Question

calculate the molality of a solution by adding 68.0 g of sodium nitrate to 300.0 ml of water.

Answer 1

Molality (m) of a solution is given by the relation,

Molality (m ) = No. of moles of solute / mass of solvent in Kg

Considering density of water = 1.00 gm / ml

Mass of 300ml of water = 300 .00 gms

...................................or = 0.300Kgs

and moles of solute ( ie. NaNO3) = 68.0 / 84.995...........[ Molar mass of NaNO3 is 84.995]

.................................................... = 0.800

Molality of NaNO3 solution .... = 0.800 / 0.300

...................................................... = 2.67 m

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