If the boiling point elevation of an aqueous solution containing a non-volatile nonelectrolyte is 0.48◦C, what is the molality of the solution?

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If the boiling point elevation of an aqueous solution containing a non-volatile nonelectrolyte is 0.48◦C, what is the molality of the solution?

Answer in units of molality.

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Answer 1 · 2018-01-15T23:04:21

I got $0.94 molal$ $"0.94 molal"$ to two sig figs.

Boiling point elevation is given by:

$DeltaT_b = T_b - T_b^"*" = iK_bm$ ,

where:

$T_b$ is the boiling point of the solvent in the context of the solution (in $""^@ "C"$ ), and $"*"$ indicates pure solvent.

$i$ is the van't Hoff factor, i.e. the effective number of solute particles that have dissociated from the original solute particles.

$K_b = 0.512^@ "C"cdot"kg/mol"$ is the boiling point elevation constant of water.

$m$ is the molality of the solution in $"mols solute/kg solvent"$ .

The solute is a nonelectrolyte, so $i = 1$ . It also is nonvolatile, so it does not alter the vapor pressure other than through means of concentration at the surface.

Therefore,

$DeltaT_b = 0.48^@ "C" = (1)(0.512^@ "C"cdot"kg/mol")cdot m$

$=> color(blue)(m) = (0.48^@ "C")/((1)(0.512^@ "C"cdot"kg/mol"))$

$=$ $color(blue)(0.93_8 " mols solute/kg solvent")$

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If the boiling point elevation of an aqueous solution containing a non-volatile nonelectrolyte is 0.48◦C, what is the molality of the solution?

Answer in units of molality.

1 Answer

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