I cant seem to figure out ANY of these... help when you can pls :.)

a) Calculatethe molality, m, of an aqueous solution of 1.22 M sucrose, C12H22O11. The density of the solution is 1.12 g/mL.b) Wh

Contact [7]

Answer:

a) 1,74 molal

b) 37,2 %

c) 0,03

Explanation:

We are going to define sucrose as solute, water as solvent and the mix of both, the solution.

Let´s start with the data:

$Molarity = M = \frac{1,22 mol solute}{lts solution}$

We can assume as a calculus base, 1 liter of solution. So, in 1 liter of solution we have 1,22 moles of solute:

$1 lts solution * \frac{1,22 moles solute}{lts solution}=1,22 moles solute$

Knowing that the molality (m) is defined as mol of solute/kgs solvent, we have to calculate the mass of solvent on the solution. Remember our calculus base (1 lts of solution). In 1 lts of solution we have 1120 grams of solution.

$1 lts solution * \frac{1,12 grs solution}{mL solution}*\frac{1000 mL solution}{1 lts solution} = 1120 grs of solution$

With the molecular weight of solute (Sum of: for carbon = 12*12=144; for hydrogen = 1*22=22 and for oxygen = 16*11=176. Final result = 342 grs per mol), we can obtain the mass of solute:

$1,22 mol solute*\frac{342 grs solute}{1 mol solute} = 417,24 grs solute$

Now, the mass of solvent is: mass solvent = mass of solution - mass of solute. So, we have: 1120 - 417,24 = 702,76 grs of solvent = 0,70276 Kgs of solvent

$molality = m = \frac{1,22 mol solute}{0,70276 kgs solvent}= 1,74 molal$

For b) question we have that the mass percent of solute is hte ratio between the mass of solute and the mass of solution. So,

$%(w/w) = \frac{417,24 grs solute}{1120 grs solution} = 37,2%$

For c) question we have that the mole fraction of solute is the ratio between moles of solute and moles of solution. Let's calculate the moles of solution as follows: Moles solution = moles solute + moles solvent. First we have that the moles of solvent are (remember that the molecular weight of water for this calculus is 18 grs per mol):

$702,76 grs solvent*\frac{1 mol solvent}{18 grs solvent} = 39,04 moles solvent$

So, we have the moles of solution: 1,22 moles of solute + 39,04 moles of solvent = 40,26 moles of solution

Finally, we have:

$Mol frac solute = \frac{1,22 mol solute}{40,26 mol solution}= 0,03$

6 0

3 years ago

Determine the enthalpy for this reaction: Al(OH)3(s)+3HCl(g)→AlCl3(s)+3H2O(l)

ivanzaharov [21]

Important information to solve the exercise :
Substance ΔHf (kJ/mol):
HCl(g)= −92.0 kJ/mol
Al(OH)3(s)= −1277.0 kJ/mol
 H2O(l)= −285.8 kJ/mol
AlCl3(s) =−705.6 kJ/mol

Al(OH)3(s)+3HCl(g)→AlCl3(s)+3H2O(l)
reactants products

products- reactants:

(−705.6) + (3 x −285.8) - ( −1277.0 ) - (3 x −92.0 ) = - 10.0 kJ per mole at 25°C

6 0

3 years ago

Consider the equations below. (1) Fe2O3(s) → 2Fe(s) + 3 2 O2(g) (2) Fe2O3(s) + 3CO(g) → 2Fe(s)+3CO2(g) Which equation must be ad

slega [8]

Answer:

C. 3CO(g)+ $\frac{3}{2}$ O2(g)→3CO2(g)

Explanation:

7 0

3 years ago

Which best describes an element? A pure substance. A type of a mixture. A pure compound. An impure substance.

mr Goodwill [35]

a pure compound because an element is untouched and is just itself

7 0

3 years ago

Read 2 more answers

What are two things that you observed when you heated the mixture of tin and nitric acid over the Bunsen burner in the virtual l

Rus_ich [418]

1. The reaction for this would be:

Sn + 4 HNO₃ → SnO₂ + 4 NO₂ + 2 H₂O

The first observation would be bubbling of the solution and brown acrid smoke is produced due to the presence of NO₂ gas. Another observation would be the presence of a white solid which is SnO₂.

2. Heating was required to get rid of the H₂O. When all moisture is gone, you weigh the sample. Afterwhich, you further heat it to get ride of the oxygen. By doing this, you would know the individual mass of each element. Then, you can solve for the empirical formula of the oxide of tin.

3 0

3 years ago