To determine the concentration of the non-competitive inhibitor needed to yield 90% inhibition of an enzyme-catalyzed reaction, we can use the equation for the fractional activity of the enzyme:

Fractional activity = 1 - (V/V₀)

where V is the reaction rate in the presence of the inhibitor and V₀ is the reaction rate in the absence of the inhibitor.

Let's assume that the initial concentration of the enzyme is E₀ and the concentration of the inhibitor is I₀. The rate equation for the enzyme-catalyzed reaction can be written as:

V = k[E][S]/(1 + [S]/Km)

where k is the rate constant, [E] is the concentration of the enzyme, [S] is the concentration of the substrate, and Km is the Michaelis constant.

Since the inhibitor is non-competitive, it can bind to both the enzyme-substrate complex ([ES]) as well as the free enzyme ([E]). The rate equation in the presence of the non-competitive inhibitor becomes:

V = k'[E][S]/(1 + [S]/Km + [I]/Ki)

where k' is the rate constant in the presence of the inhibitor and Ki is the inhibition constant.

Now, let's substitute this expression for V into the equation for fractional activity:

Fractional activity = 1 - (k'[E₀][S₀]/(1 + [S₀]/Km + [I₀]/Ki))/ (k[E₀][S₀]/(1 + [S₀]/Km))

Simplifying this expression, we get:

Fractional activity = 1 - (k'/k)(1 + [I₀]/Ki)

Since the fractional activity is given as 90% inhibition, we can set this expression equal to 0.1:

0.1 = (k'/k)(1 + [I₀]/Ki)

Now, we can rearrange this equation to solve for [I₀]:

[I₀]/Ki = (0.1 - 1)(k/k')

[I₀] = (0.1 - 1)(k/k')Ki

Given the value of k₁ = 2.9 x 10⁻⁴ moldm⁻³ for the non-competitive inhibitor, we can substitute the known values into the equation:

[I₀] = (0.1 - 1)(k/k')Ki
[I₀] = (0.1 - 1)(k/k')(2.9 x 10⁻⁴)

To find the range of values that [I₀] can take, we need to consider the upper and lower limits of the expression (0.1 - 1)(k/k')(2.9 x 10⁻⁴). Since no values are provided for k, k', or Ki, we can assume that they are all positive constants. Therefore, the upper limit occurs when k, k', and Ki are at their maximum values, and the lower limit occurs when k, k', and Ki are at their minimum values.

Since the question specifies that the answer is between 2.50 and 2.70 mol dm⁻³, it suggests that the upper and lower limits of [I₀] fall within this range. However, without further information about the values of k, k', and Ki, it is not possible to determine the exact concentration of the non-competitive