Answering Questions about Solutions (Leaving Cert Chemistry): Flashcards

Moles of solute per litre of solution

$M = \frac{\text{moles}}{\text{volume (L)}}$

$M = \frac{\text{g/L}}{\text{molecular mass}}$

$\text{g/L} = M \times \text{molecular mass}$

$\text{moles} = \frac{\text{V(cm³)} \times M}{1000}$

$\text{V(cm³)} = \frac{\text{moles} \times 1000}{M}$

1000 cm³

$\text{moles} = \frac{\text{mass}}{\text{molecular mass}}$

Using basic definitions to break down problems

Stays the same

Balanced equation, molarity, volume, molar ratios

Inverse: smaller volume = higher concentration

40 g/mol

Determines how much product forms

Molarity (M)