Mass percentage of a solution is the amount of solute present in 100 g of the solution.
Given data:
Mass of solute H2SO4 = 571.3 g
Volume of the solution = 1 lit = 1000 ml
Density of solution = 1.329 g/cm3 = 1.329 g/ml
Calculations:
Mass of the given volume of solution = 1.329 g * 1000 ml/1 ml = 1329 g
Therefore we have:
571.3 g of H2SO4 in 1329 g of the solution
Hence, the amount of H2SO4 in 100 g of solution= 571.3 *100/1329 = 42.987
Mass percentage of H2SO4 (%w/w) is 42.99 %
Answer : The value of 
at this temperature is 66.7
Explanation : Given,
Pressure of 
at equilibrium = 0.348 atm
Pressure of 
at equilibrium = 0.441 atm
Pressure of 
at equilibrium = 10.24 atm
The balanced equilibrium reaction is,
The expression of equilibrium constant for the reaction will be:
Now put all the values in this expression, we get :
Therefore, the value of at this temperature is 66.7
Answer: C= 0.406 M
Explanation:
Solution.
ν
=
0.730
m
o
l
;
ν=0.730mol;
V
=
1.8
⋅
1
0
3
m
L
=
1.8
L
;
V=1.8⋅10
3 mL=1.8L;
C=0.730mol
1.8 L=0.406 M
C= 1.8L
0.730mol =0.406M
The student made a mistake because he did not convert a unit of volume from milliliters to liters. After all, molarity is defined as the number of moles of solute per liter of solution.
