What is the molarity of the following solutions?

a. 19.5 g NaHCO3 in 460.0 ml solution
b. 26.0 g H2SO4 in 200.0 mL solution
c. 15.0 g NaCl dissolved to make 420.0 mL solution

a) NaHCO3 = 0.504 M

b) H2SO4 = 1.325 M

c) NaCl = 0.610 M

Explanation:

Step 1: Data given

Moles = mass / molar mass

Molarity = moles / volume

a. 19.5 g NaHCO3 in 460.0 ml solution

Step 1: Data given

Mass NaHCO3 = 19.5 grams

Volume = 460.0 mL = 0.460 L

Molar mass NaHCO3 = 84.0 g/mol

Step 2: Calculate moles NaHCO3

Moles NaHCO3 = 19.5 grams / 84.0 g/mol

Moles NaHCO3 = 0.232 moles

Step 3: Calculate molarity

Molarity = 0.232 moles / 0.460 L

Molarity = 0.504 M

b. 26.0 g H2SO4 in 200.0 mL solution

Step 1: Data given

Mass H2SO4 = 26.0 grams

Volume = 200.0 mL = 0.200 L

Molar mass H2SO4 = 98.08 g/mol

Step 2: Calculate moles H2SO4

Moles H2SO4 = 26.0 grams / 98.08 g/mol

Moles H2SO4 = 0.265 moles

Step 3: Calculate molarity

Molarity = 0.265 moles / 0.200 L

Molarity =1.325 M

c. 15.0 g NaCl dissolved to make 420.0 mL solution

Step 1: Data given

Mass NaCl = 15.0 grams

Volume = 420.0 mL = 0.420 L

Molar mass NaCl = 58.44 g/mol

Step 2: Calculate moles NaCl

Moles NaCl = 15.0 grams / 58.44 g/mol

Moles NaCl = 0.256 moles

Step 3: Calculate molarity

Molarity = 0.256 moles / 0.420 L

Molarity =0.610 M

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

its first and third is your answer i did the qustion the fourth isnt right