Calculate the ebullioscopic water- The heat of vapourisation is 40-685 kJ mol-1-0-512 K kg mol-1-1-86 K kg mol-1-3-56 K kg mol-1-5-12 K kg mol-1-

Question

Calculate the ebullioscopic water- The heat of vapourisation is 40-685  kJ mol-1-0-512 K kg mol-1-1-86 K kg mol-1-3-56 K kg mol-1-5-12 K kg mol-1-

Toppr Admin · Accepted Answer

Kb-MRTb-xD7-Tb1000-xD7-DHvKb-18-xD7-8-314-xD7-273-xD7-2731000-xD7-40-685-xD7-1000Kb-20820933-10840-685-xD7-1000000Kb-0-512-xA0- -xA0-M is molecular mass of water-xA0- Tb is boiling point of water

Calculate the ebullioscopic for water. The heat of vapourisation is $40.685 k J m o l^{- 1}$ .
$0.512 K k g m o l^{- 1}$
$1.86 K k g m o l^{- 1}$
$3.56 K k g m o l^{- 1}$
$5.12 K k g m o l^{- 1}$

$K_{b} = \frac{M R T_{b} \times T_{b}}{1000 \times D H_{v}}$

$K_{b} = \frac{18 \times 8.314 \times 273 \times 273}{1000 \times 40.685 \times 1000}$

$K_{b} = \frac{20820933.108}{40.685 \times 1000000}$
$K_{b} = 0.512$ M is molecular mass of water $T_{b}$ is boiling point of water

Calculate the ebullioscopic for water. The heat of vapourisation is 40.685 kJ mol−1.0.512 K kg mol−11.86 K kg mol−13.56 K kg mol−15.12 K kg mol−1

Kb=MRTb×Tb1000×DHvKb=18×8.314×273×2731000×40.685×1000Kb=20820933.10840.685×1000000Kb=0.512 M is molecular mass of water Tb is boiling point of water

Calculate the ebullioscopic for water. The heat of vapourisation is $40.685 k J m o l^{- 1}$ .
$0.512 K k g m o l^{- 1}$
$1.86 K k g m o l^{- 1}$
$3.56 K k g m o l^{- 1}$
$5.12 K k g m o l^{- 1}$

$K_{b} = \frac{M R T_{b} \times T_{b}}{1000 \times D H_{v}}$

$K_{b} = \frac{18 \times 8.314 \times 273 \times 273}{1000 \times 40.685 \times 1000}$

$K_{b} = \frac{20820933.108}{40.685 \times 1000000}$
$K_{b} = 0.512$ M is molecular mass of water $T_{b}$ is boiling point of water