.0177D. 0.0344 |
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Answer» Correct Answer - C `m=(x_(B)xx1000)/((1-x_(B))xxm_(A))` `1=(x_(B)xx1000)/((1-x_(B))xx18)` `x_(B)=(0.018)/(1.018)=0.0177`] |
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| 1712. |
The mole fraction of solute in one molal aqueous solution is ………… . |
| Answer» Correct Answer - A | |
| 1713. |
The mole fraction of the solute in one molal aqueous solution is:A. `0.027`B. `0.036`C. `0.018`D. `0.009` |
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Answer» Correct Answer - C Molality of solution = mole of solute per kg of solvent So, 1m = mole of solute per 1000 gof solvent Hence, moles of solute in 1m aqueous solution = 1 moles of solvent 1 m aqueous solution `=(1000)/(18)=55.55` Mole fraction of solution in 1 m solution `=(1)/(1+55.55)=(1)/(56.55)` `=0.0176~~0.018` |
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| 1714. |
The vapor pressure of pure benzene at a certain temperature is `640 mmHg`. A nonvolatile nonelectrolyte solid weighing `2.0 g` is added to `39 g` of benzene. The vapor pressure of the solution is `600 mmHg`. What is the molecular mass of the solid substance. Strategy: Use Equation `(2.58)` as the concept of colligative properties is based on dilute solution. The moleculsr mass of benzene is `78u`. |
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Answer» We are given `P_(1)^(0)=640 mmHg, P=600 mmHg` `w_(1)=39 g, w_(2)=2.0 g` `M_(1)=78 g mol^(-1), M_(2)=M` where M is the molar mass of solute, According to Equation `(2.58)`, we have `(P_(1)^(0)_P)/P_(1)^(0)=(w_(2)xxM_(1))/(w_(1)xxM_(1))` Thus `((640-600) mmHg)/(640 mmHg)=((2.0 g)(78 g mol^(-1)))/((39 g)(M g mol^(-1)))` `0.0625=4/M` or `M=64 g mol^(-1)` Thus, molecular mass of nonvolatile solute is `64 u`. |
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| 1715. |
What care is generally taken during intravenous Injections and why? |
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Answer» During intravenous injection, the concentration of the solution should be same as that of blood so that they are isotonic. Because if the solution concentration is hypertonic than blood cell will shrink and if it is hypotonic than blood cell will swells / burst. |
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| 1716. |
Heptane and octane form ideal solution. At `373K`, the vapour pressure of the two liquids are `105.2 kPa` and `46.8 kPa` respectively. What will be the vapour pressure, in bar, of a mixture of `25g` of heptane and `35g` of octane? |
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Answer» (A) Heptane `{:((A) "Heptane" C_(7)H_(16),m_(A) = 100),((B) "Octane" C_(8)H_(18),m_(B) =114),(n_(A)=(w_(A))/(m_(A))=(25)/(100) =0.25,,n_(B) = (35)/(114) = 0.3),(x_(A) =(0.25)/(0.25 xx 0.30) =0.45,x_(B) =(0.3)/(0.25+0.30) =0.55),(=0.45,):}` `p = p_(A)^(0) x_(A) +p_(B)^(0)x_(B)` `= 105.2 xx 0.45 +46.8 xx 0.55` `= 47.34 +25.74 = 73.08 kPa` |
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| 1717. |
Which of the following mode of concentration is affected by temperature? Molarity, Molality, mole fraction and Normality. |
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Answer» Molarity and Normality |
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| 1718. |
Assertion: The sum of mol fractions of all the components of a solution is unity. Reason: Mole fraction is temperature dependent mode of concentrations.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - C | |
| 1719. |
Which of the following concentration factors is affected by changing in volume or volume dependent ?A. weight fractionB. Mole fractionC. MolalityD. Molarity |
