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 $m o l$ $mol$ $/$ $/$ $d m^{3}$ $dm^3$ ?

Question

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 $m o l$ $mol$ $/$ $/$ $d m^{3}$ $dm^3$ ?

1 Answer

Impact of this question

12228 views around the world

You can reuse this answer
Creative Commons License

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?

Science

Math

Humanities

... and beyond

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 $m o l$ $mol$ $/$ $/$ $d m^{3}$ $dm^3$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

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?

1 Answer

Explanation:

Related questions

Impact of this question

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 $m o l$ $mol$ $/$ $/$ $d m^{3}$ $dm^3$ ?