An aqueous KCl solution of density 1.20 g mL -1 has a molality of 3.30 mol kg -1. The molarity of the solution in mol L -1 is (Nearest integer) [ Molar mass of KCl =74.5]

Question

An aqueous KCl solution of density 1.20 g mL -1 has a molality of 3.30 mol kg -1. The molarity of the solution in mol L -1 is (Nearest integer) [ Molar mass of KCl =74.5] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

1000 kg solvent has 3.3 moles of KCl 1000 kg solvent longrightarrow 3.3 × 74.5 gm KCl longrightarrow 245.85 Weight of solution =1245.85 gm Volume of solution =(1245.85/1.2) ml So molarity =(3.3 × 1.2/1245.85) × 1000=3.17

Q. An aqueous $K Cl$ solution of density $1.20 g m L^{- 1}$ has a molality of $3.30 m o l k g^{- 1}$ . The molarity of the solution in $m o l L^{- 1}$ is _____ (Nearest integer)
$[$ Molar mass of $K Cl = 74.5]$

$1000 k g$ solvent has $3.3$ moles of $K Cl$
$1000 k g$ solvent $⟶ 3.3 \times 74.5 g m K Cl$
$⟶ 245.85$
Weight of solution $= 1245.85 g m$
Volume of solution $= \frac{1245.85}{1.2} m l$
So molarity $= \frac{3.3 \times 1.2}{1245.85} \times 1000 = 3.17$

Q. An aqueous KCl solution of density 1.20gmL−1 has a molality of 3.30molkg−1. The molarity of the solution in molL−1 is _____ (Nearest integer) [ Molar mass of KCl=74.5]

1000kg solvent has 3.3 moles of KCl 1000kg solvent ⟶3.3×74.5gmKCl ⟶245.85 Weight of solution =1245.85gm Volume of solution =1.21245.85​ml So molarity =1245.853.3×1.2​×1000=3.17

Q. An aqueous $K Cl$ solution of density $1.20 g m L^{- 1}$ has a molality of $3.30 m o l k g^{- 1}$ . The molarity of the solution in $m o l L^{- 1}$ is _____ (Nearest integer)
$[$ Molar mass of $K Cl = 74.5]$

$1000 k g$ solvent has $3.3$ moles of $K Cl$
$1000 k g$ solvent $⟶ 3.3 \times 74.5 g m K Cl$
$⟶ 245.85$
Weight of solution $= 1245.85 g m$
Volume of solution $= \frac{1245.85}{1.2} m l$
So molarity $= \frac{3.3 \times 1.2}{1245.85} \times 1000 = 3.17$