-1
$\begingroup$

10 g of NaCl is dissolved in 250 g of water. The mass fraction and molality of the solution is

For calculating mass fraction, simply divided $10/250$ to get $0.04$. My book says it’s wrong. Is mine the right answer?

Another one,

To calculate molality, no. moles of $\ce{NaCl}$ is $10/58.5$

So $$\begin{align} m&=\frac{10}{58.5\times250}\\[3pt] &=0.684\end{align}$$

Again, my book says it’s wrong, and I am starting to trust is even less, but I still need to get it checked. Am I right, or is the book the winner here?

$\endgroup$
  • 5
    $\begingroup$ No, your first answer isn't correct as 250 g isn't the total mass of the solution, and the second one is wrong because you for some reason dropped all units whereas the author of the textbook probably didn't. $\endgroup$ – andselisk Sep 1 at 19:13
5
$\begingroup$

Generally, it is preferable to write down the complete quantity equation before plugging in the numbers. If necessary, perform all mathematic operations on quantities symbolically. Do not plug in the numbers until you have only one equation for the desired result.

Always write the values with the correct units and carry the units through the calculation. Do not omit the units while performing intermediate steps and do not just reintroduce units at the end of the calculation.

When a number is given without any further information, it is generally interpreted so that the last digit is rounded with a rounding range equal to 1 in the last digit. Make sure that your result is given with the correct number of significant digits, accordingly. However, it may be necessary to retain additional digits for intermediate results in order to avoid round-off errors in subsequent calculations.


Mass fraction is defined as $$w_\mathrm B=\frac{m_\mathrm B}m$$ where
$w_\mathrm B$ is the mass fraction of substance B,
$m_\mathrm B$ is the mass of substance B, and
$m$ is the total mass of the mixture.

And you know that, in this case, the total mass of the mixture is given as $$m=m_\mathrm A+m_\mathrm B$$ where $m$ is the total mass of the mixture,
$m_\mathrm A$ is the mass of the solvent A, and
$m_\mathrm B$ is the mass of substance B.

Thus, $$\begin{align} w_\mathrm B&=\frac{m_\mathrm B}m\\[3pt] &=\frac{m_\mathrm B}{m_\mathrm A+m_\mathrm B}\\[3pt] &=\frac{10\ \mathrm g}{250\ \mathrm g+10\ \mathrm g}\\[3pt] &=0.038 \end{align}$$


Molality is defined as $$b_\mathrm B=\frac{n_\mathrm B}{m_\mathrm A}$$ where
$b_\mathrm B$ is the molality of solute B,
$n_\mathrm B$ is the amount of substance of solute B, and
$m_\mathrm A$ is the mass of the solvent A.

You also know that molar mass is defined as $$M=\frac mn$$ where
$M$ is molar mass,
$m$ is mass, and
$n$ is amount of substance.

And you know that, for $\ce{NaCl}$, $M=58.44277\ \mathrm{g\ mol^{-1}}$.

Thus, $$\begin{align} b_\mathrm B&=\frac{n_\mathrm B}{m_\mathrm A}\\[3pt] &=\frac{m_\mathrm B}{M_\mathrm B\cdot m_\mathrm A}\\[3pt] &=\frac{10\ \mathrm g}{58.44277\ \mathrm{g\ mol^{-1}}\times 250\ \mathrm g}\\[3pt] &=0.00068\ \mathrm{mol\ g^{-1}}\\[3pt] &=0.68\ \mathrm{mol\ kg^{-1}} \end{align}$$

$\endgroup$
  • 2
    $\begingroup$ That's probably a weird question, but why does the multiplication sign in the last equation for $b_\ce{B}$ undergo a change from a centered dot $(\cdot)$ to a times sign $(\times)?$ $\endgroup$ – andselisk Sep 1 at 19:44
  • 4
    $\begingroup$ @andselisk According to ISO: "If the point is used as the decimal sign, the cross and not the half-high dot should be used as the multiplication sign between numbers expressed with digits." $\endgroup$ – Loong Sep 1 at 19:49
  • $\begingroup$ Reading this the next day, I am wondering how I even made that mistake. I was really tired and wasn’t thinking straight, and perhaps out of frustration I asked if over here. $\endgroup$ – Aditya Sep 2 at 5:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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