Question cdf3b

May 3, 2017

${\text{0.6083 mol kg}}^{- 1}$

Explanation:

Every time you're looking for a solution's molality, you are looking for the number of moles of solute present for every $\text{1 kg}$ of solvent.

Your first step here will be to convert the number of grams of glucose to moles by using the compound's molar mass

0.9813 color(red)(cancel(color(black)("g"))) * "1 mole glucose"/(180.156 color(red)(cancel(color(black)("g")))) = "0.005447 moles glucose"

Next, convert the mass of water to kilograms

8.9547 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "0.0089547 kg"

Now, use the known composition of the solution--remember, solutions are homogeneous mixtures, which implies that they have the same composition throughout--to calculate the number of moles of glucose present in '1 kg" of solvent.

1 color(red)(cancel(color(black)("kg water"))) * "0.005447 moles glucose"/(0.0089547 color(red)(cancel(color(black)("kg water")))) = "0.6083 moles glucose"

Therefore, you can say that this solution has a molality of

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molality = 0.6083 mol kg}}^{- 1}}}}$

The answer is rounded to four sig figs, the number of sig figs you have for the mass of glucose.

May 3, 2017

$m = 0.614$

Explanation:

Molality is defined as

$\textcolor{w h i t e}{\forall \forall \forall \forall \forall \forall \forall A} \textcolor{m a \ge n t a}{m = \left(\text{moles of solute")/(" kg of solvent}\right)}$

The $\text{glucose}$ is our $\text{solute}$. It is being dissolved in the $\text{solvent}$, $\text{water}$.

We take our mass in grams and convert it to moles, but first, we need to find the molar mass of glucose $\left({C}_{6} {H}_{12} {O}_{6}\right)$ which we will look up or it could be memorized $\left(180 \frac{g}{\text{mole}}\right)$

color(white)(aaaaaaaa)(0.9813 cancel"g")/1 *(1" mole")/(180 cancel"g") = "0.0055 moles of Glucose"

Then we convert our $\text{grams of water}$ into $\text{kg of water}$ (the solvent)

color(white)(aaaaaaaa)(8.957 cancel"g")/(1) * (1" kg")/(1*10^3 cancel"g") = 0.008957" kg of water"

Solve

color(magenta)(m = ("moles of solute")/(" kg of solvent"))->(0.0055" moles of Glucose")/(0.008957" kg of water") = 0.614 m

$\textcolor{b l u e}{\text{Answer: m = 0.614, where m equals molal}}$

May 3, 2017

The molality of the glucose solution is $\text{0.6083 molal}$, or $6.083 \textcolor{w h i t e}{.} m$.

Explanation:

Molality$\left(m\right) = \left(\text{moles of solute")/("kg of solvent}\right)$

The solute is glucose, and the solvent is water.

Step 1: Convert the given mass of glucose into moles of glucose by multiplying by the reciprocal of its molar mass.

Molar mass glucose=180.156 g/mol
https://www.ncbi.nlm.nih.gov/pccompound?term=glucose

0.9813color(red)cancel(color(black)("g glucose"))xx(1"mol glucose")/(180.156color(red)cancel(color(black)("g glucose")))="0.0054471 mol glucose"

Step 2: Convert the mass of water from grams to kilogram: $\text{1 kg=1000 g}$.

8.9547color(red)cancel(color(black)("g water"))xx(1"kg")/(1000color(red)cancel(color(black)("g")))="0.0089547 kg water"

Step 3: Calculate molality of the solution using the equation at the top.

m=(0.0054471"mol glucose")/(0.0089547"kg water")="0.6083 molal"=0.6083 m# (rounded to four significant figures)