Question

Question #cdf3b

Answer 1

`"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 `"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

`color(darkgreen)(ul(color(black)("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.

Answer 2

`m = 0.614`

Explanation:

Molality is defined as

`color(white)(AAAAAAAAAAAAAAA)color(magenta)(m = ("moles of solute")/(" kg of solvent"))`

The `"glucose"` is our `"solute"`. It is being dissolved in the `"solvent"`, `"water"`.

We take our mass in grams and convert it to moles, but first, we need to find the molar mass of glucose `(C_6H_12O_6)` which we will look up or it could be memorized `(180 g/("mole"))`

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

Then we convert our `"grams of water"` into `"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`

`color(blue)"Answer: m = 0.614, where m equals molal"`

Answer 3

The molality of the glucose solution is `"0.6083 molal"`, or `6.083color(white)(.)m`.

Explanation:

Molality`(m)=("moles of solute")/("kg of solvent")`

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: `"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)