The density of 3M solution of NaCL is 1....

The density of 3M solution of NaCL is 1.25 g`mL^(-1)`. Calculate molality of the solution.

Text Solution

Verified by Experts

M = 3 mol `L^(-1)`
Mass of NaCl
in 1 L solution = `3xx58.5=175.5g`
Mass of 1 L solution = `1000xx1.25` = 1250 g
(since density = 1.25 g `mL^(-1)`)
Mass of water in solution = 1250 - 175.5
= 1074.5 g = 1.0745 kg.
Molality = `("No. of moles of solute")/("Mass of solvent in kg")`
`=(3mol)/(1.0745kg)=2.79m`
Often in a chemistry laboratory, a solution of a desired concentration is prepared by diluting a solution of known known higher concentration. The solution of higher concentration is also known as stock solution. Note that molality of a solution does not change with temperature since mass remains unaffected with temperature.

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2.3

0.49

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2.28 mol `kg^(-1)`

0.44 mol `kg^(-1)`

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Density of a 2.05 M solution of acetic acid in water is 1.02 gm ml^(-1) . The molality of the solution is

`1.14molkg^(-1)`

`3.28molkg^(-1)`

`2.28molkg^(-1)`

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