Question

A supersaturated sugar solution \((650.0 \mathrm{~g}\) sugar in \(150.0 \mathrm{~g}\) water) is prepared to make "rock candy"" The solution is kept at \(25^{\circ} \mathrm{C}\), where the solubility of sugar is \(220.0 \mathrm{~g} / 100 \mathrm{~g}\) water. \(\mathrm{A}\) string is dipped into the solution, and in time, large crystals of sugar (rock candy) form. How many grams of solid are obtained when the solution just reaches saturation?

Short answer

Answer: 320g of sugar will become solid.

Step-by-step solution

Find the saturation point for sugar in 150g of water

We are given that the solubility of sugar at 25°C is 220.0g per 100g of water. To find the saturation point for 150g of water, we can use the proportion: (solubility of sugar in 100g of water) / 100g of water = (solubility of sugar in 150g of water) / 150g of water

Calculate the solubility of sugar in 150g of water

Let x be the solubility of sugar in 150g of water. From the proportion in step 1, we can write the equation: (220g) / 100g = x / 150g Now, let's solve for x: x = (220g * 150g) / 100g = 330g So, at 25°C, 150g of water can dissolve a maximum of 330g of sugar.

Calculate the difference between the initial amount of sugar and saturation point

We are given that the initial amount of sugar in the supersaturated solution is 650g. We just found that the maximum amount of sugar that can dissolve in 150g of water is 330g. So, the difference between the initial amount of sugar and the saturation point is: 650g (initial amount of sugar) - 330g (saturation point) = 320g So, when the supersaturated solution reaches saturation, 320g of sugar will become solid as rock candy.