Question

what is the molality of 13%(w/v) Pb(NO3)2 (1200gm/dm^3) ?

Answer 1

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