A 20.0 mL sample of CuSO₄ was evaporated to dryness,leaving 0.967 g of residue.What was the molarity of the original solution? A)48.4 M B)0.0207 M C)0.0484 M D)0.303 M E)0.433 M

Question

Quizplus · Accepted Answer

First, we need to calculate the number of moles of the residue left behind:

m = mass of residue = 0.967 g 
M = molar mass of CuSO4 = 159.6 g/mol 
n = moles of CuSO4 left behind = m/M = 0.00606 mol

Since the residue was obtained from a 20.0 mL sample, we can calculate the original concentration (molarity) of CuSO4 as follows:

M = n/V 
V = volume of original solution = 20.0 mL = 0.0200 L

M = 0.00606 mol / 0.0200 L = 0.303 M

Therefore, the answer is D, 0.303 M.

A 200 ML Sample of CuSO4 Was Evaporated to Dryness,leaving 0

A 200 ML Sample of CuSO₄ Was Evaporated to Dryness,leaving 0