The molality of a solute is equal to the moles of solute per kg of solvent. We are given the mole fraction of I₂ in CH₂Cl₂ is <em>X</em> = 0.115. If we can an arbitrary sample of 1 mole of solution, we will have:
0.115 mol I₂
1 - 0.115 = 0.885 mol CH₂Cl₂
We need moles of solute, which we have, and must convert our moles of solvent to kg:
0.885 mol x 84.93 g/mol = 75.2 g CH₂Cl₂ x 1 kg/1000g = 0.0752 kg CH₂Cl₂
We can now calculate the molality:
m = 0.115 mol I₂/0.0752 kg CH₂Cl₂
m = 1.53 mol I₂/kg CH₂Cl₂
The molality of the iodine solution is 1.53.
Answer:
203 grams
Explanation:
<em>It is known that 1.0 mole of a compound contains Avogadro's number of molecules (6.022 x 10²³).
</em>
<em><u>Using cross multiplication:</u></em>
1.0 mol contains → 6.022 x 10²³ molecules.
??? mol contains → 7.2 x 10²⁴ molecules.
∴ The no. of moles of (6.3 x 10²⁴ molecules) of NH₃ = (1.0 mol)(7.2 x 10²⁴ molecules)/(6.022 x 10²³ molecules) = 11.96 mol.
<em>∴ The no. of grams of NH₃ present = no. of moles x molar mass </em>= (11.96 mol)(17.0 g/mol) = <em>203.3 g ≅ 203.0 g.</em>
Ksp of AgCl= 1.6×10⁻¹⁰
AgCl=Ag⁺ +Cl⁻
Ksp=[Ag⁺][Cl⁻]
Assume [Ag⁺]=[Cl⁻]=x
Ksp=x²
1.6×10⁻¹⁰=x²
x=0.000012
In FeCl₃:
FeCl₃------>Fe⁺³+ 3Cl⁻
as there is 0.010 M FeCl₃
So there will be ,
[Cl⁻]= 0.030
So
[Ag⁺]=Ksp/[Cl⁻]
=1.6×10⁻¹⁰/0.030
=5.3×10⁻⁹
so solubility of AgCl in FeCl₃ will be 5.3×10⁻⁹.
Answer:
110ml
Explanation:
<em>Using the dilution equation, C1V1 = C2V2</em>
<em>Where C1 is the initial concentration of solution</em>
<em>C2 is final concentration of solution</em>
<em>V1 is intital volume of solution</em>
<em>V2 is final volume of solution.</em>
From the question , C1=6M, C2=0.5M, V1=10ml, V2=?
volume of water added = final volume -initial volume
= 120-10
=110ml
First, you must convert 7.68cal/sec to cal/min. To do so, multiply 7.68x60(seconds, thus making a minute). 7.68x60=460.8
Next, we must convert 460.8 cal to kcal. To do so, divide 460.8 by 1000 (460.8/1000). The result is .4608 kcal/min
