Asolution is prepared using 0.125 glucose c6h12o6 is enough water to make 250 g of total solution the concentration of the solution, expressed in parts per million is

Answer : The concentration of solution in ppm is, 500 ppm

Solution : Given,

Mass of solute (glucose) = 0.125 g

Mas of solution = 250 g

Parts per million (ppm) : It is defined as the mass of a solute present in one million parts by the mass of the solution.

Formula used :

Now put all the given values in this formula, we get

Hence, the concentration of solution in ppm is, 500 ppm

The concentration of the solution, expressed in parts per million (ppm) is 500 ppm

From the question given above, the following data were obtained:

Concentration of the solution in ppm can be obtained as follow:

Concentration = (mass of solute / mass of solution) × 10⁶

Concentration = (0.125/250) × 10⁶

Concentration = 500 ppm

Therefore, the concentration of the solution in ppm is 500 ppm

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n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"