# Question 36be0

May 13, 2017

"Mass %" = b/(b + 1000/M) × 100 %

#### Explanation:

The formula for molality $b$ is

color(blue)(bar(ul(|color(white)(a/a)b = "moles of solute"/"kilograms of solvent" color(white)(a/a)|)))" "

It will be more convenient during the derivation to express the mass of solvent in grams rather than kilograms.

Then,

$b = \frac{n}{{10}^{\text{-3}} {m}_{\textrm{s}}} = \frac{1000 n}{{m}_{\textrm{s}}}$

where

$n \textcolor{w h i t e}{l l} =$ moles of solute
${m}_{\textrm{s}} =$ grams of solvent

Since

$n = \text{mass"/"molar mass} = \frac{m}{M}$

we can write

(1) color(white)(mmmm)b = (1000m)/(Mm_text(s)

where

$m =$ mass of solute
$M =$ molar mass of solute

The formula for mass percent is

color(blue)(bar(ul(|color(white)(a/a)"Mass %" = "mass"/"total mass" × 100 %color(white)(a/a)|)))" "

or

(2) color(white)(mmm)"Mass %" = (m)/(m + m_text(s)) × 100 %

where

${m}_{\textrm{s}} =$ mass of solvent

From (1),

(3) $\textcolor{w h i t e}{m m m m} m = \frac{b M {m}_{\textrm{s}}}{1000}$

Substitute (3) into (2)

"Mass %" = ((bMcolor(red)(cancel(color(black)(m_text(s)))))/1000)/((bMcolor(red)(cancel(color(black)(m_text(s)))))/1000 + stackrelcolor(blue)(1)(color(red)(cancel(color(black)(m_text(s)))))) × 100 % = ((bM)/1000)/((bM)/1000 + 1) × 100 %

Multiply numerator and denominator by $\frac{1000}{M}$. Then

color(blue)(bar(ul(|color(white)(a/a)"Mass %" = b/(b + 1000/M) × 100 %color(white)(a/a)|)))" "

Example

Calculate the mass percent of a 1.00 mol/kg solution of $\text{NaCl}$.

"Mass %" = b/(b + 1000/M) × 100 % = 1.00 /(1.00 + 1000/58.44) × 100 %

= 1.00/(1.00 + 17.11) × 100 % = 1.00/18.11 × 100 % = 5.52 %#