Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol-1.

Question

Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol-1. (A) 15.38 g (B) 38.15 g (C) 81.35 g (D) 35.81 g

Tardigrade Admin · Accepted Answer

∵ Molar mass of sodium acetate = 82.0245 g mole-1 Let mass of CH3COONa required to make 500 mL of 0.375 M be w. Molarity = (w × 1000/M × V(in mL)) ⇒ w = (0.375 × 82.0245 × 500/1000) = 15.38 g

Q. Calculate the mass of sodium acetate $(C H_{3} COON a)$ required to make $500 m L$ of $0.375$ molar aqueous solution. Molar mass of sodium acetate is $82.0245 g m o l^{- 1}$ .

15.38 g

38.15 g

81.35 g

35.81 g

$∵$ Molar mass of sodium acetate $= 82.0245 g m o l e^{- 1}$
Let mass of $C H_{3} COON a$ required to make $500 m L$ of $0.375 M$ be $w$ .
Molarity $= \frac{w \times 1000}{M \times V ( in m L )}$
$\Rightarrow w = \frac{0.375 \times 82.0245 \times 500}{1000}$
$= 15.38 g$

Q. Calculate the mass of sodium acetate (CH3​COONa) required to make 500mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245gmol−1.