Question #cdf3b

3 Answers
May 3, 2017

Answer:

#"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.

May 3, 2017

Answer:

#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"#

May 3, 2017

Answer:

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)