# A solution of rubbing alcohol is 74.3% (v/v) isopropanol in water. How many mililiters of Isopropanol are in 62.4 mL sample of the alcohol solution?

In your case, the solution contains

As you know, solutions are homogeneous mixture, which implies that they have the same composition throughout.

Set up the equation as

Rearrange and solve to find

Rounded to three sig figs, the answer will be

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To find the volume of isopropanol in the 62.4 mL sample, multiply the volume of the sample by the percentage of isopropanol in the solution:

Volume of isopropanol = 62.4 mL * (74.3/100) = 46.3536 mL

So, there are approximately 46.35 mL of isopropanol in the 62.4 mL sample of the alcohol solution.

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To find the volume of isopropanol in a 62.4 mL sample of the alcohol solution, you multiply the volume of the sample by the percentage of isopropanol in the solution:

[ \text{Volume of isopropanol} = \text{Volume of sample} \times \text{Percentage of isopropanol} ]

[ \text{Volume of isopropanol} = 62.4 , \text{mL} \times 0.743 ]

[ \text{Volume of isopropanol} = 46.354 , \text{mL} ]

Therefore, there are 46.354 milliliters of isopropanol in a 62.4 mL sample of the alcohol solution.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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