another chem question

Question

If the formula for sucrose (sugar) is C₁₂H₂₂O₁₁, what is the molarity of the solution of sugar water used to make cranberry sauce in the previous question? (be sure to consider Sig figs from the previous question, and include units with your answer)

Previous question:

To make cranberry sauce, you add 200. grams of sugar to exactly 1 cup of water. What is the concentration of the solution in g/dL, %, mg/dL, and mg%? (1 cup = 236.6mL)

Answer 1

Let's start with the "previous question" listed at the bottom of this prompt. For the concentration, we are asked to present the answer in 3 different values: grams of sugar per deciliter of water, milligrams of sugar per deciliter of water, and percent mass of sugar in the solution in milligrams. (note: there is also a "%" listed by itself in the list of prompted concentration forms, but without a unit qualifier of percent by volume, mass, or molarity, we cannot give a raw percentage of sugar.)

For g/dL, we take 200. g of sugar and divide by 236.6 mL of water (the amount in 1 cup of water). We then multiply this by 100, to change from mL of water to dL of water (the conversion rate is 10^-3 L in 1 mL, and 10^-1 L in 1 dL, so the dL is greater than the mL by a ratio of 100:1). This gives us 84.5 g sugar per 1 dL of water.

For mg/dL, we take the previous answer and convert grams sugar to milligrams sugar. To do this, we simply multiply by 10³ -- there are 10³ mg in 1 g. So, we have 84,500 mg sugar per 1 dL of water.

For percent mass in mg, we need to know the total mass of the solution -- sugar plus water -- for the denominator. We already have the mass of sugar in mg (previous answer), so we just need the mass of the water. Typically, water is listed as having a density of 1.0g/mL, although some more accurately use 0.997 g/mL (this will sometimes be equivalently listed as 997 kg/m³). Let's use the more accurate value. We simply take our volume, 236.6 mL of water, and multiply this by 0.997 g/mL to get the mass of water in grams, 235.9 g. Then we multiply again by a factor of 10³ to convert g to mg water, giving us 235,900 mg water. Our fractional mass is thus 84,500 mg sugar/(84,500 mg sugar + 235,900 mg water) = 0.2367. We have 3 SigFigs, from the mass of sugar initially given, so we have a 23.7% solution of sugar by mass.

For the beginning question, regarding molarity of the solution, we need to remember that molarity means moles of solute or substance in question divided by volume of the solution. We have grams of sugar already as 200. g, and the volume of the solution is given as 236.6 mL water. We can convert to moles of sugar by finding the molar mass using the chemical formula given, which turns out to be 324.97 g / mol. So, 200. g of sugar divided by this molar mass gives us 0.584288 mol sugar. If we divide this value by our volume of water, we get 0.0024695 mol sugar/mL water. Since molarity is usually presented as moles / Liters, let's multiply this by 10³ to indicate the conversion of 1000 mL per 1 L. Our final value for the molarity of this solution is 2.47 mol sugar/ L water.