I would love to help but it is 3 or is it 3 million?
Answer:
0.077 M
Explanation:
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 Liter = 100 mL / 1000
Volume of Solution = 0.1 L
molar mass of NaCl = 58.44 g/mol
Molarity:
Molarity is the representation of the solution. 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)
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
Solid KMnO₄ needed = 7.9 g
<h3>Further explanation</h3>
Given
MW KMnO₄ = 158 g/mol
500 mL(0.5 L) of a 0.1M stock solution of KMnO₄
Required
solid KMnO₄
Solution
Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution
Input the value :
n = M x V
n = 0.1 M x 0.5 L
n = 0.05 mol
Mass KMnO₄ :
= mol x MW
= 0.05 x 158 g/mol
= 7.9 g
<u>Given:</u>
Concentration of NaClO = 0.037 M
ka (HClO) = 4.0*10⁻⁸
<u>To determine:</u>
The pH of the NaClO solution
<u>Explanation:</u>
The hydrolysis of the weak base can be represented by the ICE table shown below-
ClO- + H2O ↔ HClO + OH-
Initial 0.037M 0 0
Change -x +x +x
Equilibrium (0.037-x) x x
kb = kw/ka = [HClO][OH-]/[ClO-]
10⁻¹⁴/4*10⁻⁸ = x²/(0.037-x)
x = [OH-] = 9.62*10⁻⁵
p[OH-] = -log[OH-] = -log [9.62*10⁻⁵] = 4.02
pH = 14-p[OH-] = 14 - 4.02 = 9.98
Ans: pH of the solution is 9.98
