Question

1.. How many atoms of O in 7.80g of Ca(ClO3)2?

2 How many liters (at STP) are 4.39 x 10 ^25 molecules of nitrogen monoxide?

3. How many grams of NaOH are in 250.0 ml of a 5.0 M solution?

Answer 1

1) 1mole of Ca(ClO3)2 is having 6 moles of O atoms

    no. of moles = 7.8/207 = 0.037 moles

0.037 moles of Ca(ClO3)2 .........................?

     = 0.037*6 = 0.23 moles

1 mole of O atoms means ..............................6.023*10^23 atoms

0.23 moles of O atoms ..................................?

    = 1.38*10^23 O atoms

no. of O atoms in Ca(ClO3)2 is 1.38*10^23 or 1.4*10^23

..................................................................................................

2)

1 mole of any gas at STP occupies 22.4 lt

at STP

6.023*10^23 NO molecules .................... 1mole of NO

no.of moles of NO = (4.39*10^25/6.023*10^23) = 72.88 moles

1 mole of any gas at STP occupies ..................... 22.4 lt of vol.

72.88 moles of NO gas ............................................?

        = 1632.51 lt.

...............................................................................................................................................

3)

Molarity = (wt/mol.wt)*(1000/vol.inml)

5 = (wt/40)*(1000/250)

weight of NaOH = 50 gm.