PLEASE SHOW CALCULATIONS 1)Dissolve .5 grams of p-hydroxybenzaldehyde in Glacial
ID: 499371 • Letter: P
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