📖 Samacheer Kalvi · 11th TN - English Medium · Chemistry Volume 1 · Page 246question

Unit - 1 Basic Concepts of Chemistry and Chemical Calculations · Part 4

Chapter 3: 7 · Chemistry Volume 1

is correct . ( ) ( ) ( )  u . Correct reason: Total number of entities present in one mole of any substance is equal to . × .

. Reaction : C + O → CO × g carbon combines with g of oxygen. Hence, Equivalent mass of carbon = Reaction : C + O → CO g carbon combines with g of oxygen. Hence, Equivalent mass of carbon = .

Let the trivalent metal be M + Equivalent mass mass of the metal / valance factor g eq - mass of the metal / eq Mass of the metal g Oxide formed M O ; Mass of the oxide = ( x ) + ( x ) g 11th Std Chemistry 11th Std Chemistry - - - - . Weight of the water drop = . g No. of moles of water in the drop = Mass of water / molar mass = .

/ = - mole No of water molecules present in mole of water = . x No. water molecules in one drop of water ( - mole) = . x x - = .

x . MgCO →MgO + CO ↑ MgCO : ( x ) + ( x ) + ( x ) = g CO : ( x ) + ( x ) = g % pure g MgCO on heating gives g CO Given that g of MgCO on heating gives . g CO Therefore, g MgCO sample on heating gives . g CO Percentage of purity of the sample = .

= % % g CO g CO Percentage of impurity = % . NaHCO +

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