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Q.

A 20.0 mL sample of ${CuSO}_{4}$ solution was evaporated to dryness, leaving 0.967 g of residue. What was the molarity of the original solution? $(Cu = 63.5)$

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a

$0.0484 M$

b

$0.0207 M$

c

$48.4 M$

d

$0.303 M$

answer is D.

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Detailed Solution

Molarity of the solution,

$M = [\frac{Number of moles}{Volume}] \times 1000 M = [\frac{\frac{0.967}{159.5}}{20}] \times 1000 = 0.303 M$

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