When 3.42 g of sucrose is dissolved in 90 g of water then the mole fraction of sucrose is approximately?

The problem gives you all the information you need in order to solve for the molality and mole fraction of the solution. In order to determine its molarity, you're going to need the solution's volume.

To get the volume, you have to know what the density of the solution is. Determine the percent concentration by mass of the solution first

#"%w/w" = m_"solute"/m_"solution" * 100#

In your case, the mass of the solution will be

#m_"solution" = m_"glucose" + m_"water"#

#m_"solution" = 20 + 150 = "170 g"#

This means that you get

#"%w/w" = (20cancel("g"))/(170cancel("g")) * 100 = "11.8%"#

The density of this solution will thus be

http://us.mt.com/us/en/home/supportive_content/application_editorials/D_Glucose_de_e.html

#rho = "1.045 g/mL"#

Use glucose's molar mass to determine how many moles you have

#20cancel("g") * "1 mole glucose"/(180.16cancel("g")) = "0.111 moles glucose"#

The solution's volume will be

#170cancel("g") * "1 mL"/(1.045cancel("g")) = "162.7 mL"#

This means that its molarity is - do not forget to convert the volume to liters!

#C = n/V = "0.111 moles"/(162.7 * 10^(-3)"L") = color(green)("0.68 M")#

A solution's molality is defined as the number of moles of solute divided by the mass of the solvent - in kilograms! This means that you have

#b = n/m_"water" = "0.111 moles"/(150 * 10^(-3)"kg") = color(green)("0.74 molal")#

To get the mole fraction of sucrose, you need to know how many moles of water you have present. Once again, use water's molar mass

#150cancel("g") * "1 mole water"/(18.02cancel("g")) = "8.24 moles water"#

The total number of moles the solution contains is

#n_"total" = n_"glucose" + n_"water"#

#n_"total" = 0.111 + 8.24 = "8.351 moles"#

This means that the mole fraction of sucrose, which is defined as the number of moles of sucrose divided by the total number of moles in the solution, will be

#chi_"sucrose" = n_"sucrose"/n_"total" = (0.111cancel("moles"))/(8.351cancel("moles")) = color(green)("0.013")#

SIDE NOTE I've left the values rounded to two sig figs, despite the fact that you only gave one sig fig for the mass of glucose.

3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution are a 6.68 × 1023 b 6.09 × 1022 c 6.022 × 1023 d 6.022 × 1021

Open in App