Answer:
The answer to your question is 0.22
Explanation:
Data
Acetonitrile (CH₃CN) density = 0.786 g/ml
Methanol (CH₃OH) density = 0.791 g/ml
Volume of CH₃OH = 22 ml
Volume of CH₃CN = 98.4 ml
Process
1.- Calculate the mass of Acetonitrile and the mass of Methanol
density = mass/ volume
mass = density x volume
Acetonitrile
mass = 0.786 x 98.4
= 77.34 g
Methanol
mass = 0.791 x 22
= 17.40 g
2.- Calculate the moles of the reactants
Acetonitrile molar mass = (12 x 2) + (14 x 1) + (3 x 1)
= 24 + 14 + 3
= 41 g
Methanol molar mass = (12 x 1) + (4 x 1) + (16 x 1)
= 12 + 4 + 16
= 32 g
Moles of Acetonitrile
41 g ----------------- 1 mol
77.34g ------------ x
x = (77.34 x 1) / 41
x = 1.89 moles
Moles of Methanol
32 g -------------- 1 mol
17.40 g --------- x
x = (17.40 x 1)/32
x = 0.54 moles
3.- Calculate the mole fraction of Methanol
Total number of moles = 1.89 + 0.54
= 2.43
Mole fraction = moles of Methanol / total number of moles
Mole fraction = 0.54/ 2.43
Mole fraction = 0.22
Answer:
Option A. 0.378M
Explanation:
Data obtained from the question include:
Molarity of acid (Ma) =..?
Volume of acid (Va) = 37.0 mL
Volume of base (Vb) = 56.0 mL
Molarity of base (Mb) = 0.250 M
Next, we shall write the balanced equation for the reaction. This is given below:
HCl + NaOH —> NaCl + H2O
From the balanced equation above,
The mole ratio of the acid (nA) = 1
The mole ratio of the base (nB) = 1
Finally, we can determine the molarity of the acid as shown below :
MaVa/MbVb = nA/nB
Ma x 37 / 0.25 x 56 = 1
Cross multiply
Ma x 37 = 0.25 x 56
Divide both side by 37
Ma = 0.25 x 56 /37
Ma = 0.378M
Therefore, the molarity of the acid, HCl is 0.378M
For this case, Only 1 isotope would be present, i.e. the principal element with mass M=13 and then one isotope at mass M+2. <span>We are assuming that the principal element is the one that is the lowest mass - by definition, an isotope is one where there are additional neutrons - hence the mass increases, but the proton count is the same.
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