Molality of 2.5 g of ethanoic acid (CH3COOH) in 75 g of benzene is

Question

Molality of 2.5 g of ethanoic acid (CH3COOH) in 75 g of benzene is (A)  0.565 mol kg-1  (B)  0.656 mol kg-1  (C)  0.556 mol kg-1  (D)  0.665 mol kg-1

Tardigrade Admin · Accepted Answer

Molality ( m )=( text mass of solute ( text in g ) × 1000/ text molecular weight of solute × text massofsolvent ( text in g )) =(2.5 × 1000/60 × 7.5)=0.555 mol kg -1

Q. Molality of $2.5 g$ of ethanoic acid $(C H_{3} COO H)$ in $75 g$ of benzene is

$0.565 m o l k g^{- 1}$

$0.656 m o l k g^{- 1}$

$0.556 m o l k g^{- 1}$

$0.665 m o l k g^{- 1}$

Molality $(m) = \frac{mass of solute ( in g ) \times 1000}{molecular weight of solute \times massofsolvent ( in g )}$
$= \frac{2.5 \times 1000}{60 \times 7.5} = 0.555 m o l k g^{- 1}$

Q. Molality of 2.5g of ethanoic acid (CH3​COOH) in 75g of benzene is

0.565molkg−1

0.656molkg−1

0.556molkg−1

0.665molkg−1