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

Answer: abeexplanation:

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.

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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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