Test Index

NCERT Class XII Chemistry
Chapter - Solutions
Questions with Solutions

© examsnet.com
Question : 48
Total: 53
Henry’s law constant for CO2 in water is 1.67 × 10‌8 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K.
Solution:  
Given, KH=1.67×108 Pa,
PCO2=2.5atm=2.5×101325 Pa
Applying Henry's law, P CO2=KH×x CO2
∴‌‌x CO2=
P CO2
KH
=
2.5×101325 Pa
1.67×108 Pa
=1.517×10−3
i.e., x CO2=
n CO2
n H2 O+n CO2
=
n CO2
nH2 O
=1.517×10−3
‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌[n CO2 is negligible w.r.t. n H2 O ]
For 500 mL of soda water,
Volume of water = 500 mL; mass of water = 500 g,
moles of water =
500
18
=27.78 i.e., nH2 O=27.78
∴‌
n CO2
27.78
=1.517×10−3
or n CO2=42.14×10−3 mole,
Mass of CO2=42.14×10−3×44 g
=1.854 g
© examsnet.com
Go to Question:
Prev Question
Next Question