Answer:
Molar mass of solute: 300g/mol
Explanation:
<em>Vapor pressure of pure benzene: 0.930 atm</em>
<em>Assuming you dissolve 10.0 g of the non-volatile solute in 78.11g of benzene and vapour pressure of solution was found to be 0.900atm</em>
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It is possible to answer this question based on Raoult's law that states vapor pressure of an ideal solution is equal to mole fraction of the solvent multiplied to pressure of pure solvent:
Moles in 78.11g of benzene are:
78.11g benzene × (1mol / 78.11g) = <em>1 mol benzene</em>
Now, mole fraction replacing in Raoult's law is:
0.900atm / 0.930atm = <em>0.9677 = moles solvent / total moles</em>.
As mole of solvent is 1:
0.9677× total moles = 1 mole benzene.
Total moles:
1.033 total moles. Moles of solute are:
1.033 moles - 1.000 moles = <em>0.0333 moles</em>.
As molar mass is the mass of a substance in 1 mole. Molar mass of the solute is:
10.0g / 0.033moles = <em>300g/mol</em>
Its a covalent bond formed by Potassium and Chlorine.
The product obtained is POTASSIUM CHLORIDE.
Hope this HELPS !!!
Moles of helium is required to blow up a balloon to 87.1 liters at 74 C and 3.5 atm is 021.65 mole
Mole is the unit of amount of substances of specified elementary entities
According to the ideal gas law he number of moles of a gas n can be calculated knowing the partial pressure of a gas p in a container with a volume V at an absolute temperature T from the equation
n =pV/RT
Here given data is volume = 87.1 liters
Temperature = 74 °C means 347.15 k
Pressure = 3.5 atm
R = 0.0821
Putting this value in ideal gas equation then
n =pV/RT
n = 3.5 atm×87.1 liters / 0.0821 ×347.15 k
n = 021.65 mole
Moles of helium is required to blow up a balloon to 87.1 liters at 74 C and 3.5 atm is 021.65 mole
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Answer:
the theoretical yield of benzoin = 14.3165g
Explanation:
The steps is as shown in the attachment.
Answer:
k = -0.0525 s⁻¹
Explanation:
The equaiton for a first order reaction is stated below:
ln[A]=−kt+ln[A]₀.
[A] = 5.50 x 10⁻³ M
[A]₀ = 7.60 x 10⁻² M
t = 85.0 - 35.0 = 50.0 s
The rate constant is represented by k and can be calculated substituting the values given above:
k = (ln[A]₀ - ln[A])/t
k = (ln5.50 x 10⁻³ M - ln7.60 x 10⁻² M)/50.0s
k = -0.0525 s⁻¹
