$$0.141$$ g of a metal when dissolved in dil. HCl evolved $$31.1$$ mL of $$H2$$ when collected over water at $$13$$°C and $$739.2$$ mm pressure. Calculate the weight of oxygen present in $$102$$ g of the oxide of the metal (aqueous$$ tension at $$13$$°C $$=11.2$$ $$mm),

Question

$$0.141$$ g of a metal when dissolved in dil. HCl evolved $$31.1$$ mL of $$H2$$ when collected over water at $$13$$°C and $$739.2$$ mm pressure. Calculate the weight of oxygen present in $$102$$ g of the oxide of the metal (aqueous$$ tension at $$13$$°C $$=11.2$$ $$mm),

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(i) Wt. of metal $$=0.141g$$. According to gas equation :$$P1V1T1=P2V2T2$$  or  $$V2=P1V1T2T1P2Where$$ at NTP,                      $$T2=273K$$, $$P2=760mm$$,$$V2=$$?$$P1=739.2$$−$$11.2=728mm$$,$$V1=31.1mL$$,$$T1=13+273=286K$$.∴ $$V2=728mm$$×$$31.1mL$$×$$273K286K$$×$$760mm=28.4mL$$.We know,               wt. of metal  wt. of $$H2$$ at $$NTP=$$ Eq.wt. of metal  Eq. wt. of hydrogen $$(=1.008)=V2$$×$$0.00009$$∴ Eq. wt. of metal $$=$$ wt. of metal ×$$1.00828.4$$×$$0.00009g=0.141g$$×$$1.00828.4$$×$$0.00009g=55.6(ii)$$ Let wt. of metal $$=$$ x g; wt. of oxygen                               $$=(102$$−$$x)g$$ Eq. wt. of metal $$=$$ wt. of metal ×$$8$$ wt. of $$O2$$;$$55.6=8x102$$−x∴(55.6$$×$$102)−$$55.6x=8x$$;$$63.6x=5671.2$$∴ $$x=5671.2/63.6=89.17$$∴  wt. of $$O2=102$$−$$89.17=12.83g$$ Ans.

0.141 g of a metal when dissolved in dil. HCl evolved 31.1 mL of H2 when collected over water at 13°C and 739.2 mm pressure. Calculate the weight of oxygen present in 102 g of the oxide of the metal (aqueous tension at 13°C =11.2 mm),

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