How much water has boiled off when precipitation begins?
One expects 25 g of $\ce{NaCl}$ to saturate 62.5 g of water at 100 °C. However, 25 g of $\ce{NaCl}$, 58.443 g/mol, is 0.4278 mol in 62.5 ml of water at 100 °C. Colligative property boiling point elevation is +0.52 (°C kg)/mol. 0.8556 mol of particles in 62.5 ml of water is 13.69 mol/kg or 7.1 °C elevated.

$\ce{NaCl}$ solubility/temp increases from about 36 g/100 g at 0 °C to 40 g/100 g at 100 °C or 0.04 g/°C. This gives first approximation 0.2847 g more solubility. "25.28" g $\ce{NaCl}$ in 62.5 g water at 107 °C is then 25 g $\ce{NaCl}$ in 61.8 g water. 88.2 g water boiled off (with some decimal trim for Zeno’s paradox).

Repeat the problem with ammonium cyanate.