Ionic salt dissociate completely in water particularly in water at low concentration.
The molal freezing point of depression constant for water is 1.85kg/k/mol
therefore depression of freezing point =1.853 x 0.100 x 2.32=0.429 degrees celsius
hence solute freeze at - 0.429 degree celsius
Answer:
The atomic mass of second isotope is 7.016
Explanation:
Given data:
Average Atomic mass of lithium = 6.941 amu
Atomic mass of first isotope = 6.015 amu
Relative abundance of first isotope = 7.49%
Abundance of second isotope = ?
Atomic mass of other isotope = ?
Solution:
Total abundance = 100%
100 - 7.49 = 92.51%
percentage abundance of second isotope = 92.51%
Now we will calculate the mass if second isotope.
Average atomic mass of lithium = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
6.941 = (6.015×7.49)+(x×92.51) /100
6.941 = 45.05235 + (x92.51) / 100
6.941×100 = 45.05235 + (x92.51)
694.1 - 45.05235 = (x92.51)
649.04765 = x
92.51
x = 485.583 /92.51
x = 7.016
The atomic mass of second isotope is 7.016
Answer:
52 amu
Explanation:
To get the relative atomic mass of the element, we need to take into consideration, the atomic masses of the different isotopes and their relative abundances. We simply multiply the percentages with the masses. This can be obtained as follows:
[89/100 * 52] + [8/100 * 49] + [3/100 * 50]
46.28 + 3.92 + 1.5 =51.7 amu
The approximate atomic mass of element x is 52 amu
Answer:- 
Solution:- It is a volume unit conversion problem where we are asked to convert the volume from 
to microliters.
We know that:
= 1 mL
and, 
Let's use these conversions factors for the desired conversion using dimensional as:
=
So, the answer is .
