Please Help - Molality?

Question

A 15.7 g sample of anhydrous sodium sulfate, Na2SO4, is dissolved in 4.93 L of water. The molar mass of sodium sulfate is 142.1 g/mol. Use the equation given to calculate the molality (m) of the solution, in mol/kg.

For this situation, assume the density of water is 1.00 g/mL.

m = moles of solute
mass of solvent (kg)

0.200 mol/kg (Incorrect: This answer was incorrect.)

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Answer 1 · 2018-01-29T08:52:04

$molality = 0.0224 \cdot m o l \cdot k g^{- 1}$ $"molality"=0.0224*mol*kg^-1$ ....

We want the quotient.....

$"molality"="moles of solute"/"kilograms of solvent"...$

And so... $m_"sodium sulfuate"=((15.7*g)/(142.04*g*mol^-1))/(4.93*Lxx1*kg*L^-1)$

$~=0.02*mol*kg^-1...$ ...so you were out by an order of magnitude...no big deal....

Answer 2 · 2018-01-29T10:54:31

Please, kindly see the step process below;

Recall: $rArr"molality" = "moles of solute"/"kilogram of solvent"$

$"moles of solute" = "mass"/"molar mass"$

$"density" = "mass"/"volume"$

Making mass the subject..

$"mass" = "density" xx "volume"$

Parameters

$"mass" = 15.7g$

$"molar mass" = 142.1gmol^-1$

But; $rArr "moles of solute" = "mass"/"molar mass"$

$"moles of solute" = (15.7cancelg)/(142.1cancelgmol^-1)= 0.111mols$

$"density" = (1.00g)/(mL) = (1.00kg)/L = 1.00kgL^-1$ (since molality answers are always in $molkg^-1$ )

Well for conversion rate?

$1000g = 1kg$

$1000mL = L$

Hence;

$(1000g)/(1000mL) = (1kg)/(1L) = 1kgL^-1$ (that's how we got this)

$"volume" = 4.93L$

Hence;

$"mass" = d xx v = 1.00kgcancelL^-1 xx 4.93cancelL = 4.93kg$

Now...

$"molality" = "moles of solute"/"kilogram of solvent"$

$"molality" = (0.111mols)/(4.93kg) = 0.225molskg^-1$