Question

molarity of 98% by mass of H2SO4 whose density is 1.8g/ml

Answer 1

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

Answer 2

Answer: The molarity of the solution will be 18.18 M

Explanation:

Molarity can be written as:

$Molarity=\frac{\text{Given mass of solute}\times 1000}{\text{Molecular mass of solute}\times \text{Volume (in mL)}}$..(1)

Molar mass of $H_2SO_4$ = 98 g/mol

We are given 98% of $H2SO_4$ by mass which means that 98 grams of sulfuric acid is present in 100 grams of solution.

$Density=\frac{Mass}{Volume}$

$1.8g/mL=\frac{100g}{Volume}$

on solving for we get volume

$Volume=55.55 mL=0.055 L$

For molarity using equation (1)

$molarity=\frac{98 g}{98 g/mol\times 0.055 L}=18.18 M$