Question

A geochemist in the field takes a small sample of the crystals of mineral compound X from a rock pool lined with more crystals of X. He notes the temperature of the pool, 26.° C, and caps the sample carefully. Back in the lab, the geochemist dissolves the crystals in 3.00 L of distilled water. He then filters this solution and evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries and weighs the crystals. They weigh 0.36 kg1) Using only the information above can you calculate the solubility of X in water at 26 degrees Celsius? yes or no2) If yes calculate the solubility. Round answer to 2 signifacnt digits

Answer 1

Answer:

The solubility is 0.13 g/mL

Explanation:

Step 1: Data given

Temperature = 26.0 °C =299 K

Volume = 3.00 L

The mass of the crystals, after washing and drying = 0.36 kg = 360 grams

step 2: Calculate the solubility

3.00 L of water contains 360 grams of crystals

For 1.00L of water we'll have 360 / 3 = 130 grams of crystals

This means we have 130 grams of crystals in 1 L, this gives us a solubility of 130g/L

In 1000 mL we have 130 grams crystal

in 1 mL we have 130/1000 = 0.130 grams of crystals

The solubility is 0.13 g/mL