Answer:
Okay, we first need to calculate the amount of heat energy required to raise the temperature of water to its boiling point(100°C) using the formula Q= mcθ where Q is the amount of heat energy required, m is the mass of the sample of water, c is the specific heat capacity of water and θ is the change in temperature.
Q= (20.0)(0.999043976)(100-25)
= 1.498565965x10³ cal
We now need to calculate the amount of heat energy required to convert the sample of water to steam at its boiling point using the formula Q= mL where Q is the amount of heat energy required, m is the mass of the sample of water and L is the latent heat of vapourisation of water.
Q= (20.0)(540)
= 1.08x10⁴ cal
Now, we shall calculate the amount of heat energy required to raise the temperature of the steam to 150°C using the formula Q= mcθ.
Q= (20.0)(0.485)(150-100)
= 485 cal
Finally, we add up the amount of heat energy required at each of the three stages to determine the net amount of heat energy required, Qₙ.
Qₙ= (1.498565965x10³)+(1.08x10⁴)+(485)
= 1.278356597x10⁴ cal
= 1.28x10⁴ cal correct to 3 significant figures.
The working equation for this is:
Tbp,soln - Tbp,water = i*Kb*m
where
Kb for water is 0.512 °C/molal
m is the molality (mol solute/kg solvent)
i is the van't hoff factor which represents the number of ions dissociated for strong electrolytes
Tbp,water is the boiling point of water which is 100°C
1. <span>1.50 moles of lioh (strong electrolyte) and 3.00 moles of koh (strong electrolyte) each in 1.0 kg of water
i = 2 for LiOH and 2 for KOH
Then,
</span>Tbp,soln - 100 = (2+2)(0.512)((1.5+3)/1 kg)
Tbp,soln = 109.22°C
<span>
2. </span><span>0.40 mole of al(no3)3 (strong electrolyte) and 0.40 mole of cscl (strong electrolyte) each in 1.0 kg of water
</span>
i = 4 for al(no3)3 and 2 for cscl
Then,
Tbp,soln - 100 = (4+2)(0.512)((0.4+0.4)/1 kg)
Tbp,soln = 102.46°C
<em>Thus, the first solution will have a higher boiling point.</em>
C. Yes; a false hypothesis gives a scientist new information to use.