Question

1- An ideal gas occupies a volume of 0.224 m3 at STP (i.e 0°C; 1 atm). How many gas particles are their in the container?

2- When 100 mL pH=12 NaOH solution and 900 mL water are mixed ...?

Answer 1

at STP volume occupied by 1 mol of gas= 22.4 L

given volume = 0.224 m³= 0.000224 L

so no. of moles of gas= 0.000224/22.4

= 10^-5 moles

1 mole = 6.02* 10²³ particles

10^-5 moles = 6.02 * 10¹⁸ particle.

2).

pH=12

pOH = 14-12= 2

[OH^-] = 0.01 M

mmoles of OH^- = 0.01*100 = 1 mmol

total volume = 100+900 mL = 1000mL

concentration of OH^-= 1/1000 = 10^-3 M

pOH = - log[OH^-]

pOH= 3

pH= 14-3=11