How do you convert from $% w/v$ $%"w/v"$ to molarity?

Question

How do you convert from $% w/v$ $%"w/v"$ to molarity?

1 Answer

Impact of this question

108257 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-15T08:47:36

We define $% w/v$ $%"w/v"$ as:

$% w/v = \frac{solute mass in g}{solution volume in mL} \times 100 %$ $%"w/v" = "solute mass in g"/"solution volume in mL" xx 100%$

By unit conversion, we have that $g \times \frac{mol}{g} \to mol$ $"g" xx "mol"/"g" -> "mol"$ , and that $mL \times \frac{1 L}{1000 mL} \to L$ $"mL" xx "1 L"/"1000 mL" -> "L"$ . Define:

$m_{s o l u t e}$ $m_(solute)$ for the solute mass in $g$ $"g"$
$V_{s o ln}$ $V_(soln)$ for the solution volume in $mL$ $"mL"$
$M_{s o l u t e}$ $M_(solute)$ for the molar mass of the solute in $g/mol$ $"g/mol"$

Therefore:

$Molarity = (\frac{m_{s o l u t e}}{V_{s o ln}} \times 100 %) \times \frac{1000 mL/L}{100 % \times M_{s o l u t e}}$ $"Molarity" = (m_(solute)/V_(soln) xx 100%) xx (1000 "mL/L")/(100% xx M_(solute))$

Or, rewriting in terms of $% w/v$ $%"w/v"$ and implied unit cancellation, we have:

$Molarity (\frac{mol}{L}) = \frac{1000}{M_{s o l u t e}} \frac{% w/v}{100 %}$ $bb("Molarity" ("mol"/"L") = 1000/(M_(solute))(%"w/v")/(100%))$

As an example, if we have a $37 %$ $37%$ $w/v$ $"w/v"$ aqueous $HCl$ $"HCl"$ solution, then:

$color(blue)("Molarity" ("mol"/"L")) = 1000/(36.4609) (37cancel(%) "w/v")/(100cancel(%))$

$=$ $color(blue)("10.15 mol/L")$

Science

Math

Humanities

... and beyond

How do you convert from $% w/v$ $%"w/v"$ to molarity?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you convert from %"w/v"%w/v%"w/v" to molarity?

1 Answer

Related questions

Impact of this question

How do you convert from $% w/v$ $%"w/v"$ to molarity?