Question

If an element M has two isotopes 50M (15%) and 52M(85%) . Its average atomic mass is​

Answer 1

Explanation:

Average Atomic weight = F

1

M

1

+F

2

M

2

+...F

n

M

n

F

1

,F

2

,F

3

...F

n

are percentage fraction.

M

1

,M

2

,M

3

...M

n

are molecular weight of isotope.

51.7=x×(50)+(1−x)52

51.7=50x+52−52x

−2x=−0.3

x=

2

0.3

x=0.15

Abundance of M

50

in nature is 15%

and Abundance of M

52

in nature is 85%

Answer 2

Answer:

$X = 0.15$

Explanation:

Given: An element M has two isotopes 50M (15%) and 52M (85%)

Let the Atomic weight average be $F_1M_1 + F_2M_2 + ...... + F_nM_n$

Here, $F_1 , F_2 , F_3 , ........, F_n$ are considered as Fractional percentages, and$M_1 , M_2 , M_3 , ........, M_n$ are considered isotopes molecular weight

Using the formula ∑$F_iM_i$ = [F × $M^{50}$] + [(100-x) × $M^{52}$

Here the average of an element M is 51.7

using the above given content we get:

51.7 = x × (50) + (1-x) 52

51.7 = 50x +52 - 52x

-2x = -0.3

x=0.15

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