How do you calculate the number of moles of potassium hydroxide that must be dissolved to make the following solution: 500 $c m^{3}$ $cm^3$ of 1 $m o l$ $mol$ $/$ $/$ $d m^{3}$ $dm^3$ ?

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How do you calculate the number of moles of potassium hydroxide that must be dissolved to make the following solution: 500 $c m^{3}$ $cm^3$ of 1 $m o l$ $mol$ $/$ $/$ $d m^{3}$ $dm^3$ ?

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Answer 1 · 2016-09-20T17:04:11

We use the relationship, $Concentration$ $"Concentration"$ $=$ $=$ $\frac{Moles of solute}{Volume of solution}$ $"Moles of solute"/"Volume of solution"$ . $0.500 \cdot m o l$ $0.500*mol$ $K O H$ $KOH$ are required.

Explanation:

We are given that $Concentration$ $"Concentration"$ $=$ $=$ $1 \cdot m o l \cdot d m^{- 3}$ $1*mol*dm^-3$ , and that the volume of this solution is $500 \cdot c m^{3}$ $500*cm^3$ $=$ $=$ $0.500 \cdot d m^{3}$ $0.500*dm^3$ .

Thus $moles of solute$ $"moles of solute"$ $=$ $=$ $concentration \times volume$ $"concentration "xx" volume"$ $=$ $=$ $1*mol*cancel(dm^-3)xx0.500*cancel(dm^3)$ $=$ $0.500*mol$ .

What is the mass of the $KOH$ used here?

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How do you calculate the number of moles of potassium hydroxide that must be dissolved to make the following solution: 500 $c m^{3}$ $cm^3$ of 1 $m o l$ $mol$ $/$ $/$ $d m^{3}$ $dm^3$ ?

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How do you calculate the number of moles of potassium hydroxide that must be dissolved to make the following solution: 500 cm^3cm3cm^3 of 1 molmolmol///dm^3dm3dm^3?

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Explanation:

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How do you calculate the number of moles of potassium hydroxide that must be dissolved to make the following solution: 500 $c m^{3}$ $cm^3$ of 1 $m o l$ $mol$ $/$ $/$ $d m^{3}$ $dm^3$ ?