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joja [24]
3 years ago
8

If a chemist has a stock solution of HBr that is 10.0 M and would like to make 450.0 mL of 3.0 M HBr, how would he do it?

Chemistry
1 answer:
asambeis [7]3 years ago
4 0

This is a dilution that requires a certain volume from the stock solution to be diluted with distilled water to make a solution of HBr with a lesser concentration than the stock solution

Following dilution formula can be used

c1v1 = c2v2

Where c1 is concentration and v1 is the volume of the stock solution

c2 is concentration and v2 is volume of the diluted solution to be prepared

Substituting these values

10.0 M x v1 = 3.0 x 450.0 mL

v1 = 135.0 mL

A volume of 135.0 mL from HBr stock solution needs to be taken and diluted with distilled water upto 450.0 mL. The resulting solution will have a concentration of 3.0 M

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Answer:

Boiling point of solution is 100.65^{0}\textrm{C}

Explanation:

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According to Raoult's law for a non-volatile solute dissolved in a solution-

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Here, K_{b}=0.51^{0}\textrm{C}.kg.mol^{-1}

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Plug in all the values in the above equation, we get-

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