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docker41 [41]
2 years ago
7

A solution of 2.00 g of para-dichlorobenzene in 50.0 g

Chemistry
1 answer:
vlada-n [284]2 years ago
7 0

Answer:

151 g/mol

Explanation:

When a nonvolatile substance is added to a solvent, the freezing point of the solvent is changed, which is called cryoscopy. When temperature change can be calculated by:

ΔT = Kf*W

Where Kf is the molal freezing point constant of the solvent and W is the molality of the solution.

For cyclohexane, Kf = 20.2 °C/molal, and the freezing point is 6.4 °C, so:

6.4 - 1.05 = 20.2 * W

20.2W = 5.35

W = 0.26485 molal

The molality is:

W = m1/m2*M1

Where m1 is the mass of the solute (in g), m2 is the mass of the solvent (in kg), and M1 is the molar mass of the solute. So:

0.26485 = 2.00/0.05M1

0.0132425M1 = 2.00

M1 = 151 g/mol

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A student chromatographs a mixture and after developing the spots with a suitable reagent he observes the following
Anika [276]

Rf value is the ratio of the distance traveled by the solute to that of the solvent front on the paper used in chromatographic separation.

From the image it is clear the distance traveled by solvent front = 7.3 cm

Distance traveled by the component -1 of the mixture = 1.4 cm

Distance traveled by the component -2 of the mixture = 3.0 cm

Distance traveled by the component -3 of the mixture = 4.5 cm

Distance traveled by the component -4 of the mixture = 6.5 cm

Rf value of component-1 = \frac{1.4 cm}{7.3 cm} =0.192

Rf value of component-2 = \frac{3.0 cm}{7.3 cm} =0.410

Rf value of component-3 = \frac{4.5 cm}{7.3 cm} =0.616

Rf value of component-4 = \frac{6.5 cm}{7.3 cm} =0.890

b) Samples can be separated from a mixture using chromatography as the relative affinities for the compounds towards the paper (stationary phase) and the solvent(mobile phase) are different. Each component spends different amounts of time on the stationary phase depending on it chemical nature. So, the components in a mixture can be separated based on their polarities and relative degrees of adsorption on the stationary phase.








6 0
3 years ago
You observe mothballs disappearing in cabinets. What do you think is the reason for this? Do all substances behave like mothball
cricket20 [7]

Answer: Mothballs have weak intermolecular forces.

No all substances do not behave like mothballs at normal conditions. Example: benzene , chloroform

Explanation:

Sublimation is a process of converting a substance from solid state to gaseous state without the formation of liquid at constant temperature.

A substance which undergoes sublimation is called as sublimating substance.

As mothballs is made of napthalene which has weak inter molecular forces of attraction between its molecules, it directly sublimes into gaseous state without leaving any residue and is called as a sublimating substance.

Not all substances behave like mothballs at normal conditions. Example: benzene , chloroform

8 0
3 years ago
2. Which of the following is the term used to describe a body's resistance to a change in motion? A. Inertia B. Acceleration C.
Vanyuwa [196]
Helpppo meeee pleasee
4 0
2 years ago
Please answer quick
JulijaS [17]
The paper will turn red
6 0
3 years ago
If 8.500 g CH is burned and the heat produced from the burning is added to 5691 g of water at 21 °C, what is the final
mojhsa [17]

The final temperature = 36 °C

<h3>Further explanation</h3>

The balanced combustion reaction for C₆H₆

2C₆H₆(l)+15O₂(g)⇒ 12CO₂(g)+6H₂O(l)  +6542 kJ

MW C₆H₆ : 78.11 g/mol

mol C₆H₆ :

\tt \dfrac{8.5}{78.11}=0.109

Heat released for 2 mol C₆H₆ =6542 kJ, so for 1 mol

\tt \dfrac{0.109}{2}\times 6542=356.539~kJ/mol

Heat transferred to water :

Q=m.c.ΔT

\tt 356.539=5.691~kg\times 4.18~kj/kg^oC\times (t_2-21)\\\\t_2-21=15\rightarrow t_2=36^oC

3 0
2 years ago
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