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2 years ago

How many liters of a 0.352 M solution of Ca(SO4) would contain 62.1 g of Ca(SO4)?

Chemistry

1 answer:

konstantin123 [22]2 years ago

4 0

Answer:

1.3 L.

Explanation:

Molarity is the no. of moles of solute per 1.0 L of the solution.

M = (no. of moles of CaSO₄)/(Volume of the solution (L))

M = 0.352 M.

no. of moles of CaSO₄ = mass/molar mass = (62.1 g / 136.14 g/mol) = 0.456 mol,

Volume of the solution = ??? L.

∴ (0.352 M) = (0.456 mol)/(Volume of the solution (L))

∴ (Volume of the solution (L) = (0.456 mol)/(0.352 M) = 1.296 L ≅ 1.3 L.

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Solution

1st we find out number of moles of each element from the molecular formula

Number of moles of carbon = 14 mol

Number of moles of hydrogen = 20 mol

Number of moles of nitrogen = 2 mol

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Now we find out the mass of each element

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mass = number of moles × molecular mass

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mass of hydrogen = 20 mol × 1 g/mol

mass of hydrogen = 20 g

mass of nitrogen = 2 mol × 14 g/mol

mass of nitrogen = 28 g

mass of oxygen = 4 mol × 16 g/mol

mass of oxygen = 64 g

mass of sulfur = 1 mol × 32 g/mol

mass of sulfur = 32 g

now we find out the mass percent of each element

mass percent = ( mass ÷ total mass ) × 100

mass percent of carbon = ( 168 g ÷ 312.39 g/mol ) × 100

mass percent of carbon = 57.78 %

mass percent of hydrogen = ( 20 g ÷ 312.39 g/mol ) × 100

mass percent of hydrogen = 6.40 %

mass percent of nitrogen = ( 28 g ÷ 312.39 g/mol ) × 100

mass percent of nitrogen = 8.96 %

mass of oxygen =( 64 g ÷ 312.39 g/mol ) × 100

mass percent of oxygen = 20.49 %

mass percent of sulfur = ( 32 g ÷ 312.39 g/mol ) × 100

mass percent of sulfur = 10.24 %

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Step 2: The unbalanced equation

NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + NaNO3(aq)

Step 3: Balancing the equation

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Pb^2+(aq) + 2F-(aq) → PbF2(s)

4 0

2 years ago

How many liters of a 0.352 M solution of Ca(SO4) would contain 62.1 g of Ca(SO4)?​

How many liters of a 0.352 M solution of Ca(SO4) would contain 62.1 g of Ca(SO4)?