Question

Daisy Dominguez Solubility of Benzoic Acid in Water y = 1.479e9.0133 R-0.9955 20 40 A graph of temperature (°C) versus solubility of benzoic acid in water (g/L) has a good exponential fit, y = 1.479e0.0357x, indicated by the Goodness of Fit R? = 0.9955. Given that you used 19.9 ml of water to dissolve 0.786 g of benzoic acid, what was the actual temperature of your recrystallization setup according to this exponential equation? Don't forget that there are 1000 ml in a liter Round to a whole number answer without any decimal point. Answer:

Answer 1

Y = 1.479 × e^0.0357x (1)

Y is solubility in g/L

x is temperature in ^oc

Now , volume of solution = 19.9 mL = (19.9/1000) L= 0.0199 L

Mass of benzoic acid = 0.786 g

given solubility is

= (0.786/0.0199)

= 39.5 g/L

Putting the value in Eq.1

39.5 = 1.479 × e^0.0357x

Or, e^0.0357x = (39.5/1.479)

Or, e^0.0357x = 26.7

Or, 0.0357x = ln(26.7)

Or, 0.0357x = 3.285

Or, x = (3.28/0.0357)

Or, x = 92 ⁰ c.

Hence, temperature is 92⁰ c.