through an experiment 1.35 g of solute Tb were dissolved in 72.3 cm^3 of water and the solution was found to boil at 100.62 degrees Celsius at 1 atm. if the boiling point constant for 100 g of water is 5.2 degrees Celsius, find the molecular formula of the solute Tb.
Answers
Answer:
given
W.B = 6.8 g
W.A=100g
Tb=100.11Kb = 0.52kg/mol
MB = WT x RT
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PIE INTO ' v '
MB = KB x WB x 1000 ( for ' g ' changing into kg )
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ATB x WA
= 6.8 x 0.58 x 1000
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100.11 x 100
100 con cut by 1000 and form 10
Wb = n = 6.8
v = 100g = 0.1 Liter
R = 0.0822 atm/ k mol
T = 208 K
Explanation:
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