What is the weight to volume $ppm$ of $0.05 \% w/v$ $CaCl_2$ aqueous solution?

The density of a $3 \, M \, NaCl$ solution is $1.25 \, g \, mL^{-1}$. Calculate the molality of the solution. (in $, m$)

$V_1 \ \text{c.c.}$ of a solution having molarity $M_1$ is diluted to have molarity $M_2$. The volume of water required to be added will be $:-$

Sea water,which can be considered as a $6 \ M$ solution of $NaCl$,has a density of $2 \ g \ mL^{-1}$. The concentration of dissolved oxygen $(O_2)$ in sea water is $5.8 \ ppm$. Then the concentration of dissolved oxygen $(O_2)$ in sea water is $x \times 10^{-4} \ m$. $x = ............$ $(Nearest \ integer)$
Given: Molar mass of $NaCl$ is $58.5 \ g \ mol^{-1}$. Molar mass of $O_2$ is $32 \ g \ mol^{-1}$.

$4.5 \ g$ of compound $A$ $(MW = 90)$ was used to make $250 \ mL$ of its aqueous solution. The molarity of the solution in $M$ is $x \times 10^{-1}$. The value of $x$ is ............ (Rounded off to the nearest integer)

The volume of water added to $500 \ mL$ of $0.5 \ M \ NaOH$ solution so that its strength becomes $10 \ mg \ mL^{-1}$ will be .......... $mL$.