Questions: Quiz: Pre-lab Quiz 5 Week 6 Purpose Submis: If the nD of a fraction is measured to be 1.4653 at 25 deg C, the corrected nD at 20 deg C is 1.4631 1.5732 1.4653 1.4676 One of the fraction's corrected nD at 20 deg C is 1.4563 on the Reichert refractometer, the % toluene of this fraction is

Solution Steps

We need to correct the refractive index (\(n_D\)) of a fraction measured at 25°C to its value at 20°C. The options provided are:

• 1.4631
• 1.5732
• 1.4653
• 1.4676

The refractive index correction formula for temperature is: \[ n_{D, 20^\circ C} = n_{D, T} - 0.00045 \times (T - 20) \] where \( n_{D, T} \) is the refractive index at temperature \( T \).

Given:

• \( n_{D, 25^\circ C} = 1.4653 \)
• \( T = 25^\circ C \)

Substitute these values into the formula: \[ n_{D, 20^\circ C} = 1.4653 - 0.00045 \times (25 - 20) \] \[ n_{D, 20^\circ C} = 1.4653 - 0.00045 \times 5 \] \[ n_{D, 20^\circ C} = 1.4653 - 0.00225 \] \[ n_{D, 20^\circ C} = 1.46305 \]

Round 1.46305 to four significant digits: \[ n_{D, 20^\circ C} = 1.4631 \]

Final Answer