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What is the molarity of h2so4 solution that has a density of 1.84 g/cc and contains 98% by mass of h2so4?(atomic mass of s=32)? Ans=18.4M How?

What is the molarity of h2so4 solution that has a density of 1.84 g/cc and contains 98% by mass of h2so4?(atomic mass of s=32)? Ans=18.4M 
How?
 

Grade:11

3 Answers

Ashutosh Sharma
askIITians Faculty 181 Points
8 years ago
Molar mass of H2SO4 = 98g/mol = Mm
D =density = 1.84g/cc
W/w = 98%
V = 1dm3 = 1L = 1000ml = 1000cc
Mass of solution = V x D
M = 1000 x 1.84 = 1840g per 1000cc
Using w/w, 98% of 1840g = H2SO4
= (98/100) x 1840 = 1803.2g = m
Nos of moles = m/Mm = 1803.2/98 = 18.4moles per 1000cc
Thus, molarity = 18.4M (18.4mol/dm3)
Swagata
13 Points
4 years ago
98% h2so4 by weight means 98g h2so4 in 100g of the solution
Given density = 1.84g/cc
Density of solution = mass of solution /volume of solution
1.84 = 100/ volume
Volume = 100/1.84 = 54.3 ml
No.of moles of h2so4 = weight of h2so4/molar mass of h2so4 = 98/98 = 1 mole
 Molarity = mole * 1000/54.3 = 18.41M 
 
Ajeet Tiwari
askIITians Faculty 86 Points
3 years ago
hello students

we know that,
98% h2so4 by weight means 98g h2so4 in 100g of the solution
Given density = 1.84g/cc
Density of solution = mass of solution /volume of solution
1.84 = 100/ volume
Volume = 100/1.84 = 54.3 ml
No.of moles of h2so4 = weight of h2so4/molar mass of h2so4 = 98/98 = 1 mole
Molarity = mole * 1000/54.3 = 18.41M

hope it helps
Thankyou and Regards

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