ts) (A) How many grams of aluminum sulphate, Al,(sO,)j should be us 6.05 pf0.018
ID: 103991 • Letter: T
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ts) (A) How many grams of aluminum sulphate, Al,(sO,)j should be us 6.05 pf0.0180 Maluminum sulphate solution Al=26.9815; S=32.065 ; 0 ) aum sulphate, Al,(So , should be used to make on? [A1-26.9815 ; S=32.065 O=15.999 ] 3.00 L o MW: 340, IS D.0150 (B) How many millliters of the solution prepared in (A) should be diluted to make 50.00 mL of 2.00 x 104 M aluminum sulphate solution? 3 DOL 018 BONUS solution prepared in B if the absolute uncertainty of the Die 2.00 solution in B is + 0.05, and the original solution in A h000t alum of the pipe e used to draw the volume (you 50.00 mL flask used to dilute the final ett ind the bsolute uncertinn in A has a concentration and absolute ) if he aphate solution (5 points) Find the %relative uncertainty of the M ±0.0009 aluminum sulphatehasac00m: o.os 00003Explanation / Answer
6 .
A) no. of mole = molarity X volume of soltion in liter
no. of mole of Al2(SO4)3 required = 0.0180 X 3.0 = 0.054 mole
0.054 mole of Al2(SO4)3 requied
molar mass of Al2(SO4)3 = 2(26.9815) + 3(32.065) + 12(15.999) = 342.146 gm/mole
then 0.054 mole of Al2(SO4)3 = 0.054 X 342.146 = 18.48 gm
18.48 gm of Al2(SO4)3 required to make 3 liter of 0.0180 M solution of Al2(SO4)3
B) to calculate volume required to make 50 ml 2 X 10-4 M solution of Al2(SO4)3
Use the formula C1V1 =C2V2
Where, C1 = initial concentration = 0.0180M
V1 = initial volume = ?
C2 = final concentration = 2 X 10-4 M
V2 = final volume = 50 ml
C1V1 =C2V2 Thus
V1 = C2V2 /C1
Substitute the value in above equation
V1 = (2 X 10-4) X 50 / 0.0180 = 0.56 ml
0.56 ml of 0.0180 M Al2(SO4)3 required to make 50 ml of 2.0 X 10-4 M solution of Al2(SO4)3
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