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3. (5 pts) A solution was prepared by transferring 0.661 g of KSO to a 250.0-ml volumetric flask (container 1) and adding wat
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Answer #1

Answer 3-

Given,

Initial mass of K2SO4 = 0.661 g

Molar mass of K2SO4 = 174.259 g/mol

Volume = 250 ml or 0.25 L [1 ml = 0.001 L]

Dilution 1 - 1.000 ml sample and volume 500 ml or 0.5 L

Dilution 2 - 10.0 ml sample and volume 250 ml or 0.25 L  

Final Concentration = ?

We know that,

Molarity = Mole /Volume in L

Also,

Moles = Mass/Molar Mass

So,

Molarity = Mass/(Molar Mass * Volume in L)

Put the value,

Molarity = 0.661 g/(174.259 g/mol * 0.25 L)

Molarity = 0.015 M

Now,

M1V1 = M2V2

M1 andM2 are Molarity

V1 and V2 are volumes

For Dilution 1

M1V1 = M2V2

0.015 M * 0.001 L = M2 * 0.5 L

M2 = 0.00003 M

For Dilution 2

M1V1 = M2V2

0.00003 M * 0.01 L = M2 * 0.25 L

M2 = 0.0000012 M or 1.2 * 10-6 M [Answer]

The final concentration is 1.2 * 10-6 M [Answer]

Answer 4 -

Isotopes of Mo = 7

Mass Number Exact Weight Percentage abundance
92 91.906808 14.84
94 93.905085 9.25
95 94.905840 15.92
96 95.904678 16.68
97 96.906020 9.55
98 97.905406 24.13
100 99.907477 9.63

We know that,

Average atomic Mass = f1 M1 + f2 M2 + f3 M3 + . . . + fn Mn

where, f1, f2, f3, . . . , fn are fraction of abundance

M1, M2, M3, . . . , Mn are Mass of isotopes

Fraction abundance = % abundance/ 100

Mass Number Exact Weight Percentage abundance Fraction Abundance
92 91.906808 14.84 0.1484
94 93.905085 9.25 0.0925
95 94.905840 15.92 0.1592
96 95.904678 16.68 0.1668
97 96.906020 9.55 0.0955
98 97.905406 24.13 0.2413
100 99.907477 9.63 0.0963

Put the values,

Average atomic Mass = f1 M1 + f2 M2 + f3 M3 + . . . + fn Mn

Average atomic Mass = (0.1484*91.906808)+(0.0925*93.905085)+(0.1592*94.905840)+(0.1668*95.904678)+(0.0955*96.906020)+(0.2413*97.905406)+(0.0963*99.907477)

Average atomic Mass = 95.931290101 u [Answer]

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