Boiling point elevation is given as:
ΔTb=iKbm
Where,
ΔTb=elevation in the boiling point
that is given by expression:
ΔTb=Tb (solution) - Tb (pure solvent)
Here Tb (pure solvent)=118.1 °C
i for CaCO3= 2
Kb=2.93 °C/m
m=Molality of CaCO₃:
Molality of CaCO₃=Number of moles of CaCO₃/ Mass of solvent (Kg)
=(Given Mass of CaCO3/Molar mass of CaCO₃)/ Mass of solvent (Kg)
=(100.0÷100 g/mol)/0.4
= 2.5 m
So now putting value of m, i and Kb in the boiling point elevation equation we get:
ΔTb=iKbm
=2×2.93×2.5
=14.65 °C
boiling point of a solution can be calculated:
ΔTb=Tb (solution) - Tb (pure solvent)
14.65=Tb (solution)-118.1
Tb (solution)=118.1+14.65
=132.75
Molar mass:
H₂O = 18.0 g/mol
O₂ = 32.0 g/mol
C₅H₁₂ + 8 O₂ -> 5 CO₂ + 6 H₂<span>O
</span>
8 x (32 g )<span> ------------ 6 x (18 g )</span>
mass O₂ ------------ 108 g H₂O
mass O₂ = 108 x 8 x 32 / 6 x 18
mass O₂ = 27648 / 108
mass O₂ =<span> 256 g</span>
<span>hope this helps!</span>
It could be used as a buffer, since it is not a conductor. Hope this helps!!!
Answer:
58.61mL
Explanation:
V1 (initial volume) =?
T1 (initial temperature) = 320 K
V2 (final volume) = 50mL
T2 (final temperature) = 273 K
Using the Charles' law equation V1/T1 = V2/T2
The initial volume of the gas can obtained as follow:
V1/320 = 50/273
Cross multiply to express in linear form
V1 x 273 = 320 x 50
Divide both side by 273
V1 = (320 x 50)/273
V1 = 58.61mL
Therefore, the initial volume of the gas is 58.61mL