You will use 0.1 g of benzil. Calculate the correct number of milimoles of benzi

Question

You will use 0.1 g of benzil. Calculate the correct number of milimoles of benzil you will be using? You will purify product by crystallization. What is crystallization techique? technique to purify liquid from mixture of liquids technique to purify liquid from mixture of liquid and solid To make product impure Further purify solid Match the following chemicals with their correct OSHA hazards. Benzil C15H14O Dibenzyl ketone C14H10O2 Ethanol C2H6O Potassium hydroxide KOH That's the following chemicals with the correct molecular weights. Benzil Dibenzyl ketone Ethanol Potassium hydroxide Match the following chemicals with their correct OSHA hazards. Benzil Dibenzyl ketone Ethanol Potassium hydroxide Irritant No known OSHA hazards Flammable liquid, target organ effect, irritant, carcinogen Toxic by ingestion, corrosive

Explanation / Answer

Question 1 )

Given mass of benzil = 0.1 g

We have to find moles of benzil by using following formula

number of moles of benzil = Mass of benzil in g / Molar mass of benzil

molar mass of benzil = 210.23 g per mol

number of moles of benzil = 0.1 g / (210.23 g /mol)=0.00476 mol

Now lets convert moles into milimoles

= 0.00476 mol * 1000 milimoles / 1 mol

= 0.48 milimoles of benzil

so answer is option D = 0.48

Question 2 )

Crystallization is the technique by which solids are purified. It is done by dissolving the solid in the solvent in which solid is soluble and then solvent is evaporated.

so correct answer for this question is

Futher purify solid

Question 3 )

Benzil : C14H10O2

dibenzyl ketone = C15H14O

Ethanol : C2H6O

Potassium hydroxide : KOH

Question 4 )

Benzil :

Chemical formula for benzil = C14H10O2

Molecular weight = sum of atomic weights of all the atom in the chemical formula

= 14C * 12.011 g/mol + 10H * 1.0079 g/mol + 2O* 15.999 g/mol

=210.23 g/mol

so the

Benzil : 210.23 g/mol

Dibenzil ketone : C15H14O

molecular weight = 15C * 12.011 g/mol + 14H * 1.0079 g/mol + O* 15.999 g/mol

=210.27 g/mol

molecular weight of dibenzil ketone = 210.27 g/mol

Ethanol : C2H6O

molecular weight = 2C * 12.011 g/mol + 6H * 1.0079 g/mol + O* 15.999 g/mol

=46.0684 g/mol

= 46.07 g/mol

Potassium hydroxide : KOH

Molecular weight = K * 39.098 g/mol + O* 15.999 g/mol + H * 1.0079 g/mol

= 56.1049 g/mol

=56.11 g/mol

Question 5 )

Benzil : A ) Irritant

Dibenzyl ketone : B: No known OSHA hazards

Ethanol : C : flammable liquid, target organ effect, irritant, carcinogen

Potassium hydroxide : D toxic by ingestion, corrosive

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