Question

A solution is made by dissolving 0.565 g of potassium nitrate in enough water to make up 250. mL of solution. What is the molarity of this solution?
Please explain and show work.

Answer 1

Molar mass of Potassium Nitrate:-

$\\ \large\sf\longmapsto KNO_3$

$\\ \large\sf\longmapsto 39u+14u+3(16u)$

$\\ \large\sf\longmapsto 53u+48u$

$\\ \large\sf\longmapsto 101u$

$\\ \large\sf\longmapsto 101g/mol$

Now

$\boxed{\sf No\:of\:moles=\dfrac{Given\:mass}{Molar\:mass}}$

$\\ \large\sf\longmapsto No\:of\:moles=\dfrac{0.565}{101}$

$\\ \large\sf\longmapsto No\:of\:moles=0.005mol$

We know

$\boxed{\sf Molarity=\dfrac{Moles\:of\:solute}{Vol\:of\:Solution\:in\:L}}$

$\\ \large\sf\longmapsto Molarity=\dfrac{0.005}{\dfrac{250}{1000}L}$

$\\ \large\sf\longmapsto Molarity=\dfrac{0.005}{0.250}$

$\\ \large\sf\longmapsto Molarity=0.02M$

Answer 2

Explanation:

Molar mass of Potassium Nitrate:-

\begin{gathered}\\ \large\sf\longmapsto KNO_3\end{gathered}

⟼KNO

3

\begin{gathered}\\ \large\sf\longmapsto 39u+14u+3(16u)\end{gathered}

⟼39u+14u+3(16u)

\begin{gathered}\\ \large\sf\longmapsto 53u+48u\end{gathered}

⟼53u+48u

\begin{gathered}\\ \large\sf\longmapsto 101u\end{gathered}

⟼101u

\begin{gathered}\\ \large\sf\longmapsto 101g/mol\end{gathered}

⟼101g/mol

Now

\boxed{\sf No\:of\:moles=\dfrac{Given\:mass}{Molar\:mass}}

Noofmoles=

Molarmass

Givenmass

\begin{gathered}\\ \large\sf\longmapsto No\:of\:moles=\dfrac{0.565}{101}\end{gathered}

⟼Noofmoles=

101

0.565

\begin{gathered}\\ \large\sf\longmapsto No\:of\:moles=0.005mol\end{gathered}

⟼Noofmoles=0.005mol

We know

\boxed{\sf Molarity=\dfrac{Moles\:of\:solute}{Vol\:of\:Solution\:in\:L}}

Molarity=

VolofSolutioninL

Molesofsolute

\begin{gathered}\\ \large\sf\longmapsto Molarity=\dfrac{0.005}{\dfrac{250}{1000}L}\end{gathered}

⟼Molarity=

1000

250

L

0.005

\begin{gathered}\\ \large\sf\longmapsto Molarity=\dfrac{0.005}{0.250}\end{gathered}

⟼Molarity=

0.250

0.005

\begin{gathered}\\ \large\sf\longmapsto Molarity=0.02M\end{gathered}

⟼Molarity=0.02M