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Answer» Correct Answer - D On changing temperature, solvent `(H_(2)O) ` is vaporized or condensed. Thus, volume of solution and solvent changes. Thus, any concentration factor dependent on volume is affected by change in temperature. (a) Weight fraction `= w_(i)//w_("total")` (Unitless) (b) Mole fraction `= n_(i)//n_("total")` (unitless) (c) Molality `= ("moles of solute")/("kg of solvent")` (weight dependent ) (d) Molarity `=("Mole")/("Litre")` (volume dependent) |
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| 1720. |
Calculate the masses of cane sugar and water required to prepare 250g of 20% can sugar solution. |
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Answer» Mass percentage of can sugar = 25 We know that, Mass percentage `= ("Mass of solute")/("Mass of solution")xx100` So, `25=("Mass of cane sugar")/(250)xx100` or Mass of cane `= (25xx250)/(100)=62.5g` Mass of water `=(250-62.5)=187.5g`. |
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| 1721. |
Calculates the masses of cane sugar and water required to prepare ` 250` grams of `25%` cane sugar solution-A. ` 187.5 ` grams ,` 62.5 ` gramsB. ` 62.5 `grams,` 187.5 `gramsC. ` 162.5 `grams ,` 87.5 ` gramsD. None of these |
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Answer» Correct Answer - B Mass percent of sugar = `25` we know that Mass percentage `= ("Mass of solute")/("Mass of solution")xx100 ` So, `25 = ("Mass of cane sugar")/(250)xx100` or Mass of cane sugar =` (25xx250)/(100)= 62.5 ` grams. Mass of water =` (250-62.5)= 187.5 `grams |
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| 1722. |
A solution has been prepared by dissolving 5 g of urea in 95 g of water. What is the mass percent of urea in the solution ? |
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Answer» Mass precent `(Mass %) =(Mass of solute)/(Mass os solution)xx100` M of urea = fg, Mass of water=95g Mass percent of urea=`((5g))/((5g+95g))xx100=5%` |
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| 1723. |
What is the molarity of a solution containing 10 g of NaOH in 500 mL of solution ?A. `"0.25 mol L"^(-1)`B. `"0.75 mol L"^(-1)`C. `"0.5 mol L"^(-1)`D. `"1.25 mol L"^(-1)` |
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Answer» Correct Answer - C No. of moles of NaOH=`10/40`=0.25 mol `M=(0.25xx1000)/500=0.5 mol L^(-1)` |
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| 1724. |
Calculate molality of 2.5 of ethanoic acid `(CH_(3)COOH)` in 75g of benzene. |
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Answer» Moles of `C_(2)H_(4)O_(2):12xx2+1xx4+16xx2=60"g mol"^(-1)` Moles of `C_(2)H_(4)O_(2)=(2.5" g")/(60" g mol"^(-1))=0.0417" mol"` Mass of benzene is kg `=75" g"//1000" g kg"^(-1)=75xx10^(-3)" kg"` Molality of `C_(2)H_(4)O_(2)=("Moles of "C_(2)H_(4)O_(2))/("kg of benzene")=(0.0417xx" mol "xx1000" g kg"^(-1))/(75"g")` `=0.556" mol kg"^(-1)` |
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| 1725. |
Calculates the masses of cane sugar and water required to prepare ` 250` grams of `25%` cane sugar solution- |
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Answer» Mass precent=`("Mass of solute")/("Mass of solution")xx100` Mass percent=25, Mass of solution=250g `25=("Mass of cane sugar")/((250g))xx100` Mass of cane sugar=`(25xx(250g))/100=62.5g` ltbtgt Mass of water= 250-62.5=187.5g |
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| 1726. |
One litre of sea water weighs 1050 grams and conrins `6xx10^(-3)` g pf dossolced dessolved oxygen gas. Calculate the concerntration of the dissolced exygen in ppm. |
