A certain amount of NaOH is dissolved in certain kilograms of solvent and molality of the solution is '0.5 m' . When the same amount of NaOH is dissolved in "100 grams of less solvent than initial" then molality becomes '0.625 m'.
Determine the amount of NaOH and the initial mass of solvent .
Proper explanation expected. Best of luck guys ! :)
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Answer:
10grams
we all know that molality =moles(solute)/weight (solvent in kg)
Let moles of NaOH be x and weight of Solvent be Ykg
further solution in attachment
ALSO 2X=Y
X=.25
than y=.5kg
Y=500grams
⇝ Given :-
⇝ To Find :-
⇝ Solution :-
Let,
❒ We know molalaity (m) of solution is :
★ In 1st Condition Given That ;
When certain amount of NaOH is dissolved in certain kilograms of solvent and molality of the solution is '0.5 m.
❒ Hence, Using Formula of Molality :
✏ Multiplying Both Sides by 2
★ In 2nd Condition Given That ;
When the same amount of NaOH is dissolved in "100 grams of less solvent than initial" then molality becomes '0.625 m.
❒ Hence, Using Formula of Molality :
✏ Putting (1) in (2) :
✏ Putting Value of x in (1) :
As
We Know,
So,
Hence,
And,
Intial Mass of Solvent = y kg = 0.5 kg.
Hence,