Question

find out the mass of CH3COONa(sodium acetate) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of CH3COONa is 82.0245gmo

Answer 1

aqueous solution of CH3COONa

= 1000 mL of solution containing 0.375 moles of CH3COONa

Therefore, no. of moles of CH3COONa in 500 mL

= 0.3751000×1000

= 0.1875 mole

Molar mass of sodium acetate = 82.0245gmol−1

Therefore, mass that is required of CH3COONa

= (82.0245gmol−1)(0.1875mole)

= 15.38 gram

Answer 2

0.375 M solution of (CH3COONa) = 0.375 moles(grams) of (CH3COONa) dissolved in 1000 ml of solvent.

 

But according to question ,we have to make a 500 ml solution of (CH3COONa)

 

∴Number of moles of sodium acetate in 500 mL

 

$\frac{0.375}{1000} \times 500 \\ \\ = 0.1875mole$

 

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