the molar mass of the element is 81.36 g/mol
<u><em>calculation</em></u>
step 1 : multiply each %abundance of the isotope by its mass number
that is 79.95 x 29.9 =2391
81.95 x 70.1 = 5745
Step 2: add them together
2390.5 + 5744.7 =8136
Step 3: divide by 100
= 8136/100 = 81.36 g/mol
Answer:
Option C. 4.03 g
Explanation:
Firstly we analyse data.
12 % by mass, is a sort of concentration. It indicates that in 100 g of SOLUTION, we have 12 g of SOLUTE.
Density is the data that indicates grams of solution in volume of solution.
We need to determine, the volume of solution for the concentration
Density = mass / volume
1.05 g/mL = 100 g / volume
Volume = 100 g / 1.05 g/mL → 95.24 mL
Therefore our 12 g of solute are contained in 95.24 mL
Let's finish this by a rule of three.
95.24 mL contain 12 g of sucrose
Our sample of 32 mL may contain ( 32 . 12) / 95.24 = 4.03 g
Answer:
There are 0.219 mol of LINO3
