Asample contains 6.73 g of na2co3 in water to make a total of 250.0 ml of solution. what is the molarity of sodium carbonate in this solution and its mass-% ? use 1.00 g/ml as the density of the solution.

Molarity of Na₂CO₃ = 0.25M

% mass = 2.69

Explanation:

Molarity means mole of solute in 1L of solution

Molar mass of solute (Na₂CO₃) = 105,98 g/m

Moles = mass / molar mass → 6.73 g / 105.98 g/m = 0.0635 m

Mol/L = [M]

0.0635 mol/0.250L = 0.25M

Density of solution = Solution mass / Solution volume

1 g/ml = Solution mass / 250 mL → Solution mass is 250g

% mass will be:

In 250 g of solution we have 6.73 g of solute

in 100 g of solution we have (100 . 6.73)/250 = 2.69

The molarity of Na2CO3 in the solution = 0.254 M

The mass % of Na2CO3 is 2.69 %

Explanation:

Step 1: Data given

Mass of Na2CO3 = 6.73 grams

Volume of the solution = 250.0 mL

Density = 1.00g/mL

Molar mass of Na2CO3 = 105.99 g/mol

Step 2: Calculate moles of Na2CO3

Moles Na2CO3 = 6.73 grams / 105.99 g/mol

Moles Na2CO3 = 0.0635 moles

Step 3: Calculate molarity of Na2CO3

Molarity = moles / volume

Molarity = 0.0635 moles / 0.250 L

Molarity = 0.254 M

Step 4: Calculate mass of the solution

Mass = volume * density

Mass = 250 mL * 1.00g/mL

Mass = 250 grams

Step 5: Calculate mass %

mass% = (6.73 g / 250 g)*100%

mass % =  2.69 %

The mass % of Na2CO3 is 2.69 %

25 mL of 1 × 10–5 M Co(NO₃)₂ and 75 mL of 5 × 10–4 M Na₂S

500 mL of 7.5 × 10–4 M AlCl₃ and 100 mL of 1.7 × 10–5 M Hg₂(NO₃)₂.

650 mL of 0.0080 M K₂SO₄ and 175 mL of 0.15 M AgNO₃

Explanation:

When 2 compounds that produce an insoluble substance are mixed together, A precipitate will be formed if Q of reaction > Ksp

For the solutions:

1.5 L of 0.025 M BaCl₂ and 1.25L of 0.014 M Pb(NO₃)₂.

Ksp is:

PbCl₂(s) ⇄ Pb²⁺(aq) + 2Cl⁻(aq)

Ksp = 2.4x10⁻⁴ = [Pb²⁺][Cl⁻]²

Molar concentration of each ion is:

[Pb²⁺] =  1.25L ₓ (0.014mol / L) = 0.0175mol / 2.75L = 6.36x10⁻³M

[Cl⁻] = 2 ₓ 1.5L ₓ (0.025mol / L) = 0.075mol / 2.75L = 0.0273M

Replacing in Ksp expression to find Q:

Q = [6.36x10⁻³M][0.0273M]² = 4.73x10⁻⁶

As Q < Ksp, the mixture will not produce a precipitate.

25 mL of 1 × 10–5 M Co(NO₃)₂ and 75 mL of 5 × 10–4 M Na₂S

Ksp is:

CoS(s) ⇄ Co²⁺(aq) + S²⁻(aq)

Ksp = 4.0x10⁻²¹ = [Co²⁺][S²⁻]

Molar concentration of each ion is:

[Co²⁺] =  0.025L ₓ (1x10⁻⁵mol / L) = 2.5x10⁻⁷mol / 0.1L = 2.5x10⁻⁶M

[S²⁻] = 0.075L ₓ (5x10⁻⁴mol / L) = 3.75x10⁻⁵mol / 0.1L = 3.75x10⁻⁴M

Replacing in Ksp expression to find Q:

Q = [2.5x10⁻⁶M][3.75x10⁻⁴M] = 9.38x10⁻⁶

As Q > Ksp, the mixture will produce a precipitate.

500 mL of 7.5 × 10–4 M AlCl₃ and 100 mL of 1.7 × 10–5 M Hg₂(NO₃)₂.

Ksp is:

Hg₂Cl₂(s) ⇄ 2Hg⁺(aq) + 2Cl⁻(aq)

Ksp = 3.5x10⁻¹⁸ = [Hg⁺]²[Cl⁻]²

Molar concentration of each ion is:

[Hg⁺] =  2ₓ0.100L ₓ (1.7x10⁻⁵mol / L) = 3.4x10⁻⁶mol / 0.6L = 5.67x10⁻⁶M

[Cl⁻] = 3 ₓ 0.500L ₓ (7.5x10⁻⁴mol / L) = 1.125x10⁻³mol / 0.6L = 1.88x10⁻³M

Replacing in Ksp expression to find Q:

Q = [5.67x10⁻⁶M]²[1.88x10⁻³M]² = 1.14x10⁻⁶

As Q > Ksp, the reaction will produce a precipitate.

650 mL of 0.0080 M K₂SO₄ and 175 mL of 0.15 M AgNO₃

Ksp is:

Ag₂SO₄(s) ⇄ 2Ag⁺(aq) + SO₄²⁻(aq)

Ksp = 1.5x10⁻⁵ = [Ag⁺]²[SO₄²⁻]

Molar concentration of each ion is:

[Ag⁺] =  0.175L ₓ (0.15mol / L) = 0.02625mol / 0.825L = 0.0318M

[SO₄²⁻] = 0.650L ₓ (0.080mol / L) = 0.052mol / 0.825L = 0.0630M

Replacing in Ksp expression to find Q:

Q = [0.0318M]²[0.0630M] = 6.37x10⁻⁵

As Q > Ksp, the reaction will produce a precipitate.