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The balanced chemical equation is:

${\mathrm{AgNO}}_3\left(\mathrm{aq}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)鈫�\mathrm{AgCl}\left(\mathrm{s}\right)+{\mathrm{NO}}_3^{-}\left(\mathrm{aq}\right)$

The number of moles of ${\mathrm{AgNO}}_3$is:

$\begin{array}{c}\mathrm{moles}\mathrm{of}{\mathrm{AgNO}}_3=0.2970\text{鈥�}\mathrm{moles}/\mathrm{l}脳53.63\mathrm{ml}脳\frac{1\mathrm{lit}}{1000\mathrm{ml}}\\ =0.0159\text{鈥�}\mathrm{mol}\text{鈥�}{\mathrm{AgNO}}_3\end{array}$

The sea water is titrated with the silver nitrate solution.

Therefore,

$\begin{array}{l}\text{V}\left({\text{Cl}}^{\text{-}}\right)\text{脳厂}\left({\text{Cl}}^{\text{-}}\right)\text{=V}\left({\text{AgNO}}_{\text{3}}\right)\text{脳厂}\left({\text{AgNO}}_{\text{3}}\right)\\ 鈬�\text{S}\left({\text{Cl}}^{\text{-}}\right)=\frac{\text{V}\left({\text{AgNO}}_{\text{3}}\right)\text{脳厂}\left({\text{AgNO}}_{\text{3}}\right)}{\text{V}\left({\text{Cl}}^{\text{-}}\right)}\\ 鈬�\text{S}\left({\text{Cl}}^{\text{-}}\right)=\frac{\text{0.2970鈥尘辞濒别蝉/濒鈥夆赌壝�53.63尘濒}}{\text{25ml}}\\ 鈬�\text{S}\left({\text{Cl}}^{\text{-}}\right)=\text{0.637moles/l}\end{array}$

Here,

$\begin{array}{l}\text{痴=惫辞濒耻尘别辞蹿鈥塼丑别鈥夆赌塻辞濒耻迟颈辞苍}\\ \text{厂=蝉迟谤别苍驳迟丑鈥塷蹿迟丑别鈥夆赌塻辞濒耻迟颈辞苍}\end{array}$

Therefore, number of moles of ${\text{Cl}}^{\text{-}}$ present in 25 ml sea water is calculated.

$\begin{array}{l}{\text{numberofmolesofCl}}^{鈭�}\text{=0.637尘辞濒别蝉/濒鈥夆赌夆赌壝�25尘濒脳}\frac{\text{1l}}{\text{1000ml}}\\ =\text{0.0159鈥夆赌尘辞濒别蝉}\end{array}$

The mass of ${\mathrm{Cl}}^{-}$in seawater is calculated as:

$\begin{array}{c}\text{惭补蝉蝉鈥夆赌塷蹿鈥�}1{\text{鈥夆赌尘辞濒别辞蹿颁濒}}^{鈭�}=35.45\text{鈥塯}\\ {\text{惭补蝉蝉辞蹿鈥�0.0159尘辞濒别鈥塷蹿颁濒}}^{鈭�}=35.45\text{鈥塯}脳0.0159\\ =0.563\text{鈥塯}\end{array}$

The mass % of${\mathrm{Cl}}^{-}$ is calculated as:

$\begin{array}{c}\mathrm{mass}\mathrm{\%}=\frac{\mathrm{mass}\mathrm{of}{\mathrm{Cl}}^{-}}{\mathrm{mass}\mathrm{of}\mathrm{sea}\mathrm{water}}\mathrm{x}100\mathrm{\%}\\ \mathrm{mass}\mathrm{\%}=\frac{0.5637\mathrm{gm}\text{鈥夆赌�}{\mathrm{Cl}}^{-}}{(\mathrm{density}\mathrm{of}\mathrm{sea}\mathrm{water})脳(\mathrm{volume}\mathrm{of}\mathrm{sea}\mathrm{water})}\mathrm{x}100\mathrm{\%}\\ =\frac{0.5637\mathrm{gm}}{1.024\mathrm{gm}/\mathrm{ml}脳25\mathrm{ml}}\mathrm{x}100\mathrm{\%}\\ =2.20\text{鈥�}\mathrm{\%}\text{鈥�}{\mathrm{Cl}}^{-}\end{array}$