The molar mass of B(NO₃)₃ - Boron nitrate : 196.822 g/mol
<h3>Further explanation</h3>
In stochiometry therein includes
<em>Relative atomic mass (Ar) and relative molecular mass / molar mass (M) </em>
So the molar mass of a compound is given by the sum of the relative atomic mass of Ar
M AxBy = (x.Ar A + y. Ar B)
The molar mass of B(NO₃)₃ - Boron nitrate :
M B(NO₃)₃ = Ar B + 3. Ar N + 9.Ar O
M B(NO₃)₃ = 10.811 + 3. 14,0067 + 9. 15,999
M B(NO₃)₃ = 196.822 g/mol
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Answer:
P' = 41.4 mmHg → Vapor pressure of solution
Explanation:
ΔP = P° . Xm
ΔP = Vapor pressure of pure solvent (P°) - Vapor pressure of solution (P')
Xm = Mole fraction for solute (Moles of solvent /Total moles)
Firstly we determine the mole fraction of solute.
Moles of solute → Mass . 1 mol / molar mass
20.2 g . 1 mol / 342 g = 0.0590 mol
Moles of solvent → Mass . 1mol / molar mass
60.5 g . 1 mol/ 18 g = 3.36 mol
Total moles = 3.36 mol + 0.0590 mol = 3.419 moles
Xm = 0.0590 mol / 3.419 moles → 0.0172
Let's replace the data in the formula
42.2 mmHg - P' = 42.2 mmHg . 0.0172
P' = - (42.2 mmHg . 0.0172 - 42.2 mmHg)
P' = 41.4 mmHg
