Answer:
2
Explanation:
Mass of water molecule = mass of hydrated salt - mass of anhydrous salt
Mass of water molecule = 5.00 - 4.26 = 0.74g of water molecule.
Number of moles = mass / molarmass
Molar mass of water = 18.015g/mol
No. of moles of water = 0.74 / 18.015 = 0.0411 moles.
Mass of BaCl2 present =?
1 mole of BaCl2 = 208.23 g
X mole of BaCl2 = 4.26 g
X = (4.26 * 1) / 208.23
X = 0.020
0.020 moles is present in 4.26g of BaCl2
Mole ratio between water and BaCl2 =
0.0411 / 0.020 = 2
Therefore 2 molecules of water is present the hydrated salt.
Answer:
- Molar mass = 608.36 g/mol
Explanation:
It seems the question is incomplete. However a web search us shows this data:
" Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner R. B. Woodward. It is used as a tranquilizer and sedative. When 1.00 g reserpine is dissolved in 25.0 g camphor, the freezing-point depression is 2.63 °C (Kf for camphor is 40 °C·kg/mol). Calculate the molality of the solution and the molar mass of reserpine. "
The <em>freezing-point depression</em> is expressed by:
We put the data given by the problem and <u>solve for m</u>:
- 2.63 °C = 40°C·kg/mol * m
For the calculation of the molar mass:<em> Molality</em> is defined as moles of solute per kilogram of solvent:
- 0.06575 m = Moles reserpine / kg camphor
- 25.0 g camphor ⇒ 25.0/1000 = 0.025 kg camphor
We<u> calculate moles of reserpine:</u>
- 0.06575 m = Moles reserpine / 0.025 kg camphor
- Moles reserpine = 1.64x10⁻³ mol
Finally we use the mass of reserpine and the moles to calculate <u>the molar mass</u>:
- 1.00 g reserpine / 1.64x10⁻³ mol = 608.36 g/mol
<em>Keep in mind that if the data in your problem is different, the results will be different. But the solving method remains the same.</em>
Answer:
194 g/mol.
Explanation:
Hello,
In this case, one first must compute the mass of each element as shown below:

Next, the corresponding moles:

Then, each element's subscripts is found to be:

Therefore, the empirical formula is:

Nonetheless, it has a molar mass of 97bg/mol, thereby, by multiplying such formula by 2 one gets:

Which has a molar mass of 194 g/mol being correctly contained in the given interval.
Best regards.