How can I prepare 0.250 L of 0.085 M potassium dichromate?

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How can I prepare 0.250 L of 0.085 M potassium dichromate?

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Answer 1 · 2015-01-06T10:54:21

Here's how you'd go about doing this:

You know that molarity is defined as the number of moles of solute divided by the total volume of the solution. Let's try and do this intuitively first.

If you want to make 1 L of 0.085 M potassium dichromate solution, you need to add 0.085 moles of $K_{2} C r_{2} O_{7}$ $K_2Cr_2O_7$ to enough water to make 1 L of solution. This will give you

$C = \frac{n}{V} = \frac{0.085 moles}{1 L} = 0.085$ $C= n/V = ("0.085 moles")/("1 L") = 0.085$ $M$ $"M"$

Now, since you are dealing with 0.250 L, which is $1/4th$ $"1/4th"$ of a liter, you'd need 4 times less moles of $K_{2} C r_{2} O_{7}$ $K_2Cr_2O_7$ in order to keep the molarity unchanged. So, even without using the formula, you'd have an idea on how much $K_{2} C r_{2} O_{7}$ $K_2Cr_2O_7$ to use.

Using the formula will give you

$C = \frac{n}{V} = \frac{n}{0.250 L} = 0.085$ $C = n/V = n/("0.250 L") = 0.085$ $M$ $"M"$

$n = 0.085 M \cdot 0.250 L = 0.021$ $n = 0.085M * 0.250L = 0.021$ $moles$ $"moles"$

Knowing that $K_{2} C r_{2} O_{7}$ $K_2Cr_2O_7$ has a molar mass of $294.2 g/mol$ $"294.2 g/mol"$ , the mass you'd need for this solution will be

$m_{K_{2} C r_{2} O_{7}} = 0.021$ $m_(K_2Cr_2O_7) = 0.021$ $moles \cdot 294.2 \frac{g}{m o l} = 6.18$ $"moles" * 294.2 g/(mol) = 6.18$ $g$ $"g"$ .

So, adding 6.18 g of potassium dichromate to enough water to make 0.250L will get a 0.085 M solution.

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How can I prepare 0.250 L of 0.085 M potassium dichromate?

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