(i) The rate of increase of the mass of A, \(\frac{dx}{dt}\), is proportional to the mass of B. Since the total mass is constant at 50, the mass of B is \(50 - x\). Therefore, \(\frac{dx}{dt} = k(50 - x)\).

(ii) To solve \(\frac{dx}{dt} = k(50 - x)\), separate variables:

\(\frac{1}{50-x} dx = k \, dt\)

Integrate both sides:

\(-\ln|50-x| = kt + C\)

Using the initial condition \(x = 0\) when \(t = 0\), we find:

\(-\ln|50| = C\)

Thus, \(-\ln(50-x) = kt - \ln 50\).

Rearrange to find:

\(\ln\left(\frac{50}{50-x}\right) = kt\)

\(\frac{50}{50-x} = e^{kt}\)

\(50-x = 50e^{-kt}\)

\(x = 50(1 - e^{-kt})\)

Using \(x = 25\) when \(t = 10\), solve for k:

\(25 = 50(1 - e^{-10k})\)

\(0.5 = 1 - e^{-10k}\)

\(e^{-10k} = 0.5\)

\(-10k = \ln(0.5)\)

\(k = -\frac{\ln(0.5)}{10} \approx 0.0693\)

Thus, \(x = 50(1 - \exp(-0.0693t))\).