Molality of 1 M NaNO 3 solution (Density of solution is 1.25 g mL -1 and mol. wt. of NaNO 3=85 g mol -1 ) is

Question

Molality of 1 M NaNO 3 solution (Density of solution is 1.25 g mL -1 and mol. wt. of NaNO 3=85 g mol -1 ) is (A) 1.286 m (B) 4.44 m (C) 0.858 m (D) none of these.

Tardigrade Admin · Accepted Answer

ρ=1.25 g mL -1, M NaNO 3=85 g mol -1, Molarity =1 M (1/m)=(ρ/M)-(M NaNO ,/1000) ⇒ (1/m)=(1.25/1)-(85/1000) =1.25-0.085=1.165 m=0.858

Q. Molality of $1 MN a N O_{3}$ solution (Density of solution is $1.25 g m L^{- 1}$ and mol. wt. of $N a N O_{3} = 85 g m o l^{- 1}$ ) is

1.286 m

4.44 m

0.858 m

none of these.

$ρ = 1.25 g m L^{- 1}$ ,

$M_{N a N O_{3}} = 85 g m o l^{- 1}$ ,

Molarity $= 1 M$

$\frac{1}{m} = \frac{ρ}{M} - \frac{M _{N a NO,}}{1000}$

$\Rightarrow \frac{1}{m} = \frac{1.25}{1} - \frac{85}{1000}$

$= 1.25 - 0.085 = 1.165$

$m = 0.858$

Q. Molality of 1MNaNO3​ solution (Density of solution is 1.25gmL−1 and mol. wt. of NaNO3​=85gmol−1 ) is