Describe how to make 942 ml of 35% ethyl alcohol using a v:v dilution with 95% e
ID: 280622 • Letter: D
Question
Describe how to make 942 ml of 35% ethyl alcohol using a v:v dilution with 95% ethyl alcohol and deionized water. Show your work. (1 point) 1. 2. Describe how to prepare 500 ml of a 0.8M CaCl2 solution using deionized water and CaCl2 salt. The molecular weight of CaCl2 is 110.98 g/mol. Show your work. (Recall: M-mol/L) (1 point) Use the CIVI-c2V2 equation to determine how to make 600 and 5% yellow solution starting with 100% yellow dye. Do not forget to indicate the amount of diluent needed. Show your work. (1 point) 3. L of a 30%, 10%,Explanation / Answer
1. Given strength c1 : 95 % v/v ---( 95 ml of ethyl alcohol in 100 ml solution )
Required Strength c2 : 35 % and ( 35 ml ethyl alcohol in 100 ml solution )
Required volume v2 : 942 ml
c2/c1 * v2 = 35/95 x 942 = 347 ml
Take 347 ml of 95% v/v ethyl alcohol to id add deionized water to make final volume 942 ml ( 35 % v/v ethyl alcohol).
2. CaCl2 Solution :
1M -- 110.98 g in 1000 ml solution
0.8 M-- 88.78 g in 1000 ml solution == 44.39 g in 500 ml solution
take 44.39 g of CaCl2 , dissolve in about 450 ml of deionized water, add deionized water upto 500 ml (().8M CaCl2 solution)
3. 600 ml = 0.6 ml
5 % === c1=100 % v1 =? c2 = 5 % v2 = 0.6 ml 5/100 * 0.6 = 0.03 ml
10 % === c1= 100 % v1 =? c2 = 10 % v2 = 0.6 ml 10/100 * 0.6 =0.06 ml
30 % ==== c1=100 % v1 =? c2 = 30 % v2 = 0.6 ml 30/100 * 0.6 = 0.18 ml
% Dye Diluent total volume 5 % 0.03 ml = 30 ul 0.57 ml =570 ul 0.6 ml = 600 ul 10 % 0.06 ml = 60 ul 0.54 ml = 540 ul 0.6 ml = 600 ul 30 % 0.18 ml = 180 ul 0.42 ml = 420 ul 0.6 ml = 600 ul