A student weighs out a 6.64 g sample of , transfers it to a 500. mL volumetric flask, adds enough water to dissolve it and then adds water to the 500. mL tick mark. What is the molarity of cobalt(II) fluoride in the resulting solution

the molarity of cobalt(II) fluoride in the resulting solution is = 0.137  M

Explanation:

Given that :

a student dissolves 6.64 g of CoF₂  into 500 mL of water

volume of the solution(water) = 500 mL = 0.50 L

The standard  molar mass of CoF₂ is 96.93 g/mol

number of moles of CoF₂ = mass of CoF₂/molar mass of CoF₂

number of moles of CoF₂ = 6.64 g/96.93 g/mol

number of moles of CoF₂ = 0.0685 mol

The molarity of any given substance is known to be as the number of moles of solute dissolved in one litre of solution.

Thus ;

Molarity of  cobalt(II) fluoride CoF₂ in the resulting solution is = number of moles / Volume in (L)

Molarity of  cobalt(II) fluoride CoF₂  = 0.0685 mol/ 0.50 L

= 0.137  M

Thus ; the molarity of cobalt(II) fluoride in the resulting solution is = 0.137  M

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explanation:

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.