Pure sulfuric acid is a liquid of 1.84 density. What volume of it would be required to prepare 250 cm3 of 2.0 M solution?
(H = 1, S = 32, O = 16)
2.00
2.66
3.00
3.66
4.00
Explanation
Video Explanation
No video available
Post your Contribution
Discussions (19)
n=C×V/1000
n=250×2/1000
n=0.5moles
n=m/M
m=n×M
m=0.5×98=49g
p=m/v
V=m/p
V=49/1.84
V=26.6/10=2.66
p=density
given;
volume in cm³=250
molar conc. in M=0.2
density =1.84
so first we have to find the number of moles contained in the acid , then we find the mass of the acid
molar conc. = number of mole÷ volume
i.e, number of mole= (molar conc × volume ) ÷ 1000
number of mole=( 0.02×250)÷1000
number of mole= 50÷1000 =0.05mol/dm³
number of mole=mass÷molar mass molar mass of H2SO4 =
mass= number of mole×molar mass (1×2)+32+(16×4) =98g
mass = 0.05×98
mass=4.9g
density = mass÷ volume
volume = mass÷density
volume= 4.9÷1.84
volume= 2.66dm³
