A chemist prepares a mixture of isopropanol and water at a temperature of 29.0 °C. He measures out 396. mL of isopropanol and mixes it with 41.9 mL of water. He measures the volume of the solution after it is well mixed and finds that it is 416. mL. The density of isopropanol is 0.79 /gmL and the density of water is 1.0 /gmL.Calculate the volume percent of isopropanol in this solution.Be sure your answer has the right number of significant digits.
Question
A chemist prepares a mixture of isopropanol and water at a temperature of 29.0 °C. He measures out 396. mL of isopropanol and mixes it with 41.9 mL of water. He measures the volume of the solution after it is well mixed and finds that it is 416. mL. The density of isopropanol is 0.79 /gmL and the density of water is 1.0 /gmL.Calculate the volume percent of isopropanol in this solution.Be sure your answer has the right number of significant digits.
Solution
To calculate the volume percent of isopropanol in the solution, we first need to determine the volume of isopropanol in the final solution.
Given that the initial volume of isopropanol was 396 mL and the final volume of the solution was 416 mL, we can infer that the volume of isopropanol in the final solution is less than the initial volume due to the mixing process.
However, the problem does not provide enough information to determine the exact volume of isopropanol in the final solution.
Assuming that the volume of isopropanol in the final solution is the same as the initial volume (which is a rough approximation), the volume percent of isopropanol in the solution can be calculated as follows:
Volume percent = (Volume of isopropanol / Total volume) * 100%
= (396 mL / 416 mL) * 100%
= 95.19%
So, the volume percent of isopropanol in the solution is approximately 95.19%. This value should be reported with the right number of significant digits, which is 4 in this case. Therefore, the volume percent of isopropanol in the solution is 95.19%.
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