Question

Question #36be0

Answer 1

`"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 = n/(10^"-3" m_text(s)) = (1000n)/(m_text(s))`

where

`n color(white)(ll)=` moles of solute
`m_text(s) =` grams of solvent

Since

`n = "mass"/"molar mass" = 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_text(s) =` mass of solvent

From (1),

(3) `color(white)(mmmm)m = (bMm_text(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 `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 `"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 %`