A student needs to dilute a 0.34 M Pb(NO₃)₂ solution to make 81.0 mL of 0.18 M Pb(NO₃)₂. Set up the calculation by placing the values with the correct units into the equation. Then, calculate the volume, in milliliters, of the 0.34 M Pb(NO₃)₂ solution that is needed.
Homework Answers
Before
dilution
After dilution
M1 =
0.34M
M2 = 0.18M
V1
=
V2 = 81ml
from dilution law
M1V1 = M2V2
V1 = M2v2/M1
= 0.18*81/0.34
= 42.88ml >>>>answer
24.
stock
solution
diluted solution
M1 =
5.1M
M2 = 0.80M
V1
=
V2 = 50ml
M1V1 = M2V2
M1 =
M2V2/V1
=
0.80*50/5.1
=
7.84ml >>>>>answer
volume = 7.84ml >>>>answer
25.
Ba(OH)2 is strong base
Ba(OH)2(aq) --------> Ba^2+ (aq)
+ 2OH^- (aq)
0.550M------------- 0.550M
----- 2*0.550M
[Ba^2+] = 0.550M
[OH^-] = 2*0.550 = 1.1M
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A student needs to dilute a 0.34 M Pb(NO₃)₂ solution to make 81.0 mL of 0.18 M Pb(NO₃)₂
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