Question

0 ml of a 6.0 m hno3 stock solution is diluted using water to 100 ml. how many moles of hno3 are present in the dilute solution? 25.0 ml of a 6.0 m stock solution is diluted using water to 100 ml. how many moles of are present in the dilute solution? 6.0 moles 0.6 moles 0.15 moles 1.5 moles

Answer 1

When you are dealing with mixtures that are non-reactive such as HNO3 or nitric acid and water, you do a process of dilution. All you did was decrease the concentration of the HNO3 solution, but it still contains the same amount of moles for the solute. In the concept of law of conservation of mass, the amount of mass or moles (if non-reactive) are additive, no more no less. The final moles would then be the original moles of solute and the added water. 

You can solve this by multiplying the concentration in molarity with the volume. Molarity is moles of solute per liter solution. Since the volume is 100 mL or 0.1 L, then

6 moles/L * 0.1 L = 0.6 moles nitric acid

Answer 2

$\boxed{{\text{0}}{\text{.15 mol}}}$ of ${\text{HN}}{{\text{O}}_{\text{3}}}$ is present in the dilute solution.

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 ${\text{HN}}{{\text{O}}_{\text{3}}}$ solution is as follows:

${\text{Molarity of HN}}{{\text{O}}_3}\;{\text{solution}} = \frac{{{\text{Moles}}\;{\text{of}}\;{\text{HN}}{{\text{O}}_3}}}{{{\text{Volume }}\left( {\text{L}} \right){\text{ of}}\;{\text{HN}}{{\text{O}}_3}\;{\text{solution}}}}$            …… (1)

Rearrange equation (1) to calculate the moles of ${\text{HN}}{{\text{O}}_{\text{3}}}$.

${\text{Moles}}\;{\text{of}}\;{\text{HN}}{{\text{O}}_3} = \left( {{\text{Molarity of HN}}{{\text{O}}_3}\;{\text{solution}}} \right)\left( {{\text{Volume of}}\;{\text{HN}}{{\text{O}}_3}\;{\text{solution}}} \right)$    …… (2)

The volume of ${\text{HN}}{{\text{O}}_{\text{3}}}$ solution is to be converted into L. The conversion factor for this is,

${\text{1 mL}} = {10^{ - 3}}\;{\text{L}}$  

So the volume of ${\text{HN}}{{\text{O}}_{\text{3}}}$ solution is calculated as follows:

$\begin{aligned}{\text{Volume of HN}}{{\text{O}}_{\text{3}}}\;{\text{solution}}&=\left( {{\text{25 mL}}}\right)\left({\frac{{{{10}^{ - 3}}\;{\text{L}}}}{{{\text{1 mL}}}}} \right)\\&=0.02{\text{5 L}}\\\end{aligned}$

The molarity of ${\text{HN}}{{\text{O}}_{\text{3}}}$ solution is 6M.

The volume of ${\text{HN}}{{\text{O}}_{\text{3}}}$ solution is 0.025 L.

Substitute these values in equation (2).

$\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{HN}}{{\text{O}}_3}&=\left( {{\text{6 M}}} \right)\left( {0.02{\text{5 L}}} \right)\\&=0.1{\text{5 mol}} \\ \end{aligned}$

The number of moles of ${\text{HN}}{{\text{O}}_{\text{3}}}$ in 25 mL solution is 0.15 mol.

Dilution is the conversion of a concentrated solution into a dilute solution with the addition of extra solvent but the amount of solute is unaltered. The change that arises is an increase in the volume of the solution.

In the given solution, dilution is done and the concentration of solution decreases during the process. But the number of moles of solute remains unaltered. Therefore the number of ${\text{HN}}{{\text{O}}_{\text{3}}}$ in the dilute solution is also 0.15 mol.

Learn more:

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

2. Determine the moles of water produced: https://brainly.com/question/1405182

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: molarity, HNO3, dilution, moles of HNO3, volume, solution, 0.15 mol, concentration, 25 mL, 6 M, concentrated solution, dilute solution.