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Lapatulllka [165]
3 years ago
9

A chemist makes 940. mL of sodium carbonate (Na_2CO_3) working solution by adding distilled water to 200. mL of a 0.171 m stock

solution of sodium carbonate in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.
Chemistry
1 answer:
Alexxandr [17]3 years ago
6 0

Answer:

The answer is 0.0698 M

Explanation:

The concentration was prepared by a serial dilution method.

The formula for the preparation I M1V1 = M2V2

M1= the concentration of the stock solution = 0.171 M

V1= volume of the stock solution taken = 200 mL

M2 = the concentration produced

V2 = the volume of the solution produced = 940 mL

Substitute these values in the formula

0.171 × 200 = 490 × M2

34.2 = 490 × M2

Make M2 the subject of the formula

M2 = 34.2/490

M2 = 0.069795

M2 = 0.0698 M ( 3 s.f)

The concentration of the Chemist's working solution to 3 significant figures is 0.0698M

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\Delta T_f=i\times K_f\times m\\\\\Delta T_f=i\times K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of water in Kg}}

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