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3 years ago

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What volume of a 2.5 M NaOH solution is required to prepare 2.0 liters of a 0.75M NaOH? A. 2.5 liters B. 6.7 liters C. 0.60 lite

r D. 1.0 liters E. 0.75 liter

1 answer:

puteri [66]3 years ago

8 0

Answer:

C. 0.60

Explanation:

Mi x Vi = Mf x Vf

2.5 x Vi = 0.75 x 2.0

2.5 x Vi = 1.5

Vi = 1.5 / 2.5

Vi = 0.6 L

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<u>Answer: </u>One isotope has a percentage abundance of 75.75 % and the percentage abundance of another isotope is 24.24%.

<u>Explanation:</u>

We are given the two stable isotopes of chlorine with their respective masses. The average atomic mass of chlorine is also given.

Average atomic mass of chlorine = 35.45 amu.

Let us assume the fractional abundance of one isotope be 'x' and the fractional abundance for another isotope will be (1 - x) because the total fractional abundance is always equal to 1.

For Chlorine-35 isotope:

Fractional abundance = x

Mass = 34.97 amu

For Chlorine-37 isotope:

Fractional abundance = 1 - x

Mass = 36.95 amu

The formula for the calculation of average atomic mass is given by:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

Putting values in above equation, we get:

$35.45=34.97(x)+36.95(1-x)\\\\35.45=34.97x+36.95-36.95\\\\-1.5=-1.98x\\\\x=\frac{-1.5}{-1.98}=0.7575$

$1-x=1-0.7575=0.2424$

Converting these two fractional abundances into percentage abundances by multiplying it with 100.

$x=0.7575\times 100=75.75\%\\\\1-x=0.2424\times 100=24.24\%$

Hence, one isotope has a percentage abundance of 75.75 % and the percentage abundance of another isotope is 24.24%.

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