Look at the solubility graph in Figure 13.7, and estimate which member of each of the following pairs has the higher molar solubility at the given temperature. (b) CH3CO2Na or glucose at 20 °C

Question

Solubility graph showing solubility of various salts in water at different temperatures.

Accepted Answer

Hello. In this problem, we are asked to consider the scalability of common solid shown in the graph below. We are asked to determine which substance will have a greater say liability at 50°C, ammonium chloride or potassium bromine. So looking at the graph, we have potassium bromide which is shown in purple and we have ammonium chloride which is shown in turquoise. We're asked to consider the cell viability of 50°C. So going to the temperature scale 50°C would be right here and we're gonna follow that up and find where it intersects with the purple line, which again is potassium bromine and where it intersects with the blue or turquoise line which is ammonium chloride. And then we're gonna follow those over to where they intersect with the Buy access for potassium bromide. Then this liability is 80 g of potassium bromide in the leaders of water. And then following the intersection over for William chloride, It's a little bit above 50. So we'll call it 52. So the ability for ammonium chloride is 52 g. Good morning chloride In 100 ml of water. And so when we say sorry, ability here we mean Mueller scalability. So we need to convert the mass of each of these potassium bromide and ammonium chloride to moles. We're going to come up with the concentration in terms of polarity. So beginning with the potassium bromide, We have 80 g of potassium bromide In 100 ml of water. First convert the volume for milliliters to liters, 1000 mL of water is equal to one l water And then we're gonna use the molar mass of potassium bromine, go from mass to moles one mole Of potassium bromide as a mass of 119.002 g. So we should make sure that we set up our conversion factors correctly and that our units are canceling some milliliters of water cancels and grams of potassium bromide cancels. So we're left with moles per liter. This works out to 9.7 moles per liter for the potassium bromine. Do similar things for the ammonium chloride. So we have 52 g ammonium Clyde In 100 ml of water. Again, let's begin by converting our volume from milliliters to liters mL. Water is equal to one liter of water And then we'll use the molar mass of ammonium card to go from mass to moles one mole ammonium chloride as a mass of 53.491 g. So again, we should ensure that our units are canceling programs of ammonium chloride cancels milliliters of water cancels. And we're left with 6.7 moles of ammonium chloride per leader. So the question again asked us which one would have the greater soluble it e at 50°C. So that would be the larger of the two numbers. So money implied, I'm sorry, potassium bromide then has a solid ability of 9.7 moles per liter. So potassium bromide has a greater scalability at 50°C as compared to ammonium chloride. Thanks for watching. Hope. This helps.

Look at the solubility graph in Figure 13.7, and estimate which member of each of the following pairs has the higher molar solubility at the given temperature. (b) CH3CO2Na or glucose at 20 °C

Verified Solution

Key Concepts

Molar Solubility

Solubility Graphs

Comparative Analysis