Question

Q5. Calculate the mass of sodium acetate (CH3​COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol–1.​

Answer 1

Answer:

Mass Of CH₃COONa required = 15.38 g

Explanation:

We Have :-

Sodium Acetate

To Find :-

Mass of Sodium Acetate required to make 500 mL of 0.375 molar aqueous solution

Formula :-

Molarity ( M ) = No. of moles of Solute / Volume of Solution ( L )

Number of moles = mass / molar mass

Solution :-

Volume in L = 500 mL = 0.5 L

Number of moles = 0.375 × 0.5 = 0.1875 mol

Molar mass of sodium acetate = 82.0245 g mole⁻¹

Mass of sodium acetate = ( 82.0245 g mol⁻¹ ) ( 0.1875 mole ) = 15.38 g

Mass Of CH₃COONa required = 15.38 g

Answer 2

Answer:

0.375M solution means 0.375 moles in 1 L of solution.

1000mL of solution contains 0.375 moles

1mL of solution⟶

1000

0.375

moles

500mL of solution⟶

1000

0.375

×500

⟶0.1875

No. of moles=

Molar Mass

Mass

0.1875mol=

82.0245 gmol

−1

Mass

∴ Mass of the compound = 15.37 g.