Heat of vap. of water @ 100 °C is 40.66 kJ mol-1. Calc. quantity of heat that is absorbed/released ...

1. Q = moles steam x molar heat of vaporization
Q = 6g / 18g/mole x 40.66kJ/mole = 13.55kj

2. this requires the heat to warm liquid CCl4 from 25 to 76.8ºC and then to vaporize all CCl4 at this temperature
Q1 = mass CCl4 x specific heat CCl4 x ∆T = J
Q2 = moles CCl4 x molar heat of vaporization = kJ
watch that you convert J to kJ or kJ to J. in other words, watch your units

3. Claussius-Clapeyron equation
lnPº = (-∆Hvap / R)(1/T2 - 1/Tbpº)
convert mmHg to atm 55.1mmHg / 760mmHg/atm = 0.0725atm
ln(0.0725atm) = (-32100J/mole/0.0821L-atm/mole-K)(1/309 - 1/Tbp)
-2.624 = -3.91x10^5K(0.00324 - 1/Tbp)
6.71x10^-6 = 0.00324 - 1/Tbp
3.23x10^-3 = 1/Tbp
309.28K = bp = 36.28ºC

4. Q1 = heating ice from -50ºC to 0ºC = mass x specific heat ice x ∆T1
Q2 = melting ice at 0ºC = moles ice x molar heat of fusion
Q3 = heating water from 0ºC to 70ºC = mass water x specific heat water x ∆T2
add Q1, Q2, and Q3, watch the units