1.calculate the number of moles and the number of grams of solute in 1.0 l of 0.50m nacl solution. 2. calculate the number of grams of solute required to make 2.5 l of normal saline solution (0.90% nacl (m/ 3. what is the concentration in percent (m/v) of 75 g k2so4 in 1500 ml of solution? 4. you are asked to make 1.50 l of 0.250m hno3 by diluting concentrated 16.0m hno3. what volume of the concentrated acid will be required to make the dilution

1) The answer is: the number of moles is 0.5 mol and the number of grams of solute is 29.22 grams.

V(NaCl) = 1.0 L; volume of the solution.

c(NaCl) = 0.50 M; molarity of the solution.

n(NaCl) = V(NaCl) · c(NaCl).

n(NaCl) = 1 L · 0.5 mol/L.

n(NaCl) = 0.50 mol; amount of natrium chloride.

m(NaCl) = 0.50 mol · 58.44 g/mol.

m(NaCl) = 29.22 g; mass of sodium chloride.

2) The answer is: the number of grams is 22.5 g.

V(NaCl) = 2.5 L · 1000 mL/L.

V(NaCl) = 2500 mL; volume of solution.

m/v(NaCl) = 0.90%; 0.90 g of NaCl for every 100 mL of solution.

m(NaC) = 0.9 g · 2500 ml ÷ 100 mL.

m(NaC) = 22.5 g; mass of sodium chloride.

The mass/volume percent is used to express the concentration of a solution when the mass of the solute and volume of the solution is given.

3) The answer is: the concentration in percent is 5%.

m(K₂SO₄) = 75 g; mass of potassium sulfate.

V(K₂SO₄) = 1500 mL; volume of the solution.

mass/volume percent = m(K₂SO₄) ÷ V(K₂SO₄) · 100%.

mass/volume percent = 75 g ÷ 1500 mL · 100%.

mass/volume percent = 5%.

percentage concentration is mass/volume percent, which measures the mass or of solute in grams divided by the volume of solution in mililiters.

4) The answer is: volume of the concentrated acid is 0.0234 L (22.4 mL).

c₁ = 16.0 M; original concentration of the solution, before it gets diluted.

c₂ = 0.250 M; final concentration of the solution, after dilution.

V₁ = ?; volume to be diluted.

V₂ = 1.5 L; final volume after dilution.

c₁ · V₁ = c₂ · V₂.

V₁ = c₂ · V₂ ÷ c₁.

V₁ = 0.25 M · 1.5 L ÷ 16 M.

V₁ = 0.0234 L = 23.4 mL.

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.