solution ) 147 e) 1.75 M 1.93 M d) 1.21 M e) 1.33 AM 5, The vapor of pure water
ID: 705277 • Letter: S
Question
solution ) 147 e) 1.75 M 1.93 M d) 1.21 M e) 1.33 AM 5, The vapor of pure water st 25°C is 23.8mm Hg. Determine the vapor pressure of water above a solution containing 51 g of urea (CHAN O,a nonvolatile, nanelectrolyte, molar mass-60.0 g/mol) dissolved in 75 g of HO (molar ms 18.0 g/mol) a) 22 mm Hg b) 16 mm Hg e) 4.0 mm Hg d) 20, mm Hg e) 4.9 mm Hg 6. Which one of the following substances exhibits hydrogen bonding as an intermolecular force? a) H2s b) CoHi c) CH NH d) CHiF ) CH OCH 7. What is the correct expression for K, for the following reaction? Fe(s) +4H20(g) Fe,04(s)+4H2(8) d) [Fe][H20] [Fel'[H201 Fe 0410H2 [H2l bFeOullHy) H2l [H2Explanation / Answer
4) The answer is a (1.47 M)
total mass = 24.0 g + 76.0 g = 100 g
density = Mass / Volume
volume = Mass / density = 100 g / 1.10 g /mol = 90.90 mL
Molarity = moles of fructose / volume of solution in Litre
moles = weight in gram / molar mass
M= 24 / 180.2 x 90.90 x 10-3 = 1.47 M
5) The answer is d
According to Roult's law
Pi = Xi x pi *
Pi = vapour pressure of solution
Xi = mole fraction of solute
pi * = vapour pressure of pure solvent
mole of urea = 51 / 60 = 0.85 moles
moles of water = 75 / 18 = 4.1 moles
so Xi = moles of urea / total moles
Xi = 4.1 / 0.85 + 4.1 = 0.82
putiing them in formula
Pi = 0.828 x 23.8 = 19.7 mmHg = 20 mmHg
6) The answer is c.
a) H2S = dipole - dipole interaction
b) C6H14 = dispersion force
e) CH3OCH3 = dispersion forces and dipole - dipole forces.
7) The answer is b.
Kc = [PRODUCT ] / [REACTANT]
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