Question

A mysterious white powder could be powdered sugar (C12H22O11), cocaine (C17H21NO4), codeine (C18H21NO3), norfenefrine (C8H11NO2), or fructose (C6H12O6). When 82 mg of the powder is dissolved in 1.50 mL of ethanol (d = 0.789 g/cm3, normal freezing point −114.6 ∘C, Kf = 1.99 ∘C/m), the freezing point is lowered to −115.5 ∘C. What is the identity of the white powder?

Answer 1

Norfenefrine (C₈H₁₁NO₂).

Further explanation

We will solve a case related to one of the colligative properties, namely freezing point depression.

The freezing point of the solution is the temperature at which the solution begins to freeze. The difference between the freezing point of the solvent and the freezing point of the solution is called freezing point depression.

$\boxed{ \ \Delta T_f = T_f(solvent) - T_f(solution) \ } \rightarrow \boxed{ \ \Delta T_f = K_f \times molality \ }$

Given:

A mysterious white powder could be,

• powdered sugar (C₁₂H₂₂O₁₁) with a molar mass of 342.30 g/moles,
• cocaine (C₁₇H₂₁NO₄) with a molar mass of 303.35 g/moles,
• codeine (C₁₈H₂₁NO₃) with a molar mass of 299.36 g/moles,
• norfenefrine (C₈H₁₁NO₂) with a molar mass of 153.18 g/moles, or
• fructose (C₆H₁₂O₆) with a molar mass of 180.16 g/moles.

When 82 mg of the powder is dissolved in 1.50 mL of ethanol (density = 0.789 g/cm³, normal freezing point −114.6°C, Kf = 1.99°C/m), the freezing point is lowered to −115.5°C.

Question: What is the identity of the white powder?

The Process:

Let us identify the solute, the solvent, initial, and final temperatures.

• The solute = the powder
• The solvent = ethanol
• The freezing point of the solvent = −114.6°C
• The freezing point of the solution = −115.5°C

Prepare masses of solutes and solvents.

• Mass of solute = 82 mg = 0.082 g
• Mass of solvent = density x volume, i.e., $\boxed{ \ 0.789 \ \frac{g}{cm^3} \times 1.50 \ cm^3 = 1.1835 \ g = 0.00118 \ kg \ }$

We must prepare the solvent mass unit in kg because the unit of molality is the mole of the solute divided by the mass of the solvent in kg.

The molality formula is as follows:

$\boxed{ \ m = \frac{moles \ of \ solute}{kg \ of \ solvent} \ } \rightarrow \boxed{ \ m = \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }$

Now we combine it with the formula of freezing point depression.

$\boxed{ \ \Delta T_f = K_f \times \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }$

It is clear that we will determine the molar mass of the solute (denoted by Mr).

We enter all data into the formula.

$\boxed{ \ -114.6^0C - (-115.5^0C) = 1.99 \frac{^0C}{m} \times \frac{0.082 \ g}{Mr \times 0.00118 \ kg} \ }$

$\boxed{ \ 0.9 = \frac{1.99 \times 0.082}{Mr \times 0.00118} \ }$

$\boxed{ \ Mr = \frac{0.16318}{0.9 \times 0.00118} \ }$

We get $\boxed{ \ Mr = 153.65 \ }$

These results are very close to the molar mass of norfenefrine which is 153.18 g/mol. Thus the white powder is norfenefrine.

Learn more

1. The molality and mole fraction of water brainly.com/question/10861444
2. About the mass and density of ethylene glycol as an  antifreeze brainly.com/question/4053884
3. About the solution as a homogeneous mixture  brainly.com/question/637791

Keywords: a mysterious white powder, sugar, cocaine, codeine, norfenefrine, fructose, the solute, the solvent, dissolved, ethanol, normal freezing point, the freezing point depression, the identity

Answer 2

The identity of the white powder is $\boxed{\text{norfenefrine}(\text{C}_{8}\text{H}_{11}\text{NO}_{2})}$ .

Further Explanation:

Colligative properties

These are the properties that depend on the number of solute particles and not on their mass or identities. Following are the four colligative properties:

1. Relative lowering of vapor pressure

2. Elevation in boiling point

3. Depression in freezing point

4. Osmotic pressure

The temperature where a substance in its liquid form is converted into the solid state is known as freezing point. Freezing point depression is a colligative property because it depends on the number of moles of solute particles.

The expression for the freezing point depression is as follows:

$\Delta\text{T}_\text{f}=\text{K}_\text{f}\,\text{m}$       …… (1)

                                                                                   

Here,

$\Delta\text{T}_\text{f}$ is the depression in freezing point.

$\text{K}_\text{f}$ is the cryoscopic constant.

m is the molality of the solution.

