$解：(1)所得溶液溶質(zhì)質(zhì)量分?jǐn)?shù)\frac {\text{80g×0.1+20g}}{(80+20)\text{g}}×100\%=28\%\ \ \ 答：所得溶液溶質(zhì)的質(zhì)量分?jǐn)?shù)是28\%。\ \ \ \ \ $
$(2)所得溶液溶質(zhì)質(zhì)量分?jǐn)?shù)：\frac {\text{80g×0.1}}{(80+20)\text{g}}×100\%=8\%\ \ \ \ 答：所得溶液質(zhì)量分?jǐn)?shù)是8\%。\ \ \ \ \ $
$(3)所得溶液質(zhì)量分?jǐn)?shù)：\frac {\text{80g×0.1}}{(80-20)\text{g}}×100\%=13.3\%\ \ \ \ 答：所得溶液質(zhì)量分?jǐn)?shù)是13.3\%。\ \ \ \ $
$\ (4)設(shè)加入的硝酸鈉質(zhì)量為x。$
$\mathrm{\frac {\text{80g×0.1}+x}{80\text{g}+x}×100\%=20\%}\ \ \ 解得：x=10\mathrm{g}$
$答：應(yīng)該再加入10\mathrm{g}硝酸鈉。 $
