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>
<em />
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>
<span>1. Group 1 (Alkali Metals)Group 2 (Alkaline Earth Metals)Group 13 (Boron Group)Group 14 (Carbon Group)Group 15 (Nitrogen Group)Group 16 (Chalcogens)Group 17 (Halogens)<span>Group 18 (Noble Gases)
2.</span></span><span>The elements in each </span>group<span> have the same number of electrons in the outer orbital
3.</span><span>Group 2 </span>elements<span> share common characteristics. Each metal is naturally occurring and quite </span>reactive<span>.
4.</span><span>There are 18 numbered </span>groups<span> in the </span>periodic table<span>
5.</span><span>many characteristics are common throughout the group</span>
Answer : The mass of 
precipitate produced will be, 9.681 grams.
Explanation : Given,
Molarity of NaI = 0.210 M
Volume of solution = 0.2 L
Molar mass of = 461.01 g/mole
First we have to calculate the moles of 
.
Now we have to calculate the moles of .
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 moles of react to give 1 mole of
So, 0.042 moles of react to give 
moles of
Now we have to calculate the mass of .
Therefore, the mass of precipitate produced will be, 9.681 grams.
Answer:
152 kPa = Partial pressure O₂
Explanation:
Data by percent is the molar fraction . 100.
Molar fraction of Helium = 32/ 100 → 0.32
Molar fraction of O₂ = 68/100 → 0.68
Sum of molar fractions in a mixture = 1
0.68 + 0.32 = 1
If we apply the molar fraction, we can determine the partial pressure.
Mole fraction = Partial pressure / Total pressure
0.32 = Partial pressure O₂ / 475kPa → 0.32 . 475 kPa = Partial pressure O₂
152 kPa = Partial pressure O₂
Δt = i Kf m
2.86 °C = (1) (1.86 °C kg mol-1) (x / 0.750 kg)
2.86 °C = (2.48 °C mol-1) (x)
x = 1.1532 mol
33.7 g / 1.1532 mol = 29.2 g/mol