The formula to calculate the density of substance is as follows:

$\text{Density of substance}=\dfrac{\text{Mass of substance}}{\text{Volume of substance}}$                                                      …… (2)

Rearrange equation (2) for the mass of substance.

$\text{Mass of substance}=(\text{Density of substance })(\text{Volume of substance})$                                 …… (3)

The volume of the substance is to be converted into  . The conversion factor for this is,

$1\,\text{ml}=1\,\text{cm}^3$  

Therefore the volume of the substance can be calculated as follows:

$\begin{aligned}\text{Volume}&=(1.50\,\text{mL})\left(\frac{1\,\text{cm}^3}{1\,\text{mL}}\right)\\&=1.50\,\text{cm}^3\end{aligned}$  

Substitute $1.50\,\text{cm}^3$ for the volume of substance and $0.789\,\text{g/gm}^3$ for the density of the substance in equation (3) to calculate the mass of solvent.

$\begin{aligned}\text{Mass of solvent}&=\left(\dfrac{0.789\,\text{g}}{1\,\text{cm}^3}\right)(1.50\,\text{cm}^3)\\&=1.1835\,\text{g}\end{aligned}$  

This mass is to be converted into kg. The conversion factor for this is,

$1\,\text{g}=10^{-3}\,\text{kg}$  

Therefore the mass of solvent can be calculated as follows:

$\begin{aligned}\text{Mass of solvent}&=(1.1835\,\text{g})\left(\dfrac{10^{-3}\,\text{kg}}{1\,\text{g}}\right)\\&=0.0011835\,\text{kg}\end{aligned}$  

The mass of solute is to be converted into g. The conversion factor for this is,

 $1\,\text{mg}=10^{-3}\,\text{g}$

Therefore the mass of solute can be calculated as follows:

$\begin{aligned}\text{Mass of solute}&=(82\,\text{mg})\left(\dfrac{10^{-3}\,\text{g}}{1\,\text{g}}\right)\\&=0.082\,\text{g}\end{aligned}$  

The freezing point depression can be calculated as follows:

$\begin{aligned}\Delta\text{T}_\text{f}&=-114.6\,^\circ\text{C}-(-115.5\,^\circ\text{C})\\&=0.9\,\circ\text{C}\end{aligned}$

The formula to calculate the molality of solution is as follows:

$\text{Molality of solution}=\dfrac{\text{Amount (mol) of solute}}{\text{Mass (kg) of solvent}}$                                                        …… (4)

The formula to calculate the amount of solute is as follows:

$\text{Amount of solute}=\dfrac{\text{Mass of solute}}{\text{Molar mass of solute}}$                                                              …… (5)

Incorporating equation (5) into equation (4),

$\text{Molality of solution}=\dfrac{\text{Mass of solute}}{(\text{Mass of solvent})(\text{Molar mass of solute})}$                               …… (6)

Incorporating equation (6) into equation (1),

$\Delta\text{T}_\text{f}=\text{k}_\text{f}\left(\dfrac{\text{Mass of solute}}{(\text{Mass of solvent})(\text{Molar mass of solute})}\right)$                                           …… (7)

Rearrange equation (7) to calculate the molar mass of solute.

$\text{Molar mass of solute}=\dfrac{\text{k}_\text{f}}{\Delta\text{T}_\text{f}}\left(\dfrac{\text{Mass of solute}}{\text{Mass of solvent}}\right)$                                                        …… (8)

Substitute $1.99\,^\circ\text{C/m}$ for $\text{k}_\text{f}$, $0.9\,^\circ\text{C}$ for $\Delta\text{T}_\text{f}$, 0.082 g for the mass of solute and 0.0011835 kg for the mass of solvent in equation (8).

$\begin{aligned}\text{Molar mass of solute}&=\left(\dfrac{1.99}{0.9}\right)\left(\dfrac{0.082}{0.0011835}\right)\\&=153.199\,\text{g/mol}\\&=\approx153.2\,\text{g/mol}\end{aligned}$  

The molar mass of powdered sugar is 342.3 g/mol.

The molar mass of cocaine is 303.4 g/mol.

The molar mass of codeine is 299.4 g/mol.

The molar mass of norfenefrine is 153.2 g/mol.

The molar mass of fructose is 180.2 g/mol.

The calculated molar mass of solute is similar to that of norfenefrine. So the identity of the white powder is norfenefrine.

Learn more:

1. What is the concentration of alcohol in terms of molarity? brainly.com/question/9013318
2. What is the molarity of the stock solution? brainly.com/question/2814870

Answer details:

Grade: Senior School

Chapter: Solutions

Subject: Chemistry

Keywords: colligative properties, depression in freezing point, cryoscopic constant, freezing point, 153.199 g/mol, norfenefrine.