1.

29.9% (w/w) HCl stock solution has a density of "1.25 g mL"^(-1). The molecular weight of HCl is "6.5 g mol"^(-1). The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is

Answer»

SOLUTION :Stock solution of HCl `=29.2%(w//w)`
Thus, 29.2 G of HCl are present in 100 g of the solution
As density of the solution `="1.25 g mL"^(-1)`,
`"volume of 100 g of the solution "=(100)/(1.25)"mL"`
As molar mass of HCl `="36.5 g mol"^(-1)`, molarity of the solution
`=(29.2)/(36.5)xx(1.25)/(100)xx1000=10M`
`UNDERSET("(stock solution)")(M_(1)V_(1))=underset("(solution required)")((M_(2)V_(2)))`
`10xxV_(1)=0.4xx200"or"V_(1)="8 mL"`


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