Question

Calculate the mass of sodium acetate ($CH_{3}COONa$) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g $mol^{-1}$.

Answer 1

0.375 M aqueous solution of sodium acetate

≡ 1000 mL of solution containing 0.375 moles of sodium acetate

∴Number of moles of sodium acetate in 500 mL

Molar mass of sodium acetate = 82.0245 g mole–1 (Given)

∴ Required mass of sodium acetate = (82.0245 g mol–1) (0.1875 mole)

= 15.38 g

Answer 2

"Given, Aqueous solution = 0.375 Molar

Since "molar" represent "molarity" of the solution, hence, in the given solution "molarity" of solution = 0.375M

$Molarity (M) \quad =\quad \frac { Number\quad of\quad moles\quad of\quad solute }{ Volumes\quad of\quad solution\quad in\quad litre }$

Volume of solution in litre = 500 ml = 0.5 litre  

$Number of moles =\quad 0.375\quad \times\quad 0.5\quad =\quad 0.1875mol$

Given:

Molar mass of sodium acetate $=\quad 82.024\quad g.{ mol }^{ -1 }$

$Number\quad of\quad moles\quad =\quad \frac { Mass }{ Molar\quad Mass }$

$Mass\quad =\quad Number\quad of\quad moles\quad \times\quad molar\quad mass$

$Mass\quad of\quad sodium\quad acetate\quad =\quad (82.0245\quad g.{ mol }^{ -1 })\quad \times\quad (0.1875\quad mol)\quad =\quad 15.38\quad g$"