Two types of stoichiometry are; molar mass and coefficients from balanced equation.
The correct answer is C. Colligative properties only depend upon the number of solute particles in a solution but not on the identity or nature of the solute and solvent particles. I hope this anwers your question.
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol
15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol 2 mol
0.100 mol 0.200 mol
We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M
Concentration of NO3⁻ is 0.667 M.
Answer: The pH of a 4.4 M solution of boric acid is 4.3
Explanation:
at t=0 cM 0 0
at eqm 
So dissociation constant will be:
Give c= 4.4 M and 
= ?
Putting in the values we get:
![[H^+]=c\times \alpha](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Ctimes%20%5Calpha)
![[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D4.4%5Ctimes%200.000011%3D4.8%5Ctimes%2010%5E%7B-5%7DM)
Also ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
![pH=-log[4.8\times 10^{-5}]=4.3](https://tex.z-dn.net/?f=pH%3D-log%5B4.8%5Ctimes%2010%5E%7B-5%7D%5D%3D4.3)
Thus pH of a 4.4 M 
solution is 4.3
