"If 500ml Of Nh3 Contains 12g Nh3, What Is The Volume Of 10g Nh3? N=14g H=1g (Avogadros Law)"

If 500mL of NH3 contains 12g NH3, what is the volume of 10g NH3? N=14g H=1g (Avogadros Law)

 

Good Day....

Problem:

If 500mL of NH3 contains 12g NH3, what is the volume of 10g NH3? N=14g H=1g (Avogadros Law)

• The molar mass of NH3 = 17 g / mol
• To convert grams to moles, divide the given mass by the molar mass of NH3.

12 g NH3 ÷ 17 g/mol = 0.706 mol

10 g NH3 ÷ 17 g/mol = 0.588 mol

Given Data:

V1 (initial volume) = 500 ml                  

V2 (final volume) = ? unknown

n1 (initial amount of gas) = 12 g (0.706 mol)

n2 (final amount of gas) = 10 g (0.588 mol)

Solution:

• The formula of avogadros law is V1/n1 = V2/n2

• To find V2 we can use the formula V2 = V1 × n2 / n1

V2 = V1 × n2 / n1

     = 500 ml × 0.588 mol / 0.706 mol   (cancel the unit mol)

    = 294 ml / 0.706

     = 416.43 ml

Answer:

volume will be 416.43 ml

Hope it helps....=)