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Answer to Question #11890 in Inorganic Chemistry for Tamim
Question #11890
what is the molality of 13%(w/v) Pb(NO3)2 (1200gm/dm^3) ?
Expert's answer
b=n/m
n -amounth of solute
m - mass of solvent
w/100%=(m1/(m +
m1))
m1-mass of solute
0.13=
130/1000
m1=130
m=1000-130=870ml=0.87
L
n=m/Mr
Mr=331
n=130/331=0.393
b=0.393/0.87=0.4514
mol/L
n -amounth of solute
m - mass of solvent
w/100%=(m1/(m +
m1))
m1-mass of solute
0.13=
130/1000
m1=130
m=1000-130=870ml=0.87
L
n=m/Mr
Mr=331
n=130/331=0.393
b=0.393/0.87=0.4514
mol/L
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