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If the solubility product of CuS is 6 x 10–16, calculate the maximum molarity of CuS in aqueous solution.


Suppose solubility of CuS is x mol–1This would give x mol–1 of C
Suppose solubility of CuS is x mol–1
This would give x mol–1 of Cu2+ ions and x mol L–1 of S2– ions on dissociation.
[Cu2+] = x mol–1
[S2–] = x mol–1
             
Suppose solubility of CuS is x mol–1This would give x mol–1 of C
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Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 of C9H8O4 is dissolved in 400 g of CH3CN. 

Answer:

Mass of solute  = 6.5 g
Mass of solution = 450 + 6.5 = 456.6 g

Mass percentage  = Mass of soluteMass of solution×100            = 6.5456.5×100 = 650456.5 = 1.424%.

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If the density of some lake water is 1.25 g mL–1 and contains 92 g of Na+ ions per kg of water, calculate the molality of Na+ ions in the lake. 

Answer:

number of moles present in 92g of Na+ ion.
         =92/23g mol-1 = 4mol

Molality, m=No. of gm moles of solutewt. of solvent in kg
No. of gm moles of solute = 92/23 = 4
Wt. of water (solvent) = 1 kg
Molality = 41 = 4 m.

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The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroaacetic acid increases in the order given. Explain briefly.


Answer:



Among H,Cl and F, Hydrogen is least electronegtive while F is most electronegtive than Cl and H.
Thus F can withdraws more electron towards itslef more than Cl and H. So trifluoroacetic acid can easily lose the H+ ions. i.e.
trifluroacetic acid ionize to the larger extent .
Now more the ion produces the greater is the dpress ion of the freezing point 
Hence, the depression of freezing point increase in order :

Acetic acid<trichloracetic acid < trifluroacetic acid





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Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol. 

Answer:

Mol. mass of benzoic acid, C6H5COOH
= 6 x 12 + 5 x 1 + 12 + 16 + 16 + 1
= 72 + 5 + 12 + 16+16 + 1
= 122 g mol–1

by using formula;

M = xmolecular mass of given substance×1000required volume

here x= amount of substance required 

0.15M = x122×1000250

Amount of benzoic acid required

=1221000×250×0.15= 45751000 = 4.575 g.

 

 

 

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