Question

You have 3.50 l of solution that contains 90.0 g of sodium chloride, nacl. what is the molar mass of nacl? what is the molarity of that solution?

Answer 1

M(NaCl) = Ar(Na) + Ar(Cl) · g/mol.
M(NaCl) = 22,99 + 35,45 · g/mol.
M(NaCl) = 58,44 g/mol.
V(NaCl) = 3,5 L.
m(NaCl) = 90 g.
n(NaCl) = m(NaCl) ÷ M(NaCl).
n(NaCl) = 90 g ÷ 58,44 g/mol.
n(NaCl) = 1,54 mol.
c(NaCl) = n(NaCl) ÷ V(NaCl).
c(NaCl) = 1,54 mol ÷ 3,5 L.
c(NaCl) = 0,44 mol/l = 0,44 M.

Answer 2

Answer:

Molarity = 0.514 M NaCl

Explanation:

What you know: 105 grams of solute (NaCl) and 3.50 liters of solution. What you need to know: molarity of the solution. Convert the amount of solute from grams to moles before plugging it into the molarity ratio. This is done using the molar mass.

105 g NaCl × (1 mol NaCl / 58.44 g NaCl) = 1.80 mol NaCl

The 3.50 liters of solution is the correct unit.

Now that you have the amount of solute in moles and the volume of solution in liters, you can plug those values in and solve for molarity.

Molarity = moles of solute / liters of solution

Molarity = 1.80 mol NaCl / 3.50 L

Molarity = 0.514 M NaCl

Notice that the chemical formula of the solute is written after the unit for molarity.