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2 years ago

What is the molarity of a solution that contains 122g of MgSO4 n 3.5L of solution?

Chemistry

1 answer:

tatiyna2 years ago

6 0

Answer:

0.29mol/L or 0.29moldm⁻³

Explanation:

Given parameters:

Mass of MgSO₄ = 122g

Volume of solution = 3.5L

Molarity is simply the concentration of substances in a solution.

Molarity = number of moles/ Volume

>>>>To calculate the Molarity of MgSO₄ we find the number of moles using the mass of MgSO₄ given.

Number of moles = mass/ molar mass

Molar mass of MgSO₄:

Atomic masses: Mg = 24g

S = 32g

O = 16g

Molar mass of MgSO₄ = [24 + 32 + (16x4)]g/mol

= (24 + 32 + 64)g/mol

= 120g/mol

Number of moles = 122/120 = 1.02mol

>>>> From the given number of moles we can evaluate the Molarity using this equation:

Molarity = number of moles/ Volume

Molarity of MgSO₄ = 1.02mol/3.5L

= 0.29mol/L

IL = 1dm³

The Molarity of MgSO₄ = 0.29moldm⁻³

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Here, m is mass and M is molar mass, putting the values,

$n=\frac{0.4856 g}{204.22 g/mol}=0.00237 mol$

This will be number of moles of NaOH at equivalent point.

Detailed calculations:

Molarity is defined as number of moles in 1 L of solution, for 250 mL of solution, molarity will be:

$M=\frac{0.00237 mol}{250 \times 10^{-3}L}=0.009511 M$

For 25 mL, apply dilution law as follows:

$M_{1}V_{1}=M_{2}V_{2}$

Putting the values,

$0.009511\times 250=M_{2}\times 25 mL$

On rearranging,

$M_{2}=\frac{0.009511\times 250}{25}=0.09511 M$

Convert molarity into number of moles,

$n=M\times V=0.09511 mol/L\times 25\times 10^{-3}L=0.00237 mol$

At equivalent point, number of moles of KHP will be equal to NaOH, thus, number of moles of NaOH will be 0.00237 mol.

Calculation for molarity:

Volume of NaOH is 18.75 mL, thus, molarity can be calculated as follows:

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Answer:

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Explanation:

1) Balanced chemical equation:

CH₄ + 2O₂ → CO₂ + 2H₂O

2) mole ratios:
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<span />

O₂: 15.6 g / 32.0 g/mol = 0.4875 moles

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Which is what the first part of the answer says.

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Which is what the second part of the answer says.

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Which is what the chosen answer says.

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