29.2% (w//w) HCl stock, solution has a d...

`29.2% (w//w) HCl` stock, solution has a density of `1.25 g mL^(-1)`. The molecular weight of `HCl` is `36.5 g mol^(-1)`. The volume `(mL)` of stock solution required to prepare a `200 mL` solution of `0.4 M HCl` is :

Text Solution

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The correct Answer is:

`rarr 8`

Stock solution of `HCI = 29.2%(w//w)`
It means 29.2g of HCI is present in 100g of the solution Density of the solution `= 1.25 g mL^(-1)`
Volume of 100g of the solution `= (100)/(1.25)mL`
As molar mass of `HCI = 36.5 g mL^(-1)`
molarity of the solution
`= (29.2)/(36.5)( xx (1.25)/(100) xx 1000 = 10M`
`underset(("stock solution"))(M_(1)V_(1)) = underset(("solution required"))(M_(2)V_(2))`
`10 xx V_(1) = 0.4 xx 200` or `V_(1) = 8 mL`

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29.2% (W//W) HCl stock solution has density of 1.25g mL^(-1) . The molar mass of HCl is 36.5g mol^(-1) . The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4M HCl is

`8 mL`

`6 mL`

`7 mL`

`5 mL`

The density of 2.0 M solution of a solute is 1.2 g mL^(-1) . If the molecular mass of the solute is 100 g mol^(-1) , then the molality of the solution is

2.0 m

1.2 m

1.0 m

0.6 m

The density of 2.0 M solution of a solute is 1.2 g mL^(-1) . If the molecular mass of the solute is 100 g mol^(-1) , then the molality of the solution is :