A student performs a spectrophotometric analysis using external calibration and obtains a linear relationship with a slope of 0.0425 and a y-intercept of 0.011. In order to analyze an unknown solution, the student dilutes 5.00 mL of the unknown solution to 50.00 mL. If the absorbance of the dilute unknown solution is measured as 0.200, then what is the concentration of the unknown solution

4.44 M

Explanation:

The absorbance of a solution is a function of its concentration through the equation, known as Beer's law, as:

A = ε c l

where A is the absorbance, ε is is a constant of proportionality called molar absorptivity in units of L/mol·cm, c is the concentration in mol/L  (M), and l is the length of the cuvette we use to measure the absorbance, usually 1 cm.

Now the above equation is a function of c in  the form y= mx + b, which is exactly the linear relationship stated in the question.

Therefore

y = 0.0425 + 0.011

A = (εl) c

Therefore

slope = m = ε (1)  = 0.0425

Knowing ε we can calculate the concentration of the diluted unknown:

0.200 = 0.0425 c + 0.011

c = (0.200 - 0.011)/0.425 = 0.44 M

This the concentration of the diluted solution, the original concentration of the unknown solution can be calculated from the formula V₁M₁ = V₂M₂

calling M₁ our undiluted solution, we have

M₁ = V₂M₂/V₁ = 50 mL x 0.44 M / 5.00mL = 4.44 M

(Notice there is no need to convert the volume to liters in the above equation since they appear in both sides of the equation and so they cancel each other).

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.

it is stable as it is in group 0 which have full outer shells of electrons