The density of a solution that has 40% by mass of $C H_{3} C O O H$ is 1.2 g/ml. The molarity of the solution is

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Solution

Let mass of solution be 100 gm
The mass of $C H_{3} C O O H$ present is 40 gm
Given density, d = 1.2 g/ml = 1200 g/L
Using formula,
$M = \frac{n}{m} \times d$
where
n = no. of moles
m = mass of solution (in gm)
d = density (in g/L)

$M = \frac{40 / 100}{60} \times 1200 = 8 M$

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The density of a solution that has 40% by mass of CH3COOH is 1.2 g/ml. The molarity of the solution is

Let mass of solution be 100 gm The mass of CH3COOH present is 40 gm Given density, d = 1.2 g/ml = 1200 g/L Using formula, M=nm×d where n = no. of moles m = mass of solution (in gm) d = density (in g/L) M=40/10060×1200=8 M

The density of a solution that has 40% by mass of $C H_{3} C O O H$ is 1.2 g/ml. The molarity of the solution is

Let mass of solution be 100 gm
The mass of $C H_{3} C O O H$ present is 40 gm
Given density, d = 1.2 g/ml = 1200 g/L
Using formula,
$M = \frac{n}{m} \times d$
where
n = no. of moles
m = mass of solution (in gm)
d = density (in g/L)

$M = \frac{40 / 100}{60} \times 1200 = 8 M$