Answer to Question #317084 in Trigonometry for christian seit

Question

. How many grams of (g) can be stored in a 10.0 L container at 1000 kPa and 30C?

Answer 1

Solution

Data Given;

Convert degrees to K

$Temperature = 30°C = 30 + 273 = 303K$

$Pressure = 100Kpa = 100×1000 = 100000pa$

Conversion of Pascal pressure to Atmospheric pressure

Let the unknown pressure be = x

$1 atm = 1.01325×10^5pa$

x atm = 100000pa

$x = 100000/1.01325×10^5 = 0.9869 atm$

Pressure = 0.9869 atm

Therefore,

100Kpa = 100000pa = 0.9869 atm

From general gas equation,

PV = nRT

Where P = Pressure

V = Volume

n = Number of moles

$R = Constant = 0.0821 atm L/ mole K$

T = Temperature

PV = nRT

n = PV/RT

$n = 0.9869×\frac{10}{0.0821}×303$

n = 9.869/24.8763 = 0.3967 moles

Recall that,

Number of moles (n) Mass/Molar mass

Mass = n × Molar mass

But Molar mass of Cl2 $= 35.5×2$ = 71 g/mole

$Mass = 0.3967\times71$

= 28.1657 ≈ 28.17g

The mass of Chlorine that can be stored in the container is = 28.17g