2.5 g of an impure sample of sodium bicarbonate when heated strongly gave 300ml of carbon dioxide measured at 27 degree celsius and 760 mm preasure . Calculate thew percentage purity of the sample.

A sample of pure sodium carbonate 0.318g is dissolved in water and titrated with HCl solution. A volume of 60mL is required to reach the methyl orange end point. Calculate the molarity of the acid.

40 g Ba(MnO_(4))_(2) (mol.mass=375) sample containing some inert impurities in acidic medium completely reacts with 125 mL of "33.6 V" of H_(2)O_(2) . What is the percentage purity of the sample ?

A 1.025 g sample containing a weak acid HX (mol. Mass=82) is dissolved in 60 mL. water and titrated with 0.25 M NaOH. When half of the acid was neutralised the pH was found to be 5.0 and at the equivalence point the pH is 9.0. Calculate mass precentage of HX in sample :

0.80g is impure (NH_4)SO_4) was boiled with 100mL of 0.2N NaOH solution till all the NH_3 (g) evolved. The remaining solution was diluted to 250mL. 25mL of this solution was neutralized using 5mL of 0.2 NH_2 SO_4 solution. The percentage purity of the (NH_4)_2SO_4 sample is :

When 2.5g of a sample of mohr's salt reacts completely with 50mL of N/10 KMnO_4 solution. The % purity of the sample of Mohr's salt is :

1 M NaOH solution was slowly added in to 1000 mL of 183.75 g impure H_(2)SO_(4) solution and the following plot was obtained. The percentage purity of H_(2)SO_(4) sample and slope of the curve respectively are:

To a 10mL, 1M aqueous solution of Br_(2) , excess of NaOH is added so that all Br_(2) is disproportionated to Br^(-) and BrO_(3)^(-) . The resulting solution is free from Br^(-) , by extraction and excess of OH^(-) neutralised by acidifying the solution. The resulting solution is suffcient to react with 2 g of impure CaC_(2)O_(4) (M= 128g/mol) sample. The % purity of oxalate sample is :

A sample of 1.0g of solid Fe_(2)O_(3) of 80% purity is dissolved in a moderately concentrated HCl solution which is reduced by zinc dust. The resulting solution required 16.7mL of a 0.1M solution of the oxidant. Calculate the number of electrons taken up by the oxidant.

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 :

6.2g of a sample containing Na_2CO_3, NaHCO_3 and non-volatile inert impurity on gentle heating loses 5% of its mass due to reaction 2NaHCO_3 rarr Na_2 CO_3 + H_2 O + CO_2 . Residue is dissolved in water and formed 100mL solution and its 10mL portion requires 7.5 mL of 0.2M aqueous solution of BaCI_2 for complete precipitation of carbonates. Determine mass (in gram) of Na_2 CO_3 in the original sample.