91Ó°ÊÓ

We need to calculate the grams of$2.00\mathrm{L}\mathrm{of}\mathrm{a}6.00\mathrm{M}\mathrm{NaOH}$the solution.

Considering,

$$\begin{gathered} \mathrm{V}\left(\mathrm{NaOH}\right)=2.00\mathrm{L} \\ \mathrm{M}=6.00\mathrm{M}=6.00\raisebox{1ex}{$\mathrm{mole}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right. \end{gathered}$$

We know$\mathrm{n}=\mathrm{M}×\mathrm{V}$

Now,

$$\begin{gathered} \mathrm{n}\left(\mathrm{NaOH}\right)=\mathrm{M}×\mathrm{V}\left(\mathrm{NaOH}\right) \\ =6.00\raisebox{1ex}{$\mathrm{mole}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.×2.0\mathrm{L} \\ =12.0\mathrm{mole} \end{gathered}$$

Number of moles$\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}$

$\mathrm{m}=\mathrm{n}×\mathrm{M}$

So,

$$\begin{gathered} \mathrm{m}\left(\mathrm{NaOH}\right)=\mathrm{n}\left(\mathrm{NaOH}\right)×\mathrm{M}\left(\mathrm{NaOH}\right) \\ =12.0\mathrm{mole}×\left(23+1+6\right)\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right. \\ =48\mathrm{g} \end{gathered}$$

We need to calculate the grams of $5.00\mathrm{L}\mathrm{of}\mathrm{a}0.100\mathrm{M}{\mathrm{CaCl}}_2$the solution.

Considering

$$\begin{gathered} \mathrm{V}\left({\mathrm{CaCl}}_2\right)=5.00\mathrm{L} \\ \mathrm{M}=0.100\mathrm{M}=0.100\raisebox{1ex}{$\mathrm{mole}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right. \end{gathered}$$

We know$\mathrm{n}=\mathrm{M}×\mathrm{V}$

Now,

$$\begin{gathered} \mathrm{n}\left({\mathrm{CaCl}}_2\right)=\mathrm{M}×\mathrm{V}\left({\mathrm{CaCl}}_2\right) \\ =0.100\raisebox{1ex}{$\mathrm{mole}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.×5.0\mathrm{L} \\ =0.50\mathrm{mole} \end{gathered}$$

Number of moles$\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}$

$\mathrm{m}=\mathrm{n}×\mathrm{M}$

So,

$$\begin{gathered} \mathrm{m}\left({\mathrm{CaCl}}_2\right)=\mathrm{n}\left({\mathrm{CaCl}}_2\right)×\mathrm{M}\left({\mathrm{CaCl}}_2\right) \\ =0.50\mathrm{mole}×\left(2×35.45+40\right)\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right. \\ =55.45\mathrm{g} \end{gathered}$$

We need to calculate the grams of$175\mathrm{mL}\mathrm{of}\mathrm{a}3.00\mathrm{M}{\mathrm{NaNO}}_3$ the solution.

Considering

V = $175mL=0.175L$

M =$3M$

$\mathrm{n}=\mathrm{M}×\mathrm{V}$

Now,

$$\begin{gathered} \mathrm{n}\left({\mathrm{NaNO}}_3\right)=\mathrm{M}×\mathrm{V}\left({\mathrm{NaNO}}_3\right) \\ =3.000\raisebox{1ex}{$\mathrm{mole}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.×0.175\mathrm{L} \\ =0.525\mathrm{mole} \end{gathered}$$

The number of moles$\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}$

$\mathrm{m}=\mathrm{n}×\mathrm{M}$

So,

$$\begin{gathered} \mathrm{m}\left({\mathrm{NaNO}}_3\right)=\mathrm{n}\left({\mathrm{NaNO}}_3\right)×\mathrm{M}\left({\mathrm{NaNO}}_3\right) \\ =0.525\mathrm{mole}×\left(23+14+3+16\right)\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right. \\ =44.63\mathrm{g} \end{gathered}$$