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Answer» Mass of oxygen `(O_(2))=6xx10^(-3)g` Mass of sea water =1050g ppm of oxygen =`("Mass of oxygen")/("Mass of sea water")xx10^(6)=((6xx10^(-3)g))/((1050g))xx10^(6)=5.71" ppm"` |
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| 1727. |
Calculate the molarity of a solution containing `5 g` of NaOH in `450 mL` solution. |
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Answer» `M=(W_(2) xx 1000)/(Mw_(2) xx "volume of solution in mL") ` `(5g xx 1000)/(40g mol^(-1) xx 450 mL)` `=0.278 mol L^(-1)` `=0.278 mol dm^(-3)` |
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| 1728. |
Calculate molality of `2.5g` of ethanoic acid `(CH_(3)COOH)` in `75 g` of benzene. |
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Answer» Molality (m)`=("Mass of " CH_(3)COOH//"Molar mass of " CH_(3)COOH)/("Mass of bezene in kg")` Mass of `CH_(3)COOH =2.5g`. Molar mass of `CH_(3)COOH=2xx12+4xx1+2xx16=60" g mol"^(-1)`. `"Mass of solvent(benzene)"=75g=(75g)/(1000)=0.075kg`. Molality(M)`=((2.5)//(60.0mol^(-1)))/((0.075kg))=0.556" mol kg"^(-1)=0.556m` |
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| 1729. |
The density of 85% sulphuric accid is `1.7gcm^(-3)`. What is the volume of the volume of the solution which contains 17 g of sulphuric acid ? |
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Answer» `85g of H_(2SO_(2)` are present in the solution =100g 17 g of `H_(2)SO_(4)` are present in the solution`=((100g))/((85g))xx(17g)=20g` Thus, mass of solution= 20g , Density of solution =`1.7g cm ^(-3)` Volume of solution =`(Mass)/(Density)=(20g)/(1.7cm^(-3))=11.8cm^(3)` |
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| 1730. |
Calculate the molarity of a solution containng 5g of NaOH disslced in 450 mL of the solution. |
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Answer» Molarity(M)`=("Mass of" NaOH//"Molar mass of" NaOH)/("Volume of solution in Littres")` Mass of NaOH =5g,Molar mass of NaOH`=40gmol^(-1)`, Volume os solution`=450mL-540/1000=0.45L` Molarity (M)`=(5g//(40g mol^(-1)))/((0.45L))=0.278mol^(-1)=0.278M` |
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| 1731. |
What will be the mole fraction of ethanol in a sample of spirit containing 85% ethanol by mass ?A. `0.69`B. `0.82`C. `0.85`D. `0.60` |
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Answer» Correct Answer - A `x_(C_2H_5OH)=(n_(C_2H_5OH))/(n_(C_2H_5OH)+n_(H_2O))` Mass of `C_2H_5OH`=85 g Molar mass of `C_2H_5OH`=46 g/mol `n_(C_2H_5OH)=85/46`=1.85 mol Mass of water =100-85 =15 g `n_(H_2O)=15/18`=0.833 mol `x_(C_2H_5OH)=1.85/(1.85+0.833)=1.85/2.683`=0.69 |
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| 1732. |
A solution is 25% ethanol ans 50% acetic acid by mass. Calculate the mole fraction of ethanol and acetic acid in the solution. |
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Answer» Let us start with 100 g of the solutioon in which Mass of water =25g Mass of ethanol=25g Mass of acetic acid = 50g `n_(H_(2)O)=((25g))/((18gmol^(-1)))=1.388mol` `n_(C_(2)H_(5)OH)=((25g))/((46gmol^(-1)))=0.543 mol` `n_(CH_(3)COOH)=((50g))/((60gmol^(-1)))=0.833mol` Total number of moles =(1,388+0.543+0.833)=2.764 mole `x_(C_(2)COOH)" "=n_(C_(2)H_(5)OH)=((0.543 mol))/((2.764 mol))=0.196` `x_(CH_(3)COOH)" "n_(CH_(3)COOH)=((0.833 mol))/((2.764 mol))=0.301` |
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| 1733. |
Calculate the mole fractio of ethanol in a sample of spirit containing 92% ethanol by mass. |
