In: Chemistry

A .435 g sample of a nonvolatile, nonionizable solute with a molar mass of 67.4 g/mol is dissolved in 15.1 g of cyclohexane. Calculate the molar concentration of solute in the solution.

Calculate the boiling point of the solution described.

Calculate the freezing point of the solution described.

**we know that**

**moles = mass / molar mass**

**so**

**moles of solute = 0.435 / 67.4**

**moles of solute = 6.454 x 10-3**

**now**

**we know that**

**volume = mass / density**

**and**

**density of cyclohexane = 0.8 g/ml**

**so**

**volume of cyclohexane = 15.1 / 0.8**

**volume of cyclohexane = 18.875**

**assuming the volume of cyclohexane as the volume of
soltuion**

**we get**

**volume of solution = 18.875 ml**

**now**

**molarity = moles of solute / volume of solution
(L)**

**= 6.454 x 10-3 / 18.875 x 10-3**

**= 0.342**

**so**

**the molar conc is 0.342 M**

**now**

**we know that**

**molality = moles of solute / mass of solvent
(kg)**

**so**

**molality = 6.454 x 10-3 / 15.1 x 10-3**

**molality = .4274**

**now**

**1)**

**elevation in boiling point is given by**

**dTb = kb x m**

**T - Tb = kb x m**

**for**

**cyclohexane**

**kb = 2.79**

**Tb = 80.74**

**so**

**T - 80.74 = 2.79 x 0.4274**

**T = 81.93**

**so**

**the boiling point of the solution is 81.93
C**

**2)**

**depression in freezing point is given by**

**dTf = kf x m**

**T - Tf = kf x m**

**for**

**cyclohexane**

**kf = -20.2**

**Tf = 6.55**

**so**

**T - 6.55 = -20.2 x 0.4274**

**T = -2.08**

**so**

**the freezing point of the solution is -2.08
C**

A 0.520 g sample of a diprotic acid with a molar mass of 255.8
g/mol is dissolved in water to a total volume of 23.0 mL . The
solution is then titrated with a saturated calcium hydroxide
solution.
a. Assuming that the pKa values for each ionization step are
sufficiently different to see two equivalence points, determine the
volume of added base for the first and second equivalence
points.
b. The pH after adding 23.0 mL of the base was...

Chem Question - Gases
Molar Mass N = 14.01 g/mol
Molar Mass H2O = 18.016 g/mol
Vapor Pressure of Water at 25 C is 23.76 torr
Vapor Pressure of Water at 65 C is 187.54 torr
(Show all work and calculations, include units in
answer. If calculations must be used in several parts, rounding
should be made to 6 decimal points to ensure accuracy. Final
Answers can be rounded to 3 decimal points.)
A syringe filled with air can be...

A 0.5065 g solid sample containing a mixture of LaCl3 (molar
mass = 245.26 g/mol) and Ce(NO3)3 (molar mass = 326.13 g/mol) was
dissolved in water. The solution was titrated with KIO3, producing
the precipitates La(IO3)3(s) and Ce(IO3)3(s). For the complete
titration of both La3 and Ce3 , 44.81 mL of 0.1252 M KIO3 was
required. Calculate the mass fraction of La and Ce in the
sample.

A 0.1276−g sample of a monoprotic acid (molar mass = 1.10 × 102
g/mol) was dissolved in 25.0 mL of water and titrated with 0.0633 M
NaOH. After 10.0 mL of base had been added, the pH was determined
to be 4.87. What is the Ka for the acid? Answer in scientific
notation.

A 0.5072 g solid sample containing a mixture of LaCl3 (molar
mass = 245.26 g/mol) and Ce(NO3)3 (molar mass = 326.13 g/mol) was
dissolved in water. The solution was titrated with KIO3, producing
the precipitates La(IO3)3(s) and Ce(IO3)3(s). For the complete
titration of both La3 and Ce3 , 44.01 mL of 0.1289 M KIO3 was
required. Calculate the mass fraction of La and Ce in the
sample.

A 0.5070 g solid sample containing a mixture of LaCl3 (molar
mass = 245.26 g/mol) and Ce(NO3)3 (molar mass = 326.13 g/mol) was
dissolved in water. The solution was titrated with KIO3, producing
the precipitates La(IO3)3 and Ce(IO3)3. For the complete titration
of both La3+ and Ce3+, 42.10 mL of 0.1204 M KIO3 was required.
Calculate the mass fraction of La and Ce in the sample.
mass fraction La:
g Lag sample
mass fraction Ce:
g Ceg sample

An extremely small sample of an ideal diatomic gas (with a molar
mass of 28 g/mol) has the following distribution of molecular
speeds: 1 molecule moving at 100 m/s, 2 molecules at 200 m/s, 4 at
300 m/s, and 3 at 400 m/s.
What is the rms speed of the distribution? What is the average
kinetic energy of translational motion per molecule? What is the
temperature of this sample?

A solution is prepared by dissolving 32.5 g of a nonvolatile
solute in 200 g of water. The vapor pressure above the solution is
21.85 Torr and the vapor pressure of pure water is 23.76 Torr at
this temperature. What is the molecular weight of the solute?

Ethanol, C2H5OH (molar mass = 46 g/mol) is mixed with methanol,
CH3OH (molar mass = 32 g/mol) to make an ideal solution at a given
temperature. If 5.00 g of ethanol and methanol are mixed, what is
the resulting vapor pressure of the solution? (the vapor pressure
at the same temperature for pure ethanol is 44.5 mm Hg and pure
methanol is 88.7 mm Hg)

What mass of a weak acid with a molar mass of 100 g/mol is
necessary to neutralize 25 ml of 0.10 M NaOH solution?
What is the pH of 0.15 g of sodium acetate NaC2H3O2 in 100 ml
water H2O?

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