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3.15 Grams of oxalic acid crystals are p...

3.15 Grams of oxalic acid crystals are present in a 500 mL aqueous solution. Calculate (a) molarity and (b) normality.

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Molarity , `M = w/(GMW) xx (1000)/V = (3.15)/(126) xx (1000)/(500) = 6.3/126 = 0.05 mol L^(-1)`
Basicity of oxalic acid = n = 2
Normality , `N = n xx M = 2 xx 0.05 = 0.1 eq L^(-1)`.
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AAKASH SERIES-DILUTE SOLUTIONS-EXERCISE - 1.2
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  2. What are the factor on-which solubility depends on ?

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  3. Discuss on the vapour pressure of solutiosn of solids in liquids.

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  4. Aerated water bottles are kept under cold water during summer. Why ?

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  5. What is relative lowering of vapour pressure ?

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  6. Write the units of cryoscopic constant and ebullioscopic constant?

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  7. How is osmotic pressure useful in determining molar masses?

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  8. How is the principle of reverse osmosis? Used for desalination?

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  9. The molar concentration of proton obtained by mixing equal volumes of ...

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  11. A freash grape fruit dipped in brine solution shrinks. Why ?

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  12. Vapour pressure of beneze and toluene are 160 are 60 torr. What will b...

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  13. How water can be made boiling without heating?

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  14. Write the reason of abnormality in molar masses.

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  15. Relate the colligative properties with van't Hoff factor.

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  17. Calculate the normally of 10.6 % (w/v) Na2CO3 in aqueous solutions.

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  18. How many mL of 0.1 M ferrous oxalate can reduce 10 mL of 0.2 M potassi...

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  19. 250 mL of 0.2 M NaOH and 100 mL of 0.25 M Ba (OH)(2) are mixed. Calcu...

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  20. The density of 8.653% (w/v) Na2CO3 solution is 1.018 g mL^(-1). Calcul...

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  21. 100 mL of ethanol (density 0.8g/cc) is diluted to one litre using wate...

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