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A geochemist in the field takes a 36.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X. He notes the temperature of the pool, 170 C, and caps the sample carefully. Back in the lab, the geochemist first dilutes the sample with distilled water to 500. mL. Then he filters it and evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries and weighs the crystals. They weigh 3.96 g.

Using only the information above, can you calculate yes the solubility of X in water at 17.0 C? If you said yes, calculate it.

Respuesta :

Answer:

solubility of X in water at 17.0 [tex]^{0}\textrm{C}[/tex] is 0.11 g/mL.

Explanation:

Yes, the solubility of X in water at 17.0 [tex]^{0}\textrm{C}[/tex] can be calculated using the information given.

Let's assume solubility of X in water at 17.0 [tex]^{0}\textrm{C}[/tex] is y g/mL

The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.

So, solubility of X in 1 mL of water = y g

Hence, solubility of X in 36.0 mL of water = 36y g

So, 36y = 3.96

   or, y = [tex]\frac{3.96}{36}[/tex] = 0.11

Hence solubility of X in water at 17.0 [tex]^{0}\textrm{C}[/tex] is 0.11 g/mL.

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