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A sample of drinking water was found to be severely contaminated with chloroform (CHCl3), supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):
(i) express this in percent by mass
(ii) determine the molality of chloroform in the water sample.

Answer:

(i) 15 ppm of CHCl3 in water means that there is 15 g of CHCl3 in 106 g of water (1 million = 106)
Percentage of CHCl3
= 15106×100 = 0.0015 % = 1.5 × 10-4%

(ii)
Here we have already find the mass % is 15 x 10-4 

When even mass % is given take total mass of solution = 100 gram
And mass of solute (CHCl3)  will  =  15 x10-4 g
Mass of solvent = mass of solution – mass of solute
(Here mass of solute is very small as compare to total mass of solution so neglect it and we get)
Mass of solvent = mass of solution
Mass of solvent = 100 g
 

 Number of moles of solute = mass of solute / molar mass

number of moles of CHCl3 =15×10-4118.5 = 1.266×10-5


Molality, m of CHCl3 in drinking water sample
            = moles of CHCl3mass of water in kg

Now plug the value of number of moles and mass of solvent in above eqution
 
Here mass of solvent = 100 g
Convert it in Kg we divide by 100 we get
Mass of solvent = 0.1Kg 
   
molaity = 1.266 x 10-5 / 0.1 = 1.266 x 10-4 mol Kg-1



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Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure, in bar of a mixture of 26.0 g of heptane and 35.0 g of octane? 

Answer:
We have given,

Molar mass of heptane,
C7H16 = 100 g mol–1
Molar mass of octane,
C8H18 = 114 g mol–1
Moles of heptane

= Wt. of heptaneMol. mass of heptane

=26100=0.26

Similarly, Moles of octane


=35114=0.31

Mole fraction heptane

=nAnA+nB

= 0.260.26+0.31=0.456

Mole fraction of octane

= 0.310.26+0.31=0.543

Partial vapour pressure = Mole fraction x Vap.
Pressure of pure component.
Partial vapour pressure of heptane

= 0.456 x 105.2 = 47.97 kPa

Partial vapour pressure of octane
= 0.543 x 46.8 = 25.4 kPa

Total vapour pressure of solution = 73.08 kPa.

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An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL–1, then what shall be the molarity of the solution?

Answer:

Consider ethylene glycol as solute and water as a solvent.
Weight of solute, WB = 222.6 g
Molar mass, MB = 24 + 6 + 32 = 62
WB = 200 g = 0.200 kg.

Moles of solute,       nB = WBMB = 222.662 = 3.59

Molaity of ethylene glycol in H
2O

mb =Number of moles of soluteKg. of solvent       = 3.59WA = 3.590.20 = 17.95 molal       = 17.95 m


Total mass of solution
= 222.6 + 200 = 422.6 g
Density of solution
= 1.072 g/ ml Volume of solution

= MassDensity = 422.61.072 = 394.2 mL = 0.3942 L

Molarity of solution,


Mb = Moles of soluteVol. of solution in litres       = 3.590.3942 = 9.1M.

 

253 Views

The partial pressure of ethane over a solution containing 6.56 x 10–3 g of ethane is 1 bar. If the solution contains 5.00 x 10–2 g of ethane, then what shall be the partial pressure of th gas? 

By Henry's law, we can write,
P(C2H6 )= KH x mole fraction of gas in sol

1 bar = KH x conc. of gas in sol          = KH x 6.56 x 10-2g of C2H6  KH = 16.56 × 10-2


p' C2H6 = KH × conc. of gas in sol                = 16.56 × 10-2×5.00×10-2                = 56.56 = 0.762 bar.
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An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?


Consider 1 litre of solution,
Osmotic pressure, π = 1.004 bar
T = 273 K = 0.083 L atm K–1 mol–1
Volume of solution = 1 litre
Weight of solute in 1 litre = 20 g
Let the molar mas be Mb

Consider 1 litre of solution,Osmotic pressure, π = 1.004 barT = 273


Molar mass of non-volatile solute
= 616.7 g mol–1.

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