Question

4) 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

Answer 1

Answer:

An unsaturated solution.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to firstly realize we need to calculate the grams of vanillin in 0.070 moles by using its molar mass of 152.15 g/mol:

$m=0.070mol*\frac{152.15 g}{1mol} =10.65g$

Thus, since the solubility is 10.65 g per 1 L of solution, we can notice 5.00 g will complete dissolve and produce an unsaturated solution.

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