224mL of SO2(g) at 298K and 1 atm passes through 100mL of 0.1M NaOH solution. The non-volatile solute produced is dissolved is 36g of water. The lowering of vapour pressure of solution (assuming the solution is dilute)
of Hg, the value of x is-----------. (Integer answer)
224mL of SO2(g) at 298K and 1 atm passes through 100mL of 0.1M NaOH solution. The non-volatile solute produced is dissolved is 36g of water. The lowering of vapour pressure of solution (assuming the solution is dilute) of Hg, the value of x is-----------. (Integer answer)
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1 Answer
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Moles of SO2 =
=0.01 mole
Moles of NaOH = 0.1 × 0.1
= 0.01 mole
SO2+NaOHNaHSO3
0.01 mole0.01 mole-
-0.01 mole
Non-volatile solute is NaHSO3
Moles of water =
Using ; relative lowering in V.P
Where; is lowering in V.P
i for NaHSO3 = 2
here; since solution is dilute
So; x = 24
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