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Shima will mix 1 fluid ounce of fertiliz...

Shima will mix 1 fluid ounce of fertilizer in water for every 40 square feet of soil. At this rate, which of the following expressions gives the number of gallons of fertilizer that Shima will mix in water for 0.5 acres of soil?
(Note : 1 acre = 43,560 square feet, 1 gallon = 128 fluid ounces)

A

`(0.5(40)(128))/(43,560)`

B

`(40(128))/(0.5(43,560))`

C

`(0.5(43,560))/(40(128))`

D

`(43,560)/(0.5(40)(128))`

Text Solution

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The correct Answer is:
C
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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 :

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Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. one of its examples is the use of ethylene glycol adn water mixtures 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 kg mol^(-1) Freezing point depression constant of ethanol (K_(f)^("ethanol")) = 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)^("ethanol")) = 1.2 K kg mol^(-1) Standard freezing point of water = 273 K Standard freezing point of ethanol = 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 ethanol = 40 mm Hg Molecular weight of water =18 g mol^(-1) Molecular weight of ethanol = 46 g mol^(-1) In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. The vapour pressure of the solution M is

The main application of osmotic pressure measurement is in the determination of the molar mass of a substance which is either slightly soluble or has a very high molar mass such as proteins, polymers of various types and colloids.This is due to the fact that even a very small concentraion of the solution produces fairly large magnitude of osomotic pressure.In the laboratory the concentrations usually employed are of the order of 10^(-3) to 10^(-4) M.At concentration of 10^(-3) mol dm^(-3) , the magnitude of osmotic pressure of 300 K is : P=10^(-3)xx0.082xx300=0.0246 atm or 0.0246xx1.01325xx10^5=2492.595 Pa At this concentration, the values of other colligative properties such as boiling point elevation and depression in freezing point are too small to be determined experimentally. Further polymers have following two types of molar masses : (A) Number average molar mass (barM_n) , which is given by (undersetisumN_iM_i)/(undersetisumN_i) where N_i is the number of molecules having molar mass M_i . (B) Molar average molar mass (barM_m) , which is given by (undersetisumN_iM_i^2)/(undersetisumN_iM_i) Obviously the former is independent of the individual characteristics of the molecules and gives equal weightage to large and small molecules in the polymer sample.On the other hand later gives more weightage to the heavier molecules.Infact with the help of a colligative property only one type of molar mass of the polymer can be determined. One gram each of polymer A (molar mass=2000) and B(molar mass=6000) is dissolved in water to form one litre solution at 27^@C .The osmotic pressure of this solution will be :

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