Calculate the concentration of 5g dextrose in molarity

Molarity of dextrose = 0.03 mol / L

Explanation:

Given data

mass of dextrose = 5 g

Concentration (Molarity) = ?

Solution

as mass is given so we find out number of moles

Moles = mass / molar mass

Molar mass of dextrose = 180.156 g/mol

Moles = 5 g / 180.156 g/mol

Moles = 0.03 moles

Now we find out molarity

Molarity = moles of solute / Litres of solution

As volume is not given so we suppose it as 1 L

Molarity of dextrose = 0.03 mol / 1 L

Molarity of dextrose = 0.03 mol / L

introduction of a drug directly into the respiratory tract is a form of inhalation administration.

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.