Respuesta :
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
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
The molarity of [tex]{\text{NaN}}{{\text{O}}_{\text{3}}}[/tex] solution is [tex]\boxed{{\text{0}}{\text{.214 M}}}[/tex].
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 [tex]{\text{NaN}}{{\text{O}}_{\text{3}}}[/tex]solution is as follows:
[tex]{\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}}}}[/tex] …… (1)
The formula to calculate the moles of [tex]{\text{NaN}}{{\text{O}}_{\text{3}}}[/tex]is as follows:
[tex]{\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}}}}}[/tex] …… (2)
The given mass of [tex]{\text{NaN}}{{\text{O}}_{\text{3}}}[/tex] is 45.4 g.
The molar mass of [tex]{\text{NaN}}{{\text{O}}_{\text{3}}}[/tex]is 84.99 g/mol.
Substitute these values in equation (2).
[tex]\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}[/tex]
Substitute 0.5341 for the moles of [tex]{\text{NaN}}{{\text{O}}_{\text{3}}}[/tex]and 2.50 L for the volume of [tex]{\text{NaN}}{{\text{O}}_{\text{3}}}[/tex] solution in equation (1).
[tex]\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}[/tex]
The molarity of the [tex]{\mathbf{NaN}}{{\mathbf{O}}_{\mathbf{3}}}[/tex]solution is 0.214 M.
Learn more:
1. Calculation of volume of gas: https://brainly.com/question/3636135
2. Determine how many moles of water produce: https://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.
