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An-aqeous-solution-of-tetraoxosulphate-IV-has-a-density-of-1-80-g-cm-3-and-98-purity-level-what-volume-of-this-solution-must-be-diluted-to-give-250-cm-3-of-0-500-mol-dm-3-H-2-SO-4-solution-




Question Number 6583 by Tawakalitu. last updated on 04/Jul/16
An aqeous solution of tetraoxosulphate(IV) has a density   of 1.80 g/cm^3  and 98% purity level. what volume of this   solution must be diluted to give 250 cm^3  of 0.500 mol/dm^3   H_2 SO_4  solution ?
$${An}\:{aqeous}\:{solution}\:{of}\:{tetraoxosulphate}\left({IV}\right)\:{has}\:{a}\:{density}\: \\ $$$${of}\:\mathrm{1}.\mathrm{80}\:{g}/{cm}^{\mathrm{3}} \:{and}\:\mathrm{98\%}\:{purity}\:{level}.\:{what}\:{volume}\:{of}\:{this}\: \\ $$$${solution}\:{must}\:{be}\:{diluted}\:{to}\:{give}\:\mathrm{250}\:{cm}^{\mathrm{3}} \:{of}\:\mathrm{0}.\mathrm{500}\:{mol}/{dm}^{\mathrm{3}} \\ $$$${H}_{\mathrm{2}} {SO}_{\mathrm{4}} \:{solution}\:? \\ $$$$ \\ $$
Answered by sandy_suhendra last updated on 04/Jul/16
M_1 = ((1.8 × 98 ×10)/(98)) =18 mol/dm^3   M_(2 ) = 0.5 mol/dm^3   V_2  = 250 cm^3   V_1  = ?  V_1  × M_1  = V_2  × M_2   V_1  × 18 = 250 ×0.5  V_1 = 6.94 cm^3
$${M}_{\mathrm{1}} =\:\frac{\mathrm{1}.\mathrm{8}\:×\:\mathrm{98}\:×\mathrm{10}}{\mathrm{98}}\:=\mathrm{18}\:{mol}/{dm}^{\mathrm{3}} \\ $$$${M}_{\mathrm{2}\:} =\:\mathrm{0}.\mathrm{5}\:{mol}/{dm}^{\mathrm{3}} \\ $$$${V}_{\mathrm{2}} \:=\:\mathrm{250}\:{cm}^{\mathrm{3}} \\ $$$${V}_{\mathrm{1}} \:=\:? \\ $$$${V}_{\mathrm{1}} \:×\:{M}_{\mathrm{1}} \:=\:{V}_{\mathrm{2}} \:×\:{M}_{\mathrm{2}} \\ $$$${V}_{\mathrm{1}} \:×\:\mathrm{18}\:=\:\mathrm{250}\:×\mathrm{0}.\mathrm{5} \\ $$$${V}_{\mathrm{1}} =\:\mathrm{6}.\mathrm{94}\:{cm}^{\mathrm{3}} \\ $$
Commented by sandy_suhendra last updated on 04/Jul/16
M_1 =((density × % purity × 10)/(Mr))
$${M}_{\mathrm{1}} =\frac{{density}\:×\:\%\:{purity}\:×\:\mathrm{10}}{{Mr}} \\ $$
Commented by Tawakalitu. last updated on 05/Jul/16
Thanks so much
$${Thanks}\:{so}\:{much} \\ $$

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