Chemistry, 22.07.2019 19:00 tadvet5685

To what volume should you dilute 51 ml of a 13 m stock hno3 solution to obtain a 0.126 m hno3 solution?

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51 mL ${\text{HN}}{{\text{O}}_{\text{3}}}$ solution is diluted to $\boxed{{\text{5210}}{\text{.9 mL}}}$ to get the 0.126 M of solution.

Further explanation:

Molarity:

The molarity of the solution is defined as the concentration of the solution. It is equal to the number of moles of the solute dissolved in one liter of the solution.

The formula of molarity (M), volume (V), and number of moles (n) is as follows:

${\text{M}} = \frac{{\text{n}}}{{\text{V}}}$ …… (1)

Here V is a volume of solution in liters and n is a number of moles of solute.

If n is constant and is equal to k, the equation (1) will become:

${\text{MV}} = {\text{k}}$ …... (2)

If we have initial and final values of concentration and volume then the equation (2) can be modified as follows:

${{\text{M}}_{\text{1}}}{{\text{V}}_{\text{1}}}={{\text{M}}_{\text{2}}}{{\text{V}}_{\text{2}}}$ ……. (3)

Here,

${{\text{M}}_1}$ is the concentration before dilution.

${{\text{M}}_2}$ is the concentration after dilution.

${{\text{V}}_1}$ is the volume before dilution.

${{\text{V}}_2}$ is the volume after dilution.

On rearranging equation (3) for ${{\text{V}}_{\text{2}}}$ we get,

${{\text{V}}_{\text{2}}}=\frac{{{{\text{M}}_1}{{\text{V}}_1}}}{{{{\text{M}}_{\text{2}}}}}$ …… (4)

${{\text{M}}_1}$ is 13 M.

${{\text{M}}_2}$ is 0.126 M.

${{\text{V}}_1}$ is 51 mL.

Substitute the values of ${{\text{M}}_1}$ , ${{\text{M}}_2}$ and ${{\text{V}}_1}$ in equation(4).

$\begin{aligned}{{\text{V}}_{\text{2}}}&=\frac{{\left({{\text{13 M}}}\right)\times\left({{\text{51 mL}}}\right)}}{{{\text{(0}}{\text{.126}}\;{\text{M)}}}}\\&={\text{5261}}{\text{.90 mL}}\\\end{aligned}$

The volume of solution after dilution is 5261.90 mL.

The volume added to initial volume of solution is calculated as follows:

$\begin{aligned}{{\text{V}}_{{\text{added}}}}&={{\text{V}}_{\text{2}}}-{{\text{V}}_{\text{1}}} \\&={\text{5261}}{\text{.90 mL}}-{\text{51}}\;{\text{mL}}\\&={\text{5210}}{\text{.9}}\;{\text{mL}}\\\end{aligned}$

Hence, 51 mL is diluted to 5210.9 mL to get the 0.126 M of solution.

Learn more:

1. Write the equation to show how cation acts as an acid .

2. Calllculation of heat of vaporization:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Mole concept

Keywords: Number of moles, concentration, volume, volume of solution after dilution, solute, 1liter, molarity, 51 mL, 5210.9 mL, 13m, 0.126m.

Answer from: Quest

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. $hbr$ completely dissociates to $h^+$ and $br-$ ions. a is incorrect because $h_2s$ does not fully ionize to $h^+$ and $hs^-$ ions and some remain as the compound $h_2s$ . c and d are incorrect because, $li_2o$ and $libr$ do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise, if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both $li_2o$ and $libr$ have a complete octet so can not accept any electrons which disqualifies them as acids.

Answer from: Quest

Sorry i dont understand the question can you rephase it?

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To what volume should you dilute 51 ml of a 13 m stock hno3 solution to obtain a 0.126 m hno3 soluti...

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