Answer:
1. Vapor pressure of dry oxygen gas = 747.68 torr
2. Volume at STP = 39.97 mL
3. Number of oxygen gas molecules = 1.074 × 10²¹ molecules
4. Percent purity of KClO3 = 97.3 %
Explanation:
The balanced equation for the reaction is given below :
2 KClO3 (s) ------> 2 KCl (s) + 3 O2 (g)
1) Since the water level in the eudiometer was below the outside water level in the beaker,
Vapor pressure of dry oxygen gas = Total pressure + pressure due to difference in water levels - vapor pressure of water
Vapor pressure of water at 20 °C is 17.535 mm (torr).
Pressure due to difference in water level = 4.22 cm × 10mm/cm / 13.534 (13.534 is the density of mercury) = 3.118 mm (torr).
Vapor pressure of dry oxygen gas = 762.10 torr + 3.118 torr - 17.535 torr
Vapor pressure of dry oxygen gas = 747.68 torr
2) P₁ = 747.68 torr; V₁ = 43.60 ml; T1 = 20 °C + 273.15 = 293.15 K
P₂ = 760 torr; T₂ = 273.15 K; V₂ = ?
Using the general gas equation = P₁V₁/T₁ = P₂V₂/T₂
V2₂= P₁V₁T₂ / P₂T₁
V₂ = (747.68 × 43.60 × 273.15 ) / (760 × 293.15)
V₂ = 39.97 ml
Volume of dry oxygen gas at STP = 39.97 mL
3) Volume of oxygen gas at STP 39.97 mL = 0.03997 L
Number of moles of oxygen gas in 0.03997 L = volume of gas at STP /molarvolume at STP
Number of moles of oxygen gas = 0.03997/22.4 L
Number of molecules of oxygen gas = 0.03997/22.4 L × 6.03 × 10²³ molecules
Number of oxygen gas molecules = 1.074 × 10²¹ molecules
e) Number of moles of oxygen gas = 0.03997/22.4 = 0.001784 moles
From the equation, mole ratio of oxygen gas and potassium chlorate is 3 : 2
Moles KClO3 = 2/3 × 0.001784 moles = 0.001189 moles
Molar mass of KClO3 = 39 + 35.5 + 3 × 16 = 122.5 g
Actual mass of KClO3 decomposed = 122.5 grams × 0.001189 mole = 0.146 grams
Percent purity = (actual mass KClO3 decomposed / sample mass of impure KClO3) × 100%
Percent purity = (0.146/0.150) × 100% = 97.3 %
