Question #bca11

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Question #bca11

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Answer 1 · 2017-05-06T00:44:31

$1 M$ $"1 M"$

Explanation:

This is a classic example of a dilution problem.

You know that molarity, $c$ $c$ , is defined as the number of moles of solute present per liter of solution. This can be expressed as

$c = \frac{n}{V}$ $c = n/V$

where

$n$ $n$ is the number of moles of solute

$V$ $V$ is the volume of the solution

Now, your initial solution has a concentration $c_{1}$ $c_1$ , contains $n$ $n$ moles of solute, and has a volume

$V_{1} = {250 cm}^{3}$ $V_1 = "250 cm"^3$

Your mixing this solution with ${250 cm}^{3}$ $"250 cm"^3$ of water, which implies that the solution will still contain $n$ $n$ moles of solute, but this time, the volume will be

$V_{2} = {250 cm}^{3} + {250 cm}^{3} = {500 cm}^{3} = 2 \cdot V_{1}$ $V_2 = "250 cm"^3 + "250 cm"^3 = "500 cm"^3 = color(red)(2) * V_1$

This means that the molarity of the solution, $c_{2}$ $c_2$ , will be

$c_{2} = \frac{n}{V_{2}}$ $c_2 = n/V_2$

which is equal to

$c_{2} = \frac{n}{2 \cdot V_{1}} = \frac{1}{2} \cdot \frac{n}{V_{1}} = \frac{1}{2} \cdot c_{1}$ $c_2 = n/(color(red)(2) * V_1) = 1/color(red)(2) * n/V_1 = 1/color(red)(2) * c_1$

Therefore, you can say that

$c_{2} = \frac{1}{2} \cdot 2 M = 1 M$ $c_2 = 1/color(red)(2) * "2 M" = "1 M"$

Long story short, doubling the volume of the solution while keeping the number of moles of solute, i.e. diluting the solution by a dilution factor of $2$ $2$ , constant is equivalent to halving the concentration of the solution.

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Question #bca11

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