A mass of 100.0 g of NaCl is added to 100.0 mL of water, but not all of it dissolves. A mass of 59.5 grams of NaCl solid remains. Determine the molarity of the dissolved NaCl.

The molarity of the dissolved NaCl is 6.93 M

Explanation:

Step 1: Data given

Mass of NaCl = 100.0 grams

Volume of water = 100.0 mL = 0.1 L

Remaining mass NaCl = 59.5 grams

Molar mass NaCl= 58.44 g/mol

Step 2: Calculate the dissolved mass of NaCl

100 - 59. 5 = 40.5 grams

Step 3: Calculate moles

Moles NaCl = 40.5 grams / 58.44 g/mol

Moles NaCl = 0.693 moles

Step 4: Calculate molarity

Molarity = moles / volume

Molarity dissolved NaCl = 0.693 moles / 0.1 L

Molarity dissolved NaCl = 6.93 M

The molarity of the dissolved NaCl is 6.93 M

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.

your parents i think this could be it