Question

1. Complete the following table: Т. P V? T: 900K 0.4atm 300K 2atm А 760 torr 500ml 150K B 2500mL 227 C 0.6atm 1500 mL 2.4 atm .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

Ans 1 :

Using the ideal gas law , pV = nRT , we get the relation as :

P1V1T2 = P2V2T1

putting values :

For A :

0.4 atm x 2 L x 900 K = 2 atm x A x 300 K

A = 1.2 L

For B :

p1 = 760 torr = 1 atm

v1 = 500 mL = 0.5 L

T1 = 150 K

v2 = 2500 mL = 2.5 L

T2 = 227 C = 500 K

putting values :

1 atm x 0.5 L x 500 K = B x 2.5 L x 150 K

B = 0.67 atm

B in torr = 0.67 x 760 torr

= 506.7 torr

For C :

p1 = 0.6 atm

v1 = 1500 mL = 1.5 L

T1 = C

p2 = 2.4 atm

V2 = 0.75 L

T2 = 27 C = 300 K

putting values :

0.6 atm x 1.5 L x 300 K = 2.4 atm x 0.75 L x C

C = 150 K