xg of NaCl (M=58.5) put in a dry flask and water was added with continous strring to produce exactly 1L solution having a molality of 2.00. Find x using the data below: ltbr. Weight % NaCl Weight % NaCl 0 10 12 14 ltbr. Density of solution in g//mL 1.0591 1.0742 1.0895 1.1049

xg of NaCl (M=58.5) put in a dry flask and water was added with contin

1.	xg of NaCl (M=58.5) put in a dry flask and water was added with continous strring to produce exactly 1L solution having a molality of 2.00. Find x using the data below: ltbr. Weight % NaCl Weight % NaCl 0 10 12 14 ltbr. Density of solution in g//mL 1.0591 1.0742 1.0895 1.1049
Answer» Solution :Molality `=("WT of solute//molar mass of solute")/(("wt of solvent in "GM)/(1000))` `2=((x)/(58.5))/((W)/(1000))` or `(x)/(W)=(2xx58.5)/(1000)=0.117` Now we find `%` wt. of `NaCl` i.e. `(x)/(x+W)xx100=10.4` So with the help of table GIVEN. Density of solution in `gm//mL=1.0742` Now for `1` it solution wt of solution `=1074.2gm` `therefore` wt of solution `=(1074.2-X)gm=W` `because (x)/(W)=0.117` or `(x)/(1074.2-x)=0.117` `therefore x=112.5 gram`.