$(i)$ $(i)$ What mass of anhydrous $M g I_{2}$ $MgI_2$ is required to prepare a $400 \cdot m L$ $400mL$ volume of $0.0250 \cdot m o l \cdot L^{- 1}$ $0.0250mol*L^-1$ concentration?

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$(i)$ $(i)$ What mass of anhydrous $M g I_{2}$ $MgI_2$ is required to prepare a $400 \cdot m L$ $400mL$ volume of $0.0250 \cdot m o l \cdot L^{- 1}$ $0.0250mol*L^-1$ concentration?

$(i i)$ $(ii)$ Given $8.0 \cdot m o l \cdot L^{- 1}$ $8.0molL^-1$ $H C l O_{4} (a q)$ $HClO_4(aq)$ , what volume is required to prepare $4.00 \cdot L$ $4.00L$ of the acid at $0.250 \cdot m o l \cdot L^{- 1}$ $0.250mol*L^-1$ concentration?

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Answer 1 · 2017-06-05T10:21:16

Well, in both instances we use the relationship is:

$Concentration = \frac{Moles of solute}{Volume of solution}$ $"Concentration"="Moles of solute"/"Volume of solution"$

Explanation:

WE use the quotient to determine the number of moles of solute we need.......

i.e. if the solution is $0.0250 \cdot m o l \cdot L^{- 1}$ $0.0250*mol*L^-1$ with respect to $M g I_{2}$ $MgI_2$ , we needs a $400 \times 10^{- 3} L \times 0.0250 \cdot m o l \cdot L^{- 1} = 0.01 \cdot m o l$ $400xx10^-3Lxx0.0250*mol*L^-1=0.01*mol$ quantity of the anhydrous salt.

And this represents a mass of $0.01 \cdot m o l \times 278.11 \cdot g \cdot m o l^{- 1} = 2.78 \cdot g$ $0.01*molxx278.11*g*mol^-1=2.78*g$ .

And to prepare this solution, we take an accurately measured mass of magnesium iodide (if your salt is the hydrate, we change the mass appropriately), and dissolve it up in a $400 \cdot m L$ $400*mL$ volume of water.

The second solution REQUIRES that we add the ACID to water and NOT WATER to ACID (important! so get it right!).

We require $4.00 \cdot L \times 0.250 \cdot m o l \cdot L^{- 1} = 1.0 \cdot m o l$ $4.00*Lxx0.250*mol*L^-1=1.0*mol$ . And thus we START with $125 \cdot m L$ $125*mL$ of the $8.0 \cdot m o l \cdot L^{- 1}$ $8.0*mol*L^-1$ solution of $H C l O_{4} (a q)$ $HClO_4(aq)$ , and we add this to a $3.875 \cdot L$ $3.875*L$ volume of water.

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$(i)$ $(i)$ What mass of anhydrous $M g I_{2}$ $MgI_2$ is required to prepare a $400 \cdot m L$ $400mL$ volume of $0.0250 \cdot m o l \cdot L^{- 1}$ $0.0250mol*L^-1$ concentration?

$(i i)$ $(ii)$ Given $8.0 \cdot m o l \cdot L^{- 1}$ $8.0molL^-1$ $H C l O_{4} (a q)$ $HClO_4(aq)$ , what volume is required to prepare $4.00 \cdot L$ $4.00L$ of the acid at $0.250 \cdot m o l \cdot L^{- 1}$ $0.250mol*L^-1$ concentration?

1 Answer

Explanation:

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(i)(i) What mass of anhydrous MgI2MgI_2 is required to prepare a 400⋅mL400*mL volume of 0.0250⋅mol⋅L−10.0250*mol*L^-1 concentration?

(ii)(ii) Given 8.0⋅mol⋅L−18.0*mol*L^-1 HClO4(aq)HClO_4(aq), what volume is required to prepare 4.00⋅L4.00*L of the acid at 0.250⋅mol⋅L−10.250*mol*L^-1 concentration?

1 Answer

Explanation:

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$(i)$ $(i)$ What mass of anhydrous $M g I_{2}$ $MgI_2$ is required to prepare a $400 \cdot m L$ $400mL$ volume of $0.0250 \cdot m o l \cdot L^{- 1}$ $0.0250mol*L^-1$ concentration?

$(i i)$ $(ii)$ Given $8.0 \cdot m o l \cdot L^{- 1}$ $8.0molL^-1$ $H C l O_{4} (a q)$ $HClO_4(aq)$ , what volume is required to prepare $4.00 \cdot L$ $4.00L$ of the acid at $0.250 \cdot m o l \cdot L^{- 1}$ $0.250mol*L^-1$ concentration?