Determine the vapor pressure of a solution at 55 degree C that contains 34.2 g N

Question

Determine the vapor pressure of a solution at 55 degree C that contains 34.2 g NaCI in 375 mL of water vapor pressure of pure water at 55 degree C is 118.1 torr. The writ Hoff factor for NaCI is 1.9 A) 115 torr B) 92.8 torr C) 112 torr D) 87.1 torr E) 108 torr A possible mechanism for the overall reaction Br_2(g) + 2NO(g) rightarrow 2NOBr(g) is NO(g) + Br_2 NOBr_2(g) NOBr_2(g) + NO(g) 2NOBr The rate law for formation of NOBr based on this mechanism is rate = ____. A) (k_2k_1/k^-1)[NO]^2[Br_2] B) k_1[Br_2]^1/2 C) k_1[NO]^1/2 D) (k_2k_1/k^-1)[NO][Br_2]^2 E) (k_1/k^-1)^2[NO]^2

Explanation / Answer

Q26) mass of Na Cl = 34.2 g

mass of water = 375 g ( density of water = 1g/mL)

vapor pressure of pure water = 118.1 torr

van't Hoff factor = 1.9

From Raoult' s law

the relative lowering of vapor pressure = i xmole fraction of solute

moles of solute NaCl = 34.2g/58.5g/mol

= 0.5846

moles of solvent water = 375g/18g/mol

=20.83

Thus mole fraction = moles of NaCl / (moles of NaCl + moles of water)

=0.5846 /21.418

= 0.0273

Now using Raoult's law

(118.1- P )/118.1 = 1.9 x 0.0273

or P = 111.98 torr

Thus the option C is correct.

Q27) Option A is correct .

The mechanism is

NO + Br2 <------> NOBr2

Kequilibrium for this = k1/k-1 = [NOBr2]/[NO][Br2]

or the interediate concentration [NOBr2] = k1 [NO][Br2] /k-1

The rate of reaction is always decided by the slow step of mechanism which is

NOBr2 + NO ------> 2NOBr

Thus rate = k2 [NOBr2][NO]

Substituting the intermediate[NOBr2] from the previous expression we get

rate = k2 .k1 . [NO]2 [Br2] /k-1

This is same as first option

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