Question

1- Complete the following table: P Τι V T P2 2atm 0.4atm 300K 900K 500mL 760 torr 150K 2500ml 227 0.6atm 1500 mL .75 L 2- How many gram of KOH is needed for a 1.5M solution mixed with 500 mL of water? 4- A 9.8 gram H3PO4 is mixed with 250 ml of water. How can we dilute this to a 0.25M solution?

Answer 1

1. From ideal gas law : PV = nRT

we get = P₁V₁/T₁ = P₂V₂/T₂  

- A =  V₂ ;  P₁V₁/T₁ = (0.4 atm * 2L / 300 K) = (2 atm *A / 900 K)= P₂V₂/T₂  

  A =  V₂ = 1.2 L

-  B =  P₂ ;  P₁V₁/T₁ = (760 torr * 500 ml / 150 K) = (B *2500 ml / 227K)= P₂V₂/T₂  

  B=  P₂ = 230 torr

- C = T₁ ;  P₁V₁/T₁ = (0.6 atm * 1.5 L / C) = (2.4 atm *0.75 L / 310 K)= P₂V₂/T₂  

1500 ml = 1.5 L and 27C = 310 K

C=  T₂ = 155 K = -118 C

2. Molar mass of KOH = 56 g/mol

we need 500 ml of 1.5 M solution. 1.5 M = 1.5 mol /L

or in 500 ml we need 0.75 moles of KOH for 1.5 M solution.

So, mass of KOH we need to add :  56 g/mol *0.75 moles = 42 g.

4. 9.8 g of H₃PO₄ added to 250 ml water.

molar mass of H₃PO₄ =98 g /mol

So, moles of H₃PO₄ added = 9.8g / 98 g/mol = 0.1 mole

Molarity of resulting solution - 0.1 mole / 0.250 L = 0.4 mol/L = 0.4 M

Dilute it to 0.25 M : 0.25 mol/L

we have to dilute it = 0.4M / 0.25 M = 1.6 times

So, we have to add 150 ml of water in 0.4 M 250 ml solution to make it 0.25 M.