Statement-1: Baryta water becomes turbid on passing ${\mathrm{CO}}_2$ gas tbrough it but turbidity becomes clear on passing more ${\mathrm{CO}}_2$ gas. 

Statement-2: Carbonates give yellowish white precipitate with silver nitrate solution. The precipitate becomes yellow or brown on heating.

Baryta water becomes turbid on passing ${\mathrm{CO}}_2$ gas tbrough it but turbidity becomes clear on passing more ${\mathrm{CO}}_2$ gas. Because baryta water combines with carbon dioxide to form a white precipitate of barium carbonate, but when too much carbon dioxide is expelled, the carbonate ions change into bicarbonate ions.

$\mathrm{Ba}(\mathrm{OH}{)}_2+{\mathrm{CO}}_2\to {\mathrm{BaCO}}_3\downarrow$ (white ppt.)

${\mathrm{BaCO}}_3+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2\to \mathrm{Ba}{\left({\mathrm{HCO}}_3\right)}_2$

When carbonates and silver nitrate combine, a white precipitate of silver carbonates is produced. If the precipitate's solution is heated, silver oxide with a yellow colour is obtained.

${\mathrm{CO}}_3^{2-}+{\mathrm{Ag}}^{+}\to {\mathrm{Ag}}_2{\mathrm{CO}}_3\downarrow$ (white ppt.)

$\left.{\mathrm{Ag}}_2{\mathrm{CO}}_3\stackrel{\text{ boiled }}{\to }{\mathrm{Ag}}_2{\mathrm{O}}_3\downarrow \text{ (yellow/brown ppt. }\right)+{\mathrm{CO}}_2\uparrow$

So both statements are true but statement 2 is not a correct explanation for statement-1

As a result, choice B is correct.