Question

Calculate the molarity of a solution made by adding 45.4 g of nano3 to a flask and dissolving it with water to create a total volume of 2.50 l.

Answer 1

The molarity of ${\text{NaN}}{{\text{O}}_{\text{3}}}$ solution is $\boxed{{\text{0}}{\text{.214 M}}}$.

Further Explanation:

The proportion of substance in the mixture is called concentration. The most commonly used concentration terms are as follows:

1. Molarity (M)

2. Molality (m)

3. Mole fraction (X)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

Molarity is a concentration term that is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by M and its unit is mol/L.

The formula to calculate the molarity of the ${\text{NaN}}{{\text{O}}_{\text{3}}}$solution is as follows:

${\text{Molarity of NaN}}{{\text{O}}_{\text{3}}}{\text{ solution}} = \frac{{{\text{Moles}}\;{\text{of}}\;{\text{NaN}}{{\text{O}}_{\text{3}}}}}{{{\text{Volume }}\left( {\text{L}} \right){\text{ of}}\;{\text{NaN}}{{\text{O}}_{\text{3}}}{\text{ solution}}}}$      …… (1)

The formula to calculate the moles of ${\text{NaN}}{{\text{O}}_{\text{3}}}$is as follows:

${\text{Moles of NaN}}{{\text{O}}_{\text{3}}} = \frac{{{\text{Given mass of NaN}}{{\text{O}}_{\text{3}}}}}{{{\text{Molar mass of NaN}}{{\text{O}}_{\text{3}}}}}$                  …… (2)

The given mass of ${\text{NaN}}{{\text{O}}_{\text{3}}}$ is 45.4 g.

The molar mass of ${\text{NaN}}{{\text{O}}_{\text{3}}}$is 84.99 g/mol.

Substitute these values in equation (2).

$\begin{aligned}{\text{Moles of NaN}}{{\text{O}}_3}&=\left( {{\text{45}}{\text{.4 g}}} \right)\left( {\frac{{{\text{1 mol}}}}{{{\text{84}}{\text{.99 g}}}}} \right)\\&=0.5341\;{\text{mol}}\\\end{aligned}$

Substitute 0.5341 for the moles of ${\text{NaN}}{{\text{O}}_{\text{3}}}$and 2.50 L for the volume of ${\text{NaN}}{{\text{O}}_{\text{3}}}$ solution in equation (1).

$\begin{aligned}{\text{Molarity of NaN}}{{\text{O}}_{\text{3}}}{\text{ solution}}&=\frac{{{\text{0}}{\text{.5341 mol}}}}{{{\text{2}}{\text{.50 L}}}}\\&=0.21{\text{364 M}}\\&\approx{\text{0}}{\text{.214 M}} \\ \end{aligned}$

The molarity of the ${\mathbf{NaN}}{{\mathbf{O}}_{\mathbf{3}}}$solution is 0.214 M.

Learn more:

1. Calculation of volume of gas: brainly.com/question/3636135

2. Determine how many moles of water produce: brainly.com/question/1405182

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: molarity of NaNO3 solution, 2.50 L, volume of NaNO3 solution, moles of NaNO3, given mass, molar mass, 84.99 g/mol, 45.4 g, 0.214 M, NaNO3, molar mass, given mass.

Answer 2

molarity is defined as the number of moles of solute in 1 L of solution.
molar mass of NaNO₃ = 85 g/mol
             
number of moles of NaNO₃ = 45.4 g / 85 g/mol = 0.534 mol
 
there are 0.534 mol in 2.50 L solution
                         
therefore number of NaNO₃ moles in 1 L solution = 0.534 mol / 2.50 L = 0.214 M
molarity of solution is 0.214 M