`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

`112.82g`

Explanation

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`.