Question

A geochemist in the field takes a 25.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X . He notes the temperature of the pool, 15.° C , and caps the sample carefully. Back in the lab, the geochemist first dilutes the sample with distilled water to 450. 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 0.300 g . Using only the information above, can you calculate the solubility of X in water at 15.° C ? yesno If you said yes, calculate it. Be sure your answer has a unit symbol and 3 significant digits. ×10μ

Answer 1

Answer:

1- Yes, we can calculate the solubility of mineral compound X.

2- 0.012 g/mL.

Explanation:

1- Using only the information above, can you calculate the solubility of X in water at 15.0 °C?

The information available is:

The volume of water sample = 25.0 mL.

Weight of the mineral compound X after evaporation, drying, and washing = 0.30 g.

∴ Yes, we can calculate the solubility of mineral compound X.

2-  If you said yes, calculate it.

∵ 25.0 mL of water sample contains → 0.30 g of the mineral compound X.

∴ 1.0 mL  of water sample contains → ??? g of the mineral compound X.

1.0 ml of water sample will contain (0.3 g/25.0 mL) 0.012 g.

∴ The solubility of the mineral compound X in the water sample is 0.012 g/mL.

Answer 2

Solubility refers to the maximum amount of substance that dissolve in a solvent. Using the information, yes, we can calculate the solubility of compound x in water at 15° C.

The solubility of the compound x in water is 0.012 g/mL.

What do you mean by solubility?

The maximum amount of a chemical that will dissolve in a given amount of solvent at a certain temperature is known as solubility.

Given,

The volume of water is 25.0 ml

The temperature of the pool is 15° C.

The weight of x is after washing, drying, and evaporation is 0.30 g

Step1: Calculating the solubility of x

25.0 ml of water contain 0.30 g of x

So, 1.0 ml of water will contain  

$\bold{\dfrac{0.3\;g}{25.0\;ml}= 0.012\;g/ml }$

Thus, the solubility of x in water at  15° C is 0.012 g/ml.

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