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Murljashka [212]
3 years ago
8

A 71.0 mL aliquot of a 1.80 M solution is diluted to a total volume of 218 mL. A 109 mL portion of that solution is diluted by a

dding 113 mL of water. What is the final concentration? Assume the volumes are additive
Chemistry
1 answer:
larisa [96]3 years ago
4 0

Answer:

The final concentration is 0.288 M (approximately)

Explanation:

Using dilution formula:

C1V1 = C2V2

Where C1 is the initial concentration

C2 is the final concentration

V1 is the initial volume

V2 is the final volume

In the first dilution, C1 = 1.8M,

V1 =71mL, V2 = 218mL from the formula C2 = 1.8 ×71/281

=0.5862M

In the second dilution,

C1 = 0.5862M, V1 = 109mL, V2 = 222 mL (109+113)

C2 = 0.5862 × 109 ÷ 222

= 0.288 M (approximately) assuming volume is additive.

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It is given in the question that in 100 gms of CaSO4 there are 29.45 grams of Ca present and there is 25.6 gram of total CaSO4 sample present, So, to calculate the exact value of calcium in this given sample is:

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