What is the
of a 750 milliliters solution containing 54.0 grams of KBr?

0.605 M

Explanation:

What is the molarity of a 750 milliliters solution containing 54.0 grams of KBr?

Step 1: Given data

Step 2: Calculate the moles corresponding to 54.0 g of KBr

The molar mass of KBr is 119.00 g/mol.

54.0 g × 1 mol/119.00 g = 0.454 mol

Step 3: Calculate the molarity of the solution

Molarity is equal to the moles of solute divided by the liters of solution.

M = 0.454 mol / 0.750 L = 0.605 M

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

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 :