# Concentrated aqueous sulphuric acid is 98% H2SO4 by mass and has a density of 1.80 gmL . Volume of acid required to make 1 L of 0.1 M H2SO4 solution is

Dhruv
11 Points
7 years ago
98% by weight means 98 gm H2SO4 in 100 gm solution or 98 g H2SO4 in 2 g water.Number of moles of H2SO4 = 98/98 = 1Density = 1.8 g/LVolume of solution = Mass of solution/ Density = 100/1.8 = 55.55 ml = 0.055 L Applying,M1V1 = M2V218.18 x V1 = 0.1 x1V1 = 0.0055 L = 5.5 ml Hence , volume of acid required = 5.5 ml
Deepak
22 Points
6 years ago
Concentrated aqueous sulphuric acid is 98% H2SO4 by massDensity = 1.80gm/LNumber of moles in H2SO4 of Concentration 0.1 M and volume 1 l =n = C × v= 0.1 × 1= 0.1Calculate the mass required to get 0.1 mol H2SO4 n = unknown mass/Molar mass= unknown mass/98 gmol-1 0.1 × 98 gmol-1 = 9.8 g?We need 9.8 gram H2SO4Convert this mass in volume using density = mass/volume1.80 gml-1 = 9.8g /volumevolume = 9.8 g/1.8g ml-1= 5.44 ml.5.44 ml will require if solution is 100 % pure, But given it is 98%So the volume required = 100 x 5.44/98= 5.55 mlAns: The volume of acid required to make 1 litre of 0.1 M H2SO4
Ashish Kumar
26 Points
6 years ago
As we knowM1V1=M2V2 ; M=(mole of solute)÷volume of solution (in L)Mole=(mass)÷GMM. ;MASS=density(d)×volumeSo,M1V1=M2V2{{(d×v)÷98}÷v}1×v1=o.1×1d1v1÷98=0.1And,v1=9.8÷d=9.8÷1.8=5.5
11 Points
6 years ago
As we know M1V1=M2V2. On calculate the value of M1 we get 18 [98×1.80×10/98] We are given M2 = 0.1 and V2 = 1000mlOn applying M1V1= M2V2 then V1 =M2V2/M1= 5.5 [0.1×1000/18] So the required volume is 5.5 ml
Disha
15 Points
4 years ago
Given
Density of solution=1.84 g/ml
Mass% of H2SO4= 98%
98% of H2SO4 by mass means 98 g of H2SO4 is present in 100 g of water.
Mass of solute (H2SO4) =98g
Mass of solution=100g
Volume of solution = mass/density=100/1.84
= 54.35
No. Of moles of solute = given mass/molar mass
= 98/98=1
Now
Molarity = no. Of moles of solute/ volume of sol in ml *1000 =1/ 54.35* 1000=18.39
We know
M1V1=M2V2
18.39*V1=0.5*5000ml
V1=0.5*5000/18.39= 135.94 ml

Rishi Sharma
4 years ago
Hello students,
The solution of the above problem is in the attached file.
I hope the solution will solve all your doubts.
Thank You,
All the Best for the Exams.