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snow_tiger [21]
3 years ago
9

Suppose that you add 26.7 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the adde

d solute, you find that there is a freezing point depression of 2.74 oC compared to pure benzene. What is the molar mass of the unknown compound
Chemistry
1 answer:
nadya68 [22]3 years ago
4 0

Answer: The molar mass of the unknown compound is 200 g/mol

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=2.74^0C = Depression in freezing point

i= vant hoff factor = 1 (for molecular compound)

K_f = freezing point constant = 5.12^0C/m

m= molality

\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

Weight of solvent (benzene)= 0.250 kg  

Molar mass of solute = M g/mol

Mass of solute  = 26.7 g

2.74^0C=1\times 5.12\times \frac{26.7g}{Mg/mol\times 0.250kg}

M=200g/mol

Thus the molar mass of the unknown compound is 200 g/mol

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3 years ago
Determine whether each description applies to electrophilic aromatic substitution or nucleophilic aromatic substitution.
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Answer:

a. electrophilic aromatic substitution

b. nucleophilic aromatic substitution

c. nucleophilic aromatic substitution

d. electrophilic aromatic substitution

e. nucleophilic aromatic substitution

f. electrophilic aromatic substitution

Explanation:

Electrophilic aromatic substitution is a type of chemical reaction where a hydrogen atom or a functional group that is attached to the aromatic ring is replaced by an electrophile. Electrophilic aromatic substitutions can be classified into five classes: 1-Halogenation: is the replacement of one or more hydrogen (H) atoms in an organic compound by a halogen such as, for example, bromine (bromination), chlorine (chlorination), etc; 2- Nitration: the replacement of H with a nitrate group (NO2); 3-Sulfonation: the replacement of H with a bisulfite (SO3H); 4-Friedel-CraftsAlkylation: the replacement of H with an alkyl group (R), and 5-Friedel-Crafts Acylation: the replacement of H with an acyl group (RCO). For example, the Benzene undergoes electrophilic substitution to produce a wide range of chemical compounds (chlorobenzene, nitrobenzene, benzene sulfonic acid, etc).

A nucleophilic aromatic substitution is a type of chemical reaction where an electron-rich nucleophile displaces a leaving group (for example, a halide on the aromatic ring). There are six types of nucleophilic substitution mechanisms: 1-the SNAr (addition-elimination) mechanism, whose name is due to the Hughes-Ingold symbol ''SN' and a unimolecular mechanism; 2-the SN1 reaction that produces diazonium salts 3-the benzyne mechanism that produce highly reactive species (including benzyne) derived from the aromatic ring by the replacement of two substituents; 4-the free radical SRN1 mechanism where a substituent on the aromatic ring is displaced by a nucleophile with the formation of intermediary free radical species; 5-the ANRORC (Addition of the Nucleophile, Ring Opening, and Ring Closure) mechanism, involved in reactions of metal amide nucleophiles and substituted pyrimidines; and 6-the Vicarious nucleophilic substitution, where a nucleophile displaces an H atom on the aromatic ring but without leaving groups (such as, for example, halogen substituents).

3 0
3 years ago
A solution contains 1 LaTeX: \times\:×10−4 M OH– ions. Calculate the solution pH value, and determine if the solution is acidic,
oksano4ka [1.4K]

<u>Answer:</u> The pH value of the solution is 10 and the solution is basic in nature.

<u>Explanation:</u>

To calculate the pH of the solution, we need to determine pOH of the solution. To calculate pOH of the solution, we use the equation:

pOH=-\log[OH^-]

We are given:

[OH^-]=1\times 10^{-4}M

Putting values in above equation, we get:

pOH=-\log(1\times 10^{-4})\\\\pOH=4

To calculate pH of the solution, we use the equation:

pH+pOH=14\\pH=14-4=10

There are three types of solution: acidic, basic and neutral

To determine the type of solution, we look at the pH values.

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  • The pH range of basic solution is 7.1 to 14
  • The pH of neutral solution is 7.

As, the pH of the solution is 10 and is lying in the range of basic solution, so the solution is basic in nature.

5 0
3 years ago
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