2L of 0.2 M H2SO4 is reacted with 2L of 0.1 M NaOH solution, the molarity of the resulting product Na2SO4 in the solution is_________ millimolar (Nearest integer)

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6 months ago

H2SO4    +  2NaOH -> Na2SO4 + 2H2O             

Number of mole = 0.2 * 2                 2 * 0.1                 -              -

0.4 = 0.2

No. of moles after reaction =      0.4 – 0.1             0.2 – 0.2             0.1

= 0.3 = 0

molarity of Na2SO4 =   n V ( L ) = 0 . 1 4 = 0 . 0 2 5

= 25 milli molar

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Mole 1001620832

6.256.5 (HereO2isthelimitingreagent)

Hence mole of CO2 formed = 6.52

And wt of CO2  in gm = 6.52×44=143g

In 4d orbital, n = 4 and l=2

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Radial nodes = 4 – 2 – 1 = 1

And angular nodes,  l=2

 Molality=molesofsolute×1000wtofsolvent (gm)

Molality=35×1000×1.4636.5×100 = 14.0 M

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