In a leaching operation, the rate at which solute goes into solution is given by:
$$
\mathrm{d} M / \mathrm{d} t=k\left(c_s-c\right) \mathrm{kg} / \mathrm{s}
$$
where $M \mathrm{~kg}$ is the amount of solute dissolving in $t \mathrm{~s}, k\left(\mathrm{~m}^3 / \mathrm{s}\right)$ is a constant and $c_s$ and $c$ are the saturation and bulk concentrations of the solute respectively in $\mathrm{kg} / \mathrm{m}^3$. In a pilot test on a vessel $1 \mathrm{~m}^3$ in volume, $75 \%$ saturation was attained in 10 s . If 300 kg of a solid containing $28 \%$ by mass of a water soluble solid is agitated with $100 \mathrm{~m}^3$ of water, how long will it take for all the solute to dissolve assuming conditions are the same as in the pilot unit? Water is saturated with the solute at a concentration of $2.5 \mathrm{~kg} / \mathrm{m}^3$.