1. Isoamyl salicylate ( Molar mass = 208.25 g/mol) has a pleasant aroma and is u
ID: 998597 • Letter: 1
Question
1. Isoamyl salicylate ( Molar mass = 208.25 g/mol) has a pleasant aroma and is used in perfumes and soaps. Which of the following combinations gives a 0.75 msolution of isoamyl salicylate in ethyl alcohol (d = 0.7893 g/mL)? (show works and calculations)
a.117.2 g isoamyl salicylate in 950.0 mL of ethyl alcohol
b. 117.2 g isoamyl salicylate in 750.0 mL of ethyl alcohol
c. 117.2 g isoamyl salicylate in 750.0 mL of solution
d. 117.2 g isoamyl salicylate in 592.0 g of ethyl alcohol
e. none of these choices is correct
2.The chemist, Anna Lytic, must prepare 1.00 kg of 15.0% (w/w) acetic acid using a stock solution which is 36.0% (w/w) acetic acid (d = 1.045 g/mL). Which of the following combinations will give her the solution she wants? (show works and calculations)
a. 417 mL of 36% acetic acid in 583 mL of distilled water
b. 417 g of 36% acetic acid in 583 g of distilled water
c. 360 mL of 36% acetic acid in 640 mL of distilled water
d. 360 g of 36% acetic acid in 640 g of distilled water
e. 150 g of 36% acetic acid in 850 g of distilled water
3. Select the weakest electrolyte from the following set (explain why and choose a correct answer)
a. Na2SO4
b. KCl
c. CH3CH2COOH, propionic acid
d. CaCl2
e. LiOH
4. How many moles of bromide ions are present in 750.0 mL of 1.35 M MgBr2(show works and calculations)
a. 0.506 mol
b. 1.01 mol
c. 2.03 mol
d. 3.04 mol
e. none of these choices is correct
Explanation / Answer
1) The density of ethyl alcohol is d = 0.7893 g/mL. Thus the weight of 1000 g of ethyl alcohol is 1000/0.7893 = 1267mL.
Thus 1267mL of ethyl alcohol containing 208.25 g of Isoamyl salicylate is called 1 molal solution.
thus 117.2 g of Isoamyl salicylate in 1267mL of ethyl alcohol = 117.2/208.25= 0.56278 molal solution.
Thus 117.2 g of Isoamyl salicylate in 950 mL of ethyl alcohol = 1267/950 *0.56278= 0.75 molal solution
2)1.00 kg of 15.0% (w/w) acetic acid solution = 1000 g of solution have 150 g of acetic acid.
Volume for 1.00 kg of 36.0% (w/w) acetic acid is 1000/1.045= 956.94 mL
density of 36.0% (w/w) acetic acid is 1.045 g/mL , thus density of 15.0% (w/w) acetic acid = 0.8836 g/mL
It can be calculated as 360 g of acetic acid in 1000mL(or g) of water have density 1.045g/mL , total weight of solution=1360g.
then 150 g of acetic acid in 1000mL(or g) of water have density = 1150*1.045/1360=0.8836 g/mL
In 36.0% (w/w) acetic acid , 100mL have 36 gram of acetic acid.
We need 150 gram of acetic acid, thus volume of 36.0% (w/w) acetic acid needed= (150/36)*100 = 416.67 ~417 mL
Toatl volume of 1.00 kg is 1000/0.89 ~1100ml. Thus volume of water=1100mL-417 mL = 683 mL water should be added.
Option A.
3) a. Na2SO4 , It is insoluble in water. It do not dissociates in water. thus it is the weakest electrolyte .
4)750.0 mL of 1.35 M MgBr2 = 750*1.35 =1012.5 mmol of MgBr2.
1 mol of MgBr2. gives out 2 mol of Br- . thus total Br- given by 1012.5 mmol of MgBr2 = 2*1012.5 =2025 mmol= 2.025 mol of bromide ions.thus closest answer is 2.03 mol.
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