Home
Class 12
CHEMISTRY
An average person needs about 10000 kJ e...

An average person needs about 10000 kJ energy per day. The amount of glucose (molar mass = 180.0 g `"mol"^(-1)`) needed to meet this energy requirement is _________ g.
(Use : `Delta_C H ` (glucose) = `-2700 ` KJ `"mol"^(-1)`)

Text Solution

AI Generated Solution

Promotional Banner

Topper's Solved these Questions

  • JEE MAINS 2021

    JEE MAINS PREVIOUS YEAR|Exercise CHEMISRY (SECTION-A)|21 Videos
  • JEE MAINS 2021

    JEE MAINS PREVIOUS YEAR|Exercise CHEMISRY (SECTION-B)|9 Videos
  • JEE MAINS 2021

    JEE MAINS PREVIOUS YEAR|Exercise Chemistry (Section B )|10 Videos
  • JEE MAINS 2020

    JEE MAINS PREVIOUS YEAR|Exercise CHEMSITRY|25 Videos
  • JEE MAINS 2022

    JEE MAINS PREVIOUS YEAR|Exercise CHEMISTRY (SECTION-B)|10 Videos

Similar Questions

Explore conceptually related problems

36g of glucose (molar mass = 180 g//mol) is present in 500g of water, the molarity of the solution is

An average healthy man needs about 10000 kJ of enrgy per day.How much carbohydrates (in mass) he will have to consume assuming that all this energy needs are met only by carbohydrates in the form of glucose? The enthalpy of combustion of glucose is 2816 kJ "mol"^(-1) .

10000kJ energy is needed per day and heat of combustion 2700kj/mol, then find the grams of glucose needed?

A solution is prepared by dissolving 0.6 g of urea (molar mass =60 g"mol"^(-1) ) and 1.8 g of glucose (molar mass =180g"mol"^(-1) ) in 100 mL of water at 27^(@)C . The osmotic pressure of the solution is: (R=0.08206L "atm"K^(-1)"mol"^(-1))

Calculate the freezing point of a solution containing 60 g glucose (Molar mass = 180 g mol^(-1) ) in 250 g of water . ( K_(f) of water = 1.86 K kg mol^(-1) )

A solution containing 15 g urea ( molar mass = 60 g mol^(-1) ) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol^(-1) ) in water . Calculate the mass of glucose persent in one litre of its solution.