Calcualte the weight of H(2)O present in...

Calcualte the weight of `H_(2)O` present in 100 g sample of `Mg SO_(4). 7H_(2)O`

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`%W//W = 100 xx ( 51. 2)/( 100) = 51. 2 g`

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Find the wt of water present in 1.61 g of Na_(2)SO_(4).10H_(2)O

Oleum is considered as a solution of SO_(3) in H_(2)SO_(4) , which is obtained by passing SO_(3) in solution of H_(2)SO_(4) When 100 g sample of oleum is diluted with desired mass of H_(2)O then the total mass of H_(2)SO_(4) obtained after dilution is known is known as % labelling in oleum. For example, a oleum bottle labelled as ' 019% H_(2)SO_(4) ' means the 109 g total mass of pure H_(2)SO_(4) will be formed when 100 g of oleum is diluted by 9 g of H_(2)O which combines with all the free SO_(3) present in oleum to form H_(2)SO_(4) as SO_(3)+H_(2)O to H_(2)SO_(4) What is the % of free SO_(3) in an oleum that is labelled as '104.5% H_(2)SO_(4)' ?

Oleum is considered as a solution of SO_(3) in H_(2)SO_(4) , which is obtained by passing SO_(3) in solution of H_(2)SO_(4) When 100 g sample of oleum is diluted with desired mass of H_(2)O then the total mass of H_(2)SO_(4) obtained after dilution is known is known as % labelling in oleum. For example, a oleum bottle labelled as ' 019% H_(2)SO_(4) ' means the 109 g total mass of pure H_(2)SO_(4) will be formed when 100 g of oleum is diluted by 9 g of H_(2)O which combines with all the free SO_(3) present in oleum to form H_(2)SO_(4) as SO_(3)+H_(2)O to H_(2)SO_(4) 1 g of oleum sample is diluted with water. The solution required 54 mL of 0.4 N NaOH for complete neutralization. The % free SO_(3) in the sample is :

Oleum is considered as a solution of SO_(3) in H_(2)SO_(4) , which is obtained by passing SO_(3) in solution of H_(2)SO_(4) When 100 g sample of oleum is diluted with desired mass of H_(2)O then the total mass of H_(2)SO_(4) obtained after dilution is known is known as % labelling in oleum. For example, a oleum bottle labelled as ' 019% H_(2)SO_(4) ' means the 109 g total mass of pure H_(2)SO_(4) will be formed when 100 g of oleum is diluted by 9 g of H_(2)O which combines with all the free SO_(3) present in oleum to form H_(2)SO_(4) as SO_(3)+H_(2)O to H_(2)SO_(4) 9.0 g water is added into oleum sample lablled as "112%" H_(2)SO_(4) then the amount of free SO_(3) remaining in the solution is : (STP=1 atm and 273 K)

Oleum is considered as a solution of SO_(3) in H_(2)SO_(4) , which is obtained by passing SO_(3) in solution of H_(2)SO_(4) When 100 g sample of oleum is diluted with desired mass of H_(2)O then the total mass of H_(2)SO_(4) obtained after dilution is known is known as % labelling in oleum. For example, a oleum bottle labelled as ' 019% H_(2)SO_(4) ' means the 109 g total mass of pure H_(2)SO_(4) will be formed when 100 g of oleum is diluted by 9 g of H_(2)O which combines with all the free SO_(3) present in oleum to form H_(2)SO_(4) as SO_(3)+H_(2)O to H_(2)SO_(4) If excess water is added into a bottle sample labelled as "112% H_(2)SO_(4) " and is reacted with 5.3 g NaCO_(3) then find the volume of CO_(2) evolved at 1 atm pressure and 300 K temperature after the completion of the reaction :

Comprehension # 7 Oleum is considered as a solution of SO_(3) in H_(2)SO_(4) , which is obtained by passing SO_(3) in solution of H_(2)SO_(4) . When 100g sample of oleum is diluted with desired weight of H_(2)O then the total mass of H_(2)SO_(4) obtained after dilution is known as % labelling in oleum. For example, a oleum bottle labelled as 109% H_(2)SO_(4) means the 109g total mass of pure H_(2)SO_(4) will be formed when 100g of oleum is diluted by 9g of H_(2)O which combines combines with all the free SO_(3) present in oleum to form H_(2)SO_(4) as SO_(3)+H_(2)OrarrH_(2)SO_(4) 9.0 g water is added into 100g oleum sample labelled as 112%H_(2)SO_(4) then the amount of free SO_(3) remaining in the solution is :

CENGAGE CHEMISTRY-CHEMICAL ARITHMETIC -MANDATORY EXERCISE (EXERCISE SET III)

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