Playing with Solubility 1

Solubility at a temperature is defined as $K_{sp} = \dfrac{m_s}{m_w}$ , where $m_s$ the mass of solute $m_w$ mass of water can dissolve in a saturated solution.

Therefore for the $150 \text{ g } \ce{KNO3}$ solution at $60^\circ C$ , we have:

$\begin{aligned} \frac{m_s + 22}{150-m_s} & = \frac{m_s}{130-m_s} \\ 130 m_s + 2860 - m_s^2 - 22 m_s & = 150 m_s - m_s^2 \\ (150-130+22)m_s & = 2860 \\ 42 m_s & = 2860 \\ \Rightarrow m_s & = \frac{1430}{21} \\ \text{For} \quad \frac{m}{100} & = \frac{m_s}{130-m_s} \\ \Rightarrow m & = \frac{100 \times \frac{1430}{21}}{130-\frac{1430}{21}} \\ & = \frac{143000}{1300} \\ & = \boxed{110} \text{ g} \end{aligned}$

Let x be the amount of KNO3 in solution.

(x + 22)/(150 + 22) = x/(150-20)

x = 1430/21

If 1430/21 g are in 130 g of a saturated solution.

The proportion between KNO3 : H2O is 1430/21 : (130 - 1430/21) = 1430/21 : 1300/21 = 1430 : 1300 = 110 : 100.

Therefore, the solubility of KNO3 at 60 degrees C per 100 g of water is 110 g.

must add $22g$ $\ce{KNO3}$ $\quad \Rightarrow$ The amount of deficient $\ce{KNO3}$ is $22g$

must evaporate $20g$ $\ce{H2O}$ $\Rightarrow$ The amount of extra water is $20g$ .

Therefore $22g$ of $\ce{KNO3}$ dissolves in $20g$ of water and hence the solubility is $\dfrac{22}{20}\times{100} = \boxed{110}$

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The second statement in the problem was extraneous.