Question

How many grams of KOH are needed to prepare 250 milliliters of a 2.00M solution of KOH?

Answer 1

Molar mass KOH = 56.1056 g/mol

250 mL / 1000 = 0.25 L 

mass = molarity x molar mass x volume

mass = 2. 00 x 56.1056 x 0.25

mass = 28.0528 g of KOH 

hope this helps!

Answer 2

Answer : The mass of KOH needed are, 28 grams

Explanation :

Molarity : It is defined as the number of moles of solute present in one liter of solution.

In this question, the solute is KOH.

Formula used :

$Molarity=\frac{w_b}{M_b\times V}\times 1000$

where,

$w_b$ = mass of solute KOH = ?

$M_b$ = molar mass of solute KOH = 56 g/mole

V = volume of solution = 250 ml

Molarity = 2.00 M = 2.00 mole/L

Now put all the given values in the above formula, we get the mass of KOH.

$2.00mole/L=\frac{w_b}{56g/mole\times 250ml}\times 1000$

$w_b=28g$

Therefore, the mass of KOH needed are, 28 grams