01:160:162 Lecture Notes - Lecture 3: Molar Mass, Stoichiometry, Chief Operating Officer

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rubysalmon273

2 Nov 2016

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Compare the freezing points of three 0. 10 m aqueous solutions: Tf = kf m = 1. 86 (0. 10) = 0. 186. We would expect all 0. 10 m solutions to have the same freezing point depression of 0. 186: the freezing point depression for the nacl solution is almost double that of the sucrose or urea. Since freezing point depression is a colligative property, depending only on the amount of solute present, the actual amount of nacl doubles when it dissociates. One mol of nacl produces a total of 2 moles of ions. We can take this into account by using the van"t hoff factor i in our equations. Each mol of nacl produces 1 mol of na+ and 1 mol of cl-, or 2 mol of total particles. i = moles particles in solution moles solute dissolved. The i becomes a multiplier for all concentrations of solute.

Related Questions

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Need help with these 4 problems.. please show all work, thank you :)

When a solute is added to a liquid, the boiling point of the resulting solution is __________ than the freezing point of the pure liquid.

A. no different

B. higher

C. lower

When a solute is added to a liquid, the boiling point of the resulting solution is __________ than the freezing point of the pure liquid.

A. higher

B. no different

C. lower

Solution properties are defined as _______________ if the property depends solely on the number of particles of solute and solvent in the solution, not the identity of the solute.

A. vapor pressure

V.colligative

chemical

hydrogen bonding

The advantage of using ________________ rather than the more commonly used unit of ___________________ is that it is independent of temperature effects on density (and thus, volume) of the solution

A. molarity, ppb

B. molarity, % mass

C. molality, molarity

D. molarity, ppm

If a 6.75 g sample of NaCl were dissolved in 8.41 kg of methanol, the molality would be ___________________ m.

A. 0.0137

B. 1.602

C. 0.803

D. 0.0445

The only information needed to determine the freezing point depression of a non-electrolyte is ______________________________ and _________________________________. f

A. the freezing point constant for the solvent, the temperature

B. the boiling point constant, the boiling point elevation

C. the delta T, the color of the most concentrated sample

D. the molality of the solute, the Kf of the solvent

An aqueous solution boils at 100.50oC. What is the freezing point of the solution? Kb(H2O) = 0.51 oC/m; Kf(H2O) = 1.86 oC/m

A. 0.50 oC

B. -0.50 oC

C. -1.82 oC

D. -5.50 oC

Bromoform has a normal freezing point of 7.80 oC and its Kf = 14.4 oC/m. A solution of 2.58 g of an unknown in 100 g of bromoform freezes at 5.43 oC. What is the molecular weight of the unknown?

A. 18.02 g / mol

B. 0.165 g / mol

C. 156 g / mol

D. 44.01 g / mol

Calculate the freezing point of 0.200 m solution of fructose, C6H12O6, a) in water (boiling point, 0 oC), and b) in acetic acid (boiling point, 16.6 oC). Kf(water) = 1.86 oC/m and Kf(acetic acid) = 3.9 oC.

A. -0.372 oC and 15.8 oC

B. -0.958 oC and 12.3 oC

C. 3.221 oC and 0.875 oC

D. -4.124 oC and 0.0035 oC

10.

Calculate the boiling point, in Celsius, at 1atm of a solution containing 30.0 of cane sugar (molecular weight = 342 g / mol) and 150 g of water.

A. 100.30

B. 103.00

C. 99.7

D. 301.00

01:160:162 Lecture Notes - Lecture 3: Molar Mass, Stoichiometry, Chief Operating Officer

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