0
$\begingroup$

A solution is made up by dissolving $\pu{13.8 g}$ $\ce{Na2CO3 · 10 H2O}$ in $\pu{103.5 g}$ of water. What is the molality of $\ce{Na2CO3}$ in this solution?

I can't solve it. I tried converting to moles with $\pu{13.8 g mol}/\pu{286 g} = \pu{0.04825 mol}$. Then I did $\pu{0.04825 mol}/\pu{0.1035 kg} = 0.466$ but it didn't work.

$\endgroup$

closed as off-topic by Todd Minehardt, Jan, Jon Custer, Mithoron, Buck Thorn Sep 11 at 14:12

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ Because your salt contains waters of hydration, you actually have more than $\pu{103.5 g}$ of water. $\endgroup$ – a-cyclohexane-molecule Jan 11 '18 at 4:17
  • 1
    $\begingroup$ Show some respect to the units too, they deserve better than this. I always recommend to start from writing down a complete math equation where you have an unknown on one side (molality) and the other side would contain known variables and constants (masses of salt and water, molar masses). Then plug in the numbers and keep an eye on the units – they should match (if they don't, convert). $\endgroup$ – andselisk Jan 11 '18 at 12:35
1
$\begingroup$

Let's denote a hydrated salt as $\ce{M_pA_q*xH2O}$ ($\ce{M}$ – metal, $\ce{A}$ – anion). Then molality $b$ of the aqueous solution of $\ce{M_pA_q}$ is defined as

$$b=\frac{n(\ce{M_pA_q})}{m_\mathrm{tot}(\ce{H2O})}\, ,\label{eq:1}\tag{1}$$

where $n(\ce{M_pA_q})$ – amount of water-free salt; $m_\mathrm{tot}(\ce{H2O})$ – total mass of water.

Note that the amounts of the hydrated and the water-free salt are equal:

$$n(\ce{M_pA_q})=n(\ce{M_pA_q*xH2O})=\frac{m(\ce{M_pA_q*xH2O})}{M(\ce{M_pA_q*xH2O})}\, ,\label{eq:2}\tag{2}$$

where $M$ – molar mass.

Total amount of water is found as the sum of water added to dissolve the salt and the water "stored" within the hydrated salt itself:

$$ \begin{align} m_\mathrm{tot}(\ce{H2O}) &= m(\ce{H2O})+m_\mathrm{hydr}(\ce{H2O})\\ &= m(\ce{H2O})+n_\mathrm{hydr}(\ce{H2O})\cdot M(\ce{H2O})\\ &=m(\ce{H2O})+x\cdot n(\ce{M_pA_q*xH2O})\cdot M(\ce{H2O})\\ \eqref{eq:2}\quad\Rightarrow\quad&=m(\ce{H2O})+x\cdot m(\ce{M_pA_q*xH2O})\cdot \frac{M(\ce{H2O})}{M(\ce{M_pA_q*xH2O})}\label{eq:3}\tag{3} \end{align} $$

Now, putting \eqref{eq:2} and \eqref{eq:3} back in \eqref{eq:1}, one can obtain uniform expression for molality of a hydrated salt:

$$b=\frac{m(\ce{M_pA_q*xH2O})}{m(\ce{H2O})\cdot M(\ce{M_pA_q*xH2O})+x\cdot m(\ce{M_pA_q*xH2O})\cdot M(\ce{H2O})}\label{eq:4}\tag{4}$$

Plugging in the numbers for sodium carbonate decahydrate: $$b=\frac{\pu{13.8 g}}{\pu{103.5 g}\cdot\pu{286.14 g mol-1}+10\cdot\pu{13.8 g}\cdot\pu{18.02 g mol-1}}=\pu{4.3e-4mol g-1}$$

A side note regarding the notations and units [1, p. 28]:

The term molal and the symbol $\pu{m}$ should no longer be used because they are obsolete. One should use instead the term molality of solute $\ce{B}$ and the unit $\pu{mol/kg}$ or an appropriate decimal multiple or submultiple of this unit. (A solution having, for example, a molality of $\pu{1 mol/kg}$ was often called a $1$ molal solution, written $\pu{1 m}$ solution.)

Reference

  1. Thompson, A.; Taylor, B. N. Guide for the Use of the International System of Units (SI). NIST Special Publication 811, 2008.
$\endgroup$
0
$\begingroup$

Sodium trioxocarbonate (iv) pentahydrate, a.k.a washing soda, is an efflorescent crystal containing 10 molecules of water of crystallization in its molecular formula.The molar mass of washing soda comprises a: carbonate portion of 106 g + pentahydrate portion of 180 g = 286 g/mol. i.e. 1 mole or 286 g of washing soda contains 106 g of sodium trioxocarbonate (iv)

therefore, 13.8 g of washing soda contains (106/286) x 13.8 g = 5.1147 g = 5.1147/106 mol = 0.04825 mol sodium carbonate Similarly, 13.8 g of washing soda contains (180/286) x 13.8 g = 8.6853 g crystal water On dissolution of washing soda, this 8.6853 g of crystal water adds to the 103.5g of solvent water so that the total mass of water = 8.6853 + 103.5 g = 112.19 g water This implies 112.19 g of water contains 0.04825 mol sodium carbonate But, molality is the amount of substance, in moles, contained in 1000 g or 1 kg of solvent (water in this case). Thus, 112.19 g of water contains 0.04825 mol sodium carbonate therefore, 1000 g water contains (1000/112.19) x 0.04825 mol per 1000 g
= 4.300739816 E(-4) mol/kg = 4.3 E(-4) mol/kg sodium carbonate

        - John Kerry Omoti, Lagos, Nigeria.           
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.