Question

How many grams of potassium hydroxide, KOH, would be dissolved in 250 mL

a 0.10M solution of potassium hydroxide?

Answer 1

`\huge{ \boxed{1.4 \: \text{g}}}`

Step-by-step explanation:

The number of grams of KOH required can be found by using the formula;

`c = (m)/(M * v)`

c is the concentration in M , mol/dm³ or mol/L

v is the volume in L or dm³

m is the mass in grams

M is the molar mass in g/mol

Making m the subject we have;

m = c × M × v

From the question;

c = 0.10 M

Molar mass of KOH = 39 + 16 + 1 = 56 g/mol

v = 250 mL

Since the volume is in mL it has to be converted to litres (L ).

`\text{ if} \: 1000 \: ml \: = 1 \: l \\ \text{then} \: 250 \: ml \: = (250 \: ml)/(1000 \: ml) * 1 \: l \\ = 0.25 \: l \\ \therefore \: v = 0.25 \: l`

`\therefore \: m = 0.1 * 56 * 0.25 \\ = 1.4 \: g`