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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