You are given a stock solution of 0.1 M, (1) 100-mL volumetric flask; (1) 25-mL

Question

You are given a stock solution of 0.1 M, (1) 100-mL volumetric flask; (1) 25-mL volumetric flask; (1) 5- ml transfer pipet; and (1) 10-mL transfer pipet. What is the final concentration of the serial dilution you could perform with these tools? (Keep in mind that you are using each piece of volumetric glassware only once.)

a) 0.20 mM

b) 2.0 mM

c) 20 mM

d) 0.20 M

e) 20 M

Calculate the mass of a heterogeneous mixture containing 139.32 grams sand, 34.99 grams gravel, and 9.372 grams salt.

A. 183.68 g

B. 183.684 g

C. 183.7 g

D. 184 g

E. 184.0 g

Explanation / Answer

1. If you take 5ml pipette and thus transfer 5ml of 0.1M stick solution to 25 ml volumetric flask the molarity of that solution can be calculated from molarity equation as:

M1 V1= M2 V2

therefore, 0.1x5= M(after 1st dilution) x 25

i.e M(after first dilution)= .5/25= 0.02

so after first dilution it's molarity is 0.02

now if you take 10ml of this 0.02 dilution and transfer it to 100ml volumetric flask the final mp molarity after dilution will be.

from molarity equation:

0.02x10=M(after second dilution) x 100

therefore, molarity after second dilution= 0.002

or M= 2.0mM so correct answer is b.

.

For the calculation of mass heterogeneous mixture remember that your final answer can't have more significant figures than the lowest significant figure from the data. As lowest significant figure is 4 (in 34.99 and 9.372)

So your answer correct to four significant figures is C i.e 183.7 g.

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