If 50 grams of water saturated with potassium chlorate at 23*C is slowly evaporated to dryness, how many grams of the dry salt will be recovered? I know that I should know this.. But it has 50 grams of water instead of 100. ): Please, someone tell me how to do this?
Solubility=mass/molar *1000/volume
Dats d formula nd u will get, 50g is d volume
Ah, potassium chlorate, the salt that's just dying to get dry! Don't worry, I'm here to help! So, if you have 50 grams of water saturated with potassium chlorate, we need to figure out how much of the dry salt will be left behind when it's all evaporated.
To calculate this, we need to know the solubility of potassium chlorate in water at 23°C. Unfortunately, I'm not a walking solubility database, but I can make you laugh while you look it up!
To determine the amount of dry salt that will be recovered, we need to first calculate the initial amount of potassium chlorate dissolved in the water and then subtract the amount remaining after evaporation.
Step 1: Calculate the amount of potassium chlorate initially dissolved in the 50 grams of water.
The solubility of potassium chlorate at 23°C is 7.5 grams per 100 grams of water. Since you have 50 grams of water, the amount of potassium chlorate dissolved can be calculated as follows:
(7.5 grams / 100 grams) * 50 grams = 3.75 grams of potassium chlorate
Step 2: Determine the amount of potassium chlorate remaining after evaporation.
Since the water is evaporated to dryness, all the water will be lost, leaving behind only the potassium chlorate. Therefore, the entire amount of potassium chlorate initially dissolved will be recovered.
Step 3: Calculate the final amount of dry salt recovered.
Therefore, the amount of dry salt recovered will be 3.75 grams.
To summarize:
- If you have 50 grams of water saturated with potassium chlorate at 23°C, the initial amount of potassium chlorate dissolved is 3.75 grams.
- When the water is evaporated to dryness, all the water will be lost, leaving behind the entire 3.75 grams of potassium chlorate as dry salt.
I assume you have the solubility of KClO3 in water.
solubility in 50 g H2O will the just 1/2 the solubility in 100 g H2O.