What is the molal concentration of lead nitrate in 0.726 M Pb(NO_3)_2?

density = 1.202 g/ml

Dec 9, 2016

$0.755 m$

Explanation:

I originally misread the question and thought the user was asking for molar concentration.

$0.0726 \cdot m o l \cdot 331.22 g \cdot m o {l}^{-} 1$
$= 24.05 g \cdot P b N {O}_{3}$ in $100 m l$ of $0.725 M$ solution.

Subtract, density tells us its $1.202 g \cdot m {l}^{-} 1$ so,

$120.2 g - 24.05 g = 96.15 g$

$96.1534 g \cdot \frac{1 k g}{1000 g} = 0.09615 k g$

And now we have:

$m =$molality$= \frac{m o l}{k g}$

$m =$$\frac{0.0726 \cdot m o l}{0.09615 \cdot k g}$

$=$$0.755 m$