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Answer» `x_(C_(2)H_(5)OH)=(n_(C_(2)H_(5)OH))/(n_(C_(2)H_(5)OH)+n_(H_(2)O)` Mass of `C_(2)H_(5)OH=92g,Molar mass of C_(2)H_(5)OH=46g mol^(-1)` `C_(2)H_(5)OH=((92g))/((46gmol^(-1))=2 mol)` Mass of `H_(2)O=8g, Molar mass of H_(2)O=189g mol ^(-1), n_(H_(2)O)=((8g))/((18gmol^(-1))=0.44 mol` `x_(C_(2)H_(5)OH)=((2 mol))/((2mol)+(0.44mol))=0.82` |
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| 1734. |
A solution of crab hemocyanin, a pigmented protein extracted from crabs,was prepared by dissolving 0.750 g in 125`"cm"^(3)` of an aqueous medium . At `4^(@)"C"` an osmotic pressure rise of 2.6 mm of the solution was observed. The solution had a density of 1.00`"g"//"m"^(3)`. Determine the molecular weight of the protein. |
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Answer» Correct Answer - `5.4xx10^(5)"g/mol"` Solute-B, Water-A `"h"_("Hg")"P"_("Hg")"g"=hxx"P"xx"g"` `"h"_("Hg")=("h"_("solution")xx"P"_("solution"))/("P"_("Hg"))=(2.6xx1)/(13.6)"mm of Hg"` `pi=(2.6)/(13.6xx760)"atm"` `"m"_("B")=("W"_("B")"RT")/(pixx"V"_("sol")"(L)")=(0.75xx0.082xx227)/((2.6)/(13.6xx760)xx(125)/(100))=(0.75xx0.082xx277xx13.6xx760xx1000)/(2.6xx125)` `=5.4xx10^(5)"gm mol"^(-1)` |
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| 1735. |
A solution was prepared by adding `125cm^(3)` of isopropyl alcohol to water until the volume of the solution was `175 cm^(3)`. Find the volume fraction and volume percent of isopropyl alcohol in the solution. |
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Answer» volume of solute `= 125 cm^(3)` volume of solution `= 175 cm^(3)` `:.` volume fraction `= (125)/(175) = 0.714` and volume percent `= (125)/(175) xx 100 = 71.4%` |
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| 1736. |
A solution is prepared by dissolving `1.08 g` of human serum albumin, a protein obtained from blood plasma, in `50 cm^(3)` of aqueous solution. The solution has an osmotic pressure of `5.85 mm Hg` at `298 K`. a. What is the molar mass of albumin ? b. What is the height of water column placed in solution ? `d_((H_(2)O)) =1 g cm^(-3)` |
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Answer» a. The molar massof albumin can be calculated using the following relation `Mw_(B)=(W_(B)xxRT)/(piV)` ….(i) Given,`W_(B)=1.08 g`,`R=0.0821 L atm K^(-1) mol^(-1)` `T=298 K , pi=(5.85)/(760) atm,V=(50)/(1000) = 0.005 L` Substituting these values in Eq.(i), we get `Mw_(B)=(1.08xx0.0821xx298)/((5.85//760)xx0.05)=68655 g mol^(-1)` b. `pi=hdg` `(5.85)/(760)xx101325=hxx1xx10^(-3)xx9.8` [1 atm = 101325 Pa] `:. h=7.958 xx10^(-2) m =7.958 cm` |
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| 1737. |
The freezing point depression of `0.001 m K_(x)` `[Fe(CN)_(6)]` is `7.10xx10^(-3) K`. Determine the value of x. Given, `K_(f)=1.86 K kg "mol"^(-1)` for water. |
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Answer» `Deltax=ixxK_(f)xxm` `7.10xx10^(-3)=ixx1.86xx0.001` `i=3.817` `alpha=(i-1)/(n-1)` `1=(3.817-1)/((x+1)-1)` `x=2.817~~3` `:.` Molecular formula of the compound is `K_(3)[Fe(CN)_(6)]`. |
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| 1738. |
A solution of crab haemovyanin, a pigmented protein extracted from crabs,was prepared by dissolving `0.750g "in" 125cm^(3)` of aqueous medium.AT `4^(@)C` an osmotic pressure rise of `2.6`mm of the solution was observed. The solution has a density of `1.00g//cm^(3)`.Determine the molecular weight of the protein.A. `53.14xx10^(5)g//mol`B. `52.14xx10^(5)g//mol`C. `51.14xx10^(5)g//mol`D. `54.14xx10^(5)g//mol` |
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Answer» Correct Answer - A `((0.75/m))/(125)xx(8.314xx10^(5)xx10^(2))(227)=2.6xx10^(-2)(1000)` `m=((0.75)(8.314xx10^(7))(277))/(125xx26)` `m=53.14xx10^(5)g//mol` |
