Gases and Solutions (Grade 11 NSC Matric Physical Sciences): Flashcards

273 K (0°C) and 101.3 kPa

22.4 dm³

$V_g = 22.4n_g$

$V_A = \frac{a}{b}V_B$

Number in front of substance in balanced equation

Moles of solute per litre of solvent (mol·dm⁻³)

$C = \frac{n}{V}$

Solution with precisely known concentration

$\frac{C_A V_A}{a} = \frac{C_B V_B}{b}$

Divide by 1000