What observed rotation is expected when a 1.44 M solution of (R)-2-butanol is mixed with an equal volume of a 0.720 M solution of racemic 2-butanol, and the resulting solution is analyzed in a sample container that is 1 dm long? The specific rotation of (R)-2-butanol is –13.9 degrees mL g–1 dm–1.
Is it not -1.85?
To determine the observed rotation of the solution, we need to calculate the final concentration and specific rotation of the mixture.
Let's begin by calculating the final concentration of the mixture. Since we are mixing equal volumes, the resulting solution will have a total volume of 2 dm (1 dm + 1 dm).
For the (R)-2-butanol solution:
C1 = 1.44 M (concentration)
V1 = 1 dm (volume)
For the racemic 2-butanol solution:
C2 = 0.720 M (concentration)
V2 = 1 dm (volume)
We can use the formula for dilution:
C1 * V1 = C2 * V2
Substituting the values:
1.44 M * 1 dm = C2 * 1 dm
By rearranging the equation, we can find the concentration of the racemic solution:
C2 = (1.44 M * 1 dm) / (1 dm)
C2 = 1.44 M
Now that we have the final concentration, we can calculate the observed rotation using the formula:
Observed rotation = (specific rotation * concentration * path length) / 10
Specific rotation = -13.9 degrees mL g^(-1) dm^(-1)
Concentration = 1.44 M
Path length = 1 dm
Now we substitute the values into the formula:
Observed rotation = (-13.9 degrees mL g^(-1) dm^(-1) * 1.44 M * 1 dm) / 10
Calculating this will give us the observed rotation.