Question -
Shikhar was to perform a neutralization reaction in laboratory. In order to neutralize 500 ml of 2MNaOH solution he was supposed to add certain volume of 1MH3PO3 solution, but by mistake, he added the same volume of water instead. What is the molality of the resulting solution – Assume density of the original solution = 1.08g/cc Density of water = 1 g/cc.
Options:
A .  0.67
B .  1
C .  1.2
D .  2
Answer: Option B : B Wt. of original solution = Density x volume