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| 1739. |
A`0.2` molal aqueous solution of a week acid (HX) is 20 % ionised,The frrezing point of this solution is (Given `k_(f)=1.86^(@)Ckg mol^(-1)` for water) :A. `-0.45^(@)C`B. `-0.90^(@)C`C. `-0.31^(@)C`D. `-0.53^(@)C` |
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Answer» Correct Answer - A `DeltaT_(f)=0.2xx1.2xx1.86=0.45` therefore freezing point=`-0.45^(@)C` |
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| 1740. |
Phenol associates in benzene to certain extent to form a dimer. A solution containing `20 xx 10^(-3) kg` of phenol in `1.0 kg` of benzene hs its freezing point depressed by `0.69 K`. Calculate the fraction of phenol that has dimerized. (`K_(f)` for benzene is `5.12 K kg mol^(-1)`).A. `746.24`B. `646.24`C. `846.24`D. `546.23` |
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Answer» Correct Answer - A `a=0.7` `n_(r)=(1+2alpha)n_(1)+n_(2)` `=2.4n_(1)+n_(2)` `n_(1)=(7)/(164)` `n_(2)=(100)/(18)` `P_(s)=(760)(100/18)/(2.4xx(7)/(164)+(100)/(8))` `P_(s)=746.24mmHg` |
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| 1741. |
Phenol associates in benzene to certain extent to form a dimer. A solution containing `20 xx 10^(-3) kg` of phenol in `1.0 kg` of benzene hs its freezing point depressed by `0.69 K`. Calculate the fraction of phenol that has dimerized. (`K_(f)` for benzene is `5.12 K kg mol^(-1)`). |
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Answer» Onserved mol. Mass `= (1000xxK_(f)xxw)/(WxxDeltaT)` `=(1000xx5.13xx20xx10^(-3))/(1xx0.69)=1.48.4` Normal mol. Mass of phenol `(C_(6)H_(5)OH)=94` So, `("Normal mol. mass")/("Observed mol. mass")=(94)/(148.4)` `=1+((1)/(n)-1)alpha=1+((1)/(2)-1)alpha` `(94)/(148.4)=1-(alpha)/(2)` or `alpha=0.733 or73.3%`. |
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| 1742. |
To `500 cm^(3)` of water, `3 xx 10^(-3) kg` of acetic acid is added. If `23%` of axcetic acid is dissociated wha will be the depression in freezing point? `K_(f)` and density of water are `1.86 K kg^(-1) mol^(-1)` and `0.997 g cm^(-3)` respectively. |
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Answer» Correct Answer - 0.229 `W_("water")=500xx0.997"gm"=498.5"gm"` for dissociation `DeltaT_("f")=[1+(n-1)alpha]K_(f)xx(W_("solute")xx1000)/(m_("solute")xxW_("solvent"))=[1+(n-1)alpha]xx(1.86xx3xx1000)/(60xx498.5)` `DeltaT_(f)=0.229` |
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| 1743. |
Phenol associates in benzene to a certain extent to form a dimer. A solution containing `20xx10^(-1)` kg phenol in 1 kg of benzene has its freezing pint depressed by 0.69 K. Calculate the fraction of phenol that has dimerised. `"K"_("f")" for benzene"=5.12" kg mol"^(-1)"k"`. |
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Answer» Correct Answer - `a=0.7333` `Delta"T"_("f")=[1+(1/("n")-1)alpha]"K"_("f")("W"_("solute")xx1000)/("m"_("solute")xx"W"_("solvent"))` `0.69=[1+(1/("2")-1)alpha](5.12xx20xx1000)/(94xx1000)` `alpha=0.7333` |
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| 1744. |
Write two differences between ideal and non-ideal solutions. |
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Answer» The solutions which obey Raoult‘s law over the entire range of concentration are known as ideal solutions. ΔHmix = 0, ΔVmix = 0, AA, BB and AB interactions are equal. When a solution does not obey Raoult‘s law over the entire range of concentration, then it is called non-ideal solution. ΔHmix and ΔVmix are not equal to 0, AA, BB and AB interactions are not equal. |
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| 1745. |
Calculate the osmotic pressure of 5% solution of urea at 273 K. |
