In 200 milliliters of water, 208 grams of BaBr2 can be dissolved. The question is, what is the molar solubility of Barium Bromide?
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00:00In 200 milliliters of water, 208 grams of BaBr2 can be dissolved.
00:08The question is, what is the molar solubility of barium bromide?
00:13OK, the first step is identification.
00:20The volume of water, V is equal to 0.2 liters.
00:25The mass of the solute, M is equal to 208 grams.
00:31The relative molecular mass of barium bromide, 137.3 plus 2 times 79.9.
00:41This value is equal to 297.1 grams per mole.
00:46The next step is to calculate the solubility.
00:50Solubility is measured against 1 liter of solvent.
00:54To get a volume of 1 liter, 5 tubes with a volume of 200 milliliters are needed.
01:04At this volume, the mass of BaBr2 dissolved is 5 times 208, or 1070 grams.
01:12The solubility of barium bromide is 1040 grams per mole.
01:18Mathematically, solubility can be calculated through the equation, the mass of the solute
01:24divided by the volume of the solvent.
01:27Enter both values.
01:31The result is equal to 1040 grams per liter.
01:35Now, we can calculate the molar solubility.
01:40Solubility is equal to M, R multiplied by molar solubility.
01:44Enter the solubility value and the relative molecular mass.
01:49Molar solubility is equal to 3.5 moles per liter.
01:53This is the molar solubility of barium bromide.
01:56Yup, hopefully useful.
02:00And don't forget to follow this channel.