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Question #02f6e

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Feb 1, 2018

$= 60.15g M g S O_{4}$ $="60.15g " MgSO_4$

Explanation:

Find the molar mass of $M g S O_{4}$ $MgSO_4$ which relative atomic masses of elements constituting the compound is obtainable in the periodic table.
$= 120 g/mol$ $=120" g/mol"$
Given the concentration of the solution, and the assumed volume, the number of moles $(η)$ $(eta)$ of the solute can be computed as described below;
$M = \frac{η}{V}$ $M=eta/V$ ; rearrange formula and isolate $η$ $eta$
$η = M \times V$ $eta=MxxV$

$w h e r e :$ $where:$

$M = M o l a r i t y = 0.5 M = 0.5 mol/L$ $M=Molarity=0.5M=0.5" mol/L"$
$V = V o l u m e = 1 L$ $V=Volume=1L$

$η = M V$ $eta=MV$
$eta=(0.5mol)/cancel((L))xx1cancel(L)$
$eta=0.5molMgSO_4$

Now, use the molar mass of the compound to find the mass of the solute; i.e.,
$=0.5molcancel(MgSO_4)xx(120.3gMgSO_4)/(1cancel(molMgSO_4))$
$=60.15gMgSO_4$

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