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Answer» Correct Answer - 18.67 atm `pi=(W_(B)xxRxxT)/(M_(B)xxV)=((5g)xx(0.0821"L atm K"^(-1)mol^(-1))xx(273K))/((60"g mol"^(-1))xx(0.1L))` = 18.68 atm. |
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| 1746. |
If the freezing point of a 0.01 molal aqueous solution of a cobalt (iii) chloride ammonia complex (behaves as a strong electrolyte) is `-0.0558^(@)C`, What is the number of chloride (s) in the co-ordeination sphere of the complex `(K_(f)"of water "1.86" K kg mol"^(-1))`. |
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Answer» `DeltaT_(f)=ixxK_(f)xxm` `i=(DeltaT_(f))/(K_(f)xxm)=((0.0558K))/((1.86 Km^(-1))xx(0.01m))=3` The value of I indicates that in aqueous solution, the complex dissociates to give ions. `[Co(NH_(3))_(5)CI]CI_(2)overset(("aq"))to[Co(NH_(3))_(5)CI]^(+)2CI^(-)` This means that one chlorine atom is present in the cooedination sphere. For more details, consult Unit-9 (Co-ordination compounds). |
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| 1747. |
If the freezing point of a `0.01` molal aqueous solution of a cobalt (III) chloride-ammonia complex (which behaves as a strong electrolyte) is `-0.0558^(@)C`, the number of chloride (s) in the coordination sphere of the complex if `[K_(f)` of water `=1.86 K kg mol^(-1)]` |
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Answer» Correct Answer - 1 `DeltaT_(f) = iK_(f)m` `0.0558 = ixx 1.86 xx 0.01 rArr i=3` Structure should be: `[Co(NH_(3))_(5)CI] Ci_(2)` |
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| 1748. |
One volume precent measurement that appears frequently in everyday life is proof, which is equal to the volume percent of enthanol in water.A. thriceB. halfC. twiceD. one-third |
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Answer» Correct Answer - 3 Thus, a wine that is 12% alcohol by volume is 24 proof. A run that is 150 proof contains 75% alcohol by volume. A 3.2 % bear is 6.4 proof. |
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| 1749. |
A solution has been prepared by dissolving 60 g of methyI alcogol in 120 g of water. What are the mole freaction of methyI alcohol and water ? |
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Answer» Correct Answer - 0.220 0.780 `"Moles of"CH_(3)OH=("Mass of" CH_(3)OH)/("Molar mass")=((60g)/(32 g mol^(-1)))=1.875` mol `"Mass of water"=("Mass of water")/("Molar mass")=((120g)/(18 g mol^(-1)))=6.667` mol `"Mole fraction of" CH_(2)OH=((1.875"mol"))/((1.874 "mol"+6.667"mol"))=((1.875"mol"))/((8.542"mol"))=0.220`. Mole fraction of water = 1-0.220 = 0.780. |
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| 1750. |
The concentration of solutions is sometimes expressed in terms of percentage of mass or volume precent in total mass or volume of the solution. When nothing is mentioned, it satnds forA. mass percentB. volume percentC. mass by volume percentD. volume by mass percent |
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Answer» Correct Answer - 1 In case of a solid dissolved in a liquid, the percentage by mass means the mass of the solute in grams present in 100 g of the solution, i.e., it is `weight//weight (w//w)` and percentage by volume means the mass of the solute dissolved in `100 mL` or `100 cm^(3)` of the solution i.e., it is `weight//volume (w//v)`. In case of a liquid dissolved in another liquid, percentage by mass has the same meaning but percentage by volume means the volume of the liquid solute in `cm^(3)` present in `100 cm^(3)` of the solution, i.e., it is `volume//volume (w//v)`. |
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