Answer:
You will have 19.9L of Cl2
Explanation:
We can solve this question using:
PV = nRT; V = nRT/P
<em>Where V is the volume of the gas</em>
<em>n the moles of Cl2</em>
<em>R is gas constant = 0.082atmL/molK</em>
<em>T is 273.15K assuming STP conditions</em>
<em>P is 1atm at STP</em>
The moles of 63g of Cl2 gas are -molar mass: 70.906g/mol:
63g * (1mol / 70.906g) = 0.8885 moles
Replacing:
V = 0.8885mol*0.082atmL/molK*273.15K/1atm
V = You will have 19.9L of Cl2
They have to form a chemical bond in order to brake them down first
Mass of solute ( m1 ) = 50.0 g
mass of solvent ( m2 ) = 150.0 g
Therefore:
m/m = ( m1 / m1 + m2 )
m/m = ( 50.0 / 50.0 + 150.0 )
m/m = ( 50.0 / 200 )
m/m = 0.25