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12 October, 18:49

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/v)).

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

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  1. 12 October, 18:57
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    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.
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