Answer:
The solubility is [tex]S = 0.00912 \ g/mL[/tex]
Explanation:
From the question we are told that
The original volume of sample is [tex]V_o = 34.0 mL = 34 *10^{-3} \ L[/tex]
The temperature is [tex]T = 22.0 ^oC[/tex]
The new volume of sample is [tex]V_n = 750.0 mL = 750 *10^{-3} \ L[/tex]
The weight of the crystal is [tex]X = 0.31 \ g = 0.31 *10^{-3} \ kg[/tex]
Now looking at the question we see that 34.0 mL of the sample is saturated with 0.31g of the crystal X
Generally the solubility of X in the water sample at [tex]22.0 ^oC[/tex] can be mathematically evaluate as
[tex]S = \frac{0.31 }{34.0}[/tex]
[tex]S = 0.00912 \ g/mL[/tex]
