Answer:
At about 51 °C
Explanation:
Hello there!
In this case, according to the attached solubility chart of potassium nitrate, our first step for this problem is to calculate the grams of this solute in 100 g given that 23 g are soluble in 25 g of water:
[tex]\frac{23g}{25g}=\frac{x}{100g}\\\\x= \frac{23g*100g}{25g}=92g[/tex]
Thus, such solubility of 92 g in 100 g of water is exhibited at about 51 °C.
Regards!
