PLEASE SHOW CALCULATIONS 1)Dissolve .5 grams of p-hydroxybenzaldehyde in Glacial

Question

PLEASE SHOW CALCULATIONS

1)Dissolve .5 grams of p-hydroxybenzaldehyde in Glacial accetic acid ( approx 1 ml of Glacial accetic acid per mmol of p-hydroxybenzaldehyde)

How many ml of glacial acetic acid should i use?

2) add 10% of Bromine in acertic acid (pprox 1ml per mmol of p-hydroxybenzaldehyde)

How many ml of Bromine in acetic acid should i add?

p-hydroxybenzaldehyde Molecular weight = 122.12 g/mol

Glacial acetic acid Molecular weight= 60.06 g/mol

Bromine in acetic acid molecular weight= 138.948

Explanation / Answer

Molecular weight of p-hydroxybenzaldehyde = 122.12 g/mol

Number of moles in 0.5 g of p-hydroxybenzaldehyde = 0.5 /122.12 =0.0041 moles = 4.1 mmol      (1mol=1000 mmol)

1. For dissolution of 1 mmol p-hydroxybenzaldehyde, 1 mL of glacial acetic acid is required

Thus glacial acetic acid required for dissolution of 4.1 mmol of p-hydroxybenzaldehyde = 4.1 mL

2. Again, 1 mmol of p-hydroxybenzaldehyde requires 1 mL of 10% of Bromine in acetic acid solution

Thus 4.1 mmol will require 4.1 mL of 10% of Bromine in acetic acid solution

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.