Question

(f) what is the observed rotation of 100 ml of a solution that contains 0.01 mole of d and 0.005 mole of l? (assume a 1-dm path length.)

Answer 1

Observed rotation of solution that consist of 0.01 mole of D and 0.005 mole L is $\boxed{{{0}}{\text{.01299 }}{\text{degrees}}}$

Further explanation:

The specific rotation associated with a chiral molecule is defined as observed rotation concentration per unit path length of the polarimeter per unit concentration of solute.

The formula to calculate concentration from number of moles is as follows:

${\text{Concentration}}\left( {{\text{g/L}}} \right) = {\text{Molarity}}\left( {{\text{mol/L}}} \right) \times {\text{molar}}\,{\text{mass}}\left( {{\text{g/mol}}} \right)$                                            …… (1)

Substitute $0.100{\text{ M}}$ for molarity and $150{\text{ g/mol}}$ for molar mass in equation (1).

$\begin{aligned}{\text{Concentration}}\left( {{\text{g/L}}} \right) &= 0.100{\text{ M}} \times 150{\text{ g/mol}}\\&= 15{\text{ g/L}}\\\end{aligned}$

The formula to calculate specific rotation is as follows:  

$\left[ \alpha\right]=\dfrac{{\,\alpha }}{{l \times c}}$      …… (2)                                                                        

Here,

$\alpha$ represents the rotation in degrees caused by the polarimeter.

$l$ represents the path length.

$c$ represents the concentration.

Substitute $0.26^\circ$ for $\alpha$ , $1{\text{ dm}}$ for $l$ and $15{\text{ g/L}}$ for $c$.

 $\begin{aligned}\left[ \alpha\right] &= \frac{{0.26^\circ }}{{1{\text{ dm}} \times 15{\text{ g/L}}}} \\&= 0.01733{\text{ deg}} \cdot {\text{L/g}} \cdot {\text{dm}}\\\end{aligned}$

Since the D isomer is present in excess.  Its excess amount can be determined as follows:

$\begin{aligned}{\text{Excess D - isomer}}&= \left( {0.01{\text{ mol}} - {\text{0}}{\text{.005}}\;{\text{mol}}} \right)\\&= 0.005{\text{ mol}}\\\end{aligned}$

The formula to calculate molarity is as follows:

${\text{Molarity}}\left( {{\text{mol/mL}}} \right){\text{ = }}\dfrac{{{\text{Number of moles}}}}{{{\text{volume}}}} \times 1000$  

Substitute $0.005{\text{ mol}}$  for number of moles and $100{\text{ mL}}$ for volume to calculate molarity of excess D-isomer.

$\begin{aligned}(\text{Molarity}\left({mol/mL} \right) &= \frac{{0.005 mol}}{{100}} \times 1000\\&= 0.05 M\\\end{aligned}$

Substitute $0.05 M$ for molarity and $150{\text{ g/mol}}$ for molar mass in equation (1).

$\begin{aligned}{\text{Concentration}}\left( {{\text{g/L}}} \right)&= \left( {0.05 M} \right)\left( {150{\text{ g/mol}}} \right)\\&= 7.5{\text{ g/L}}\\\end{aligned}$

Rearrange equation (2) calculate the observed rotation

$\alpha = \left[ \alpha \right] \times l \times c$                           …… (3)

Substitute $0.01733{\text{ deg}} \cdot {\text{L/g}} \cdot {\text{dm}}$ for $\left[ \alpha \right]$, $1{\text{ dm}}$ for $l$ and ${\text{7}}{\text{.5 g/L}}$ for $c$ in the equation (3).

$\begin{aligned}\alpha&= \left( {0.01733{\text{ deg}} \cdot {\text{L/g}} \cdot {\text{dm}}} \right)\left( {1{\text{ dm}}} \right)\left( {{\text{7}}{\text{.5 g/L}}} \right)\\&= {\text{0}}{\text{.01299 }}{\text{degrees}}\\\end{aligned}$

 

Learn more:

1. Calculation of volume of gas: brainly.com/question/3636135

2. Determine how many moles of water produce: brainly.com/question/1405182

Answer details:  

Grade: Senior School

Subject: Chemistry

Chapter: Stereochemistry

Keywords: specific rotation, concentration, polarimeter, excess D-isomer, molar mass, molarity, chiral, enantiomeric excess, and observed rotation.

Answer 2

Answer: .01 moles of D to .005 moles of L ~ so, .01+.005 = .015 total; using this total value, divide the portions of D and L. so .01/.015 to .005/.015 ~ 67% D to 33% L. And thus, the enantiomer excess will be 34%.