In water, Vanillin, C8H8O3, has a solubility of 0.070 moles of vanillin per liter of solution at 25C. What will be produced if 5.00 g of vanillin are added to 1 L of water at 25 C?

The full amount (5.00 g) will be dissolved in 1 L of water at 25°C.

Explanation:

The molecular weight (MW) of Vanillin (C₈H₈O₃) is calculated from the chemical formula as follows:

MW(C₈H₈O₃) = (12 g/mol x 8) + (1 g/mol x 8) + (16 g/mol x 3) = 152 g/mol

If 0.070 mol of C₈H₈O₃ are soluble per liter of water at 25°C, the maximum mass that can be dissolved in 1 L is:

0.070 mol x 152 g/mol = 10.64 g

Since 5.00 g is lesser than the maximum amount that can be dissolved (10.64 g), the added amount will be completely dissolved in 1 L of water at 25°C.

answer

the ions in solute supperate from eachouthere

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.