A tonne of $20wt\%$ aqueous sodium nitrate $(NaN{O}_3)$ is concentrated to saturation by evaporation at $373K.$ The solution is then cooled to $293K$ to form a slurry of $NaN{O}_3$ crystals in a saturated aqueous solution of $NaN{O}_3.$ The crystals of sodium nitrate are removed by filtration of the slurry. The wet filter cake consists of $5\%$ by weight of mother liquor and the remainder is $NaN{O}_3$ crystals. The solubility of sodium nitrate is $1\mathrm{.}76k\frac{g}{k}g{H}_{2}O$ at $373K,$ and $0\mathrm{.}88k\frac{g}{k}g{H}_{2}O$ at $293K.$ Calculate,

$($a$)$ The amount of water evaporated to achieve saturation at $373K$

$($b$)$ The total weight of dry sodium nitrate obtained.

Please explain with steps. Part b especially