<u>Answer:</u> The mass of second isotope of indium is 114.904 amu
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
.....(1)
Let the mass of isotope 2 of indium be 'x'
Mass of isotope 1 = 112.904 amu
Percentage abundance of isotope 1 = 4.28 %
Fractional abundance of isotope 1 = 0.0428
Mass of isotope 2 = x amu
Percentage abundance of isotope 2 = [100 - 4.28] = 95.72 %
Fractional abundance of isotope 2 = 0.9572
Average atomic mass of indium = 114.818 amu
Putting values in equation 1, we get:
![114.818=[(112.904\times 0.0428)+(x\times 0.9572)]\\\\x=114.904amu](https://tex.z-dn.net/?f=114.818%3D%5B%28112.904%5Ctimes%200.0428%29%2B%28x%5Ctimes%200.9572%29%5D%5C%5C%5C%5Cx%3D114.904amu)
Hence, the mass of second isotope of indium is 114.904 amu
Answer:
1.6M
Explanation:
molarity = moles /Volume = 0.4/0.25= 1.6 M
Molar mass of methanol=12 + 1×3 + 16 + 1=32g
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Answer:
0.077 M
Explanation:
Molarity is the representation of the solution.
Molarity:
It is amount of solute in moles per liter of solution and represented by M
Formula used for Molarity
M = moles of solute / Liter of solution . . . . . . . . . . (1)
Data Given :
The concentration of half normal (NaCl) saline = 0.45g / 100 g
So,
Volume of Solution = 100 g = 100 mL
Volume of Solution in L = 100 mL / 1000
Volume of Solution = 0.1 L
molar mass of NaCl = 58.44 g/mol
Now to find number of moles of Nacl
no. of moles of NaCl = mass of NaCl / molar mass
no. of moles of NaCl = 0.45g / 58.44 g/mol
no. of moles of NaCl = 0.0077 g
Put values in the eq (1)
M = moles of solute / Liter of solution . . . . . . . . . . (1)
M = 0.0077 g / 0.1 L
M = 0.077 M
So the molarity of half-normal saline solution (0.45% NaCl) = 0.077 M