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a: What is the molality of a solution created with 36.9 g of [tex]C_6H_{12}O_6[/tex] dissolved in 482 g of water?

b: What is the molarity of a solution created with 35.2 g of [tex]C_6H_{12}O_6[/tex] dissolved in 482 g of water to make 555 mL of solution?

Answer :

Final Answer:

According to the given question,

(a) The molality of a solution created with 36.9 g of C₆H₁₂O₆ dissolved in 482 g of water to make 500. mL of solution is approximately 0.426 mol/kg.

(b) The molarity of a solution created with 35.2 g of C₆H₁₂O₆ dissolved in 482 g of water to make 555 mL of solution is approximately 0.351 M.

Explanation:

a) First, calculate the moles of C₆H₁₂O₆ (glucose) for the molality:

Calculate moles of glucose (C₆H₁₂O₆):

Molar mass of C₆H₁₂O₆= 6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol) = 180.18 g/mol

For molality (m):

we have the following formula for the Molality (m):

[tex]\[ m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} \][/tex]

Moles of C₆H₁₂O₆ = \(\frac{36.9 g}{180.18 g/mol} = 0.205 mol\)

Mass of water (solvent) = 482 g = 0.482 kg

Molality (m) = \(\frac{0.205 mol}{0.482 kg} = 0.426 mol/kg\)

So, The molality (m) of the solution is approximately 0.426 mol/kg.

b) First, calculate the moles of C₆H₁₂O₆ (glucose) for the molarity:

Molar mass of C₆H₁₂O₆= 6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol) = 180.18 g/mol

we have the following formula for the Molarity (M):

[tex]\[ M = \frac{\text{moles of solute}}{\text{volume of solution (in L)}} \][/tex]

Moles of C₆H₁₂O₆ = [tex]\(\frac{35.2 g}{180.18 g/mol} = 0.195 mol\)[/tex]

Volume of solution = 555 mL = 0.555 L

Molarity (M) = [tex]\(\frac{0.195 mol}{0.555 L} = 0.351 M\)[/tex]

So, The molarity (M) of the solution is approximately 0.351 M.

To learn more about the Molality, click here.

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