(i) Given that the rate of formation of X is proportional to the mass of Y, we have \(\frac{dx}{dt} = k(100 - x)\). Using the initial condition \(\frac{dx}{dt} = 1.9\) when \(x = 5\), we find \(1.9 = k(100 - 5)\), giving \(k = 0.02\). Thus, \(\frac{dx}{dt} = 0.02(100 - x)\).

(ii) To solve \(\frac{dx}{dt} = 0.02(100 - x)\), separate variables: \(\frac{1}{100 - x} dx = 0.02 dt\). Integrate both sides: \(-\ln|100 - x| = 0.02t + C\). Solve for x: \(x = 100 - Ce^{-0.02t}\). Use the initial condition \(x = 5\) when \(t = 0\) to find \(C = 95\). Thus, \(x = 100 - 95e^{-0.02t}\).

(iii) As \(t \to \infty\), \(e^{-0.02t} \to 0\), so \(x \to 100\).