three isotopes have mass no. M,(M+1)(M+2). If the mean mass number is (m+0.5)then which of the following ratio may be for M,(M+1)(M+2). A)1:1:1 B)4:1:1 C)3:2:1 D)2:1:1

To solve the problem regarding the isotopes with mass numbers M, (M+1), and (M+2), and to find the ratio that corresponds to a mean mass number of (m + 0.5), we need to delve into some basic principles of isotopes and averages. Let's break this down step by step.

Understanding Isotopes and Mean Mass Number

Isotopes are variants of a particular chemical element that have the same number of protons but different numbers of neutrons, resulting in different mass numbers. In this case, we have three isotopes with mass numbers M, (M+1), and (M+2).

Calculating the Mean Mass Number

The mean mass number can be calculated using the formula:

Mean Mass Number = (Σ(mass number × number of atoms)) / Total number of atoms

Let’s denote the number of atoms of the isotopes M, (M+1), and (M+2) as x, y, and z, respectively. The mean mass number can then be expressed as:

Mean = (Mx + (M+1)y + (M+2)z) / (x + y + z)

According to the problem, this mean mass number is equal to (m + 0.5). For our purposes, we can set m = M, which simplifies our calculations.

Setting Up the Equation

Substituting m with M, we have:

(Mx + (M+1)y + (M+2)z) / (x + y + z) = M + 0.5

Multiplying both sides by (x + y + z) gives us:

Mx + (M+1)y + (M+2)z = (M + 0.5)(x + y + z)

Expanding the right side, we get:

Mx + (M+1)y + (M+2)z = Mx + My + Mz + 0.5x + 0.5y + 0.5z

Now, simplifying this equation leads us to:

(M+1)y + (M+2)z = My + Mz + 0.5x + 0.5y + 0.5z

Analyzing the Ratios

To find the ratio of x, y, and z, we can rearrange the equation and analyze the coefficients. The left side represents the contributions of each isotope, while the right side includes the average contributions adjusted by the mean mass number.

Let’s consider the possible ratios provided in the options:

• A) 1:1:1
• B) 4:1:1
• C) 3:2:1
• D) 2:1:1

To find which ratio satisfies our mean mass number condition, we can substitute these ratios into our mean mass number equation and check if it equals (M + 0.5).

Testing the Ratios

For each ratio, we can calculate the left-hand side and see if it balances with the right-hand side. For example:

• For 1:1:1 (x=1, y=1, z=1): Mean = (M + (M+1) + (M+2)) / 3 = (3M + 3) / 3 = M + 1 (not valid)
• For 4:1:1 (x=4, y=1, z=1): Mean = (4M + (M+1) + (M+2)) / 6 = (6M + 3) / 6 = M + 0.5 (valid)
• For 3:2:1 (x=3, y=2, z=1): Mean = (3M + 2(M+1) + (M+2)) / 6 = (6M + 4) / 6 = M + 2/3 (not valid)
• For 2:1:1 (x=2, y=1, z=1): Mean = (2M + (M+1) + (M+2)) / 4 = (4M + 3) / 4 = M + 0.75 (not valid)

Conclusion

From our calculations, the only ratio that satisfies the condition of the mean mass number being (M + 0.5) is option B: 4:1:1. This means that for the isotopes with mass numbers M, (M+1), and (M+2), the ratio of their quantities is 4:1:1. This analysis not only helps in understanding isotopes but also reinforces the concept of weighted averages in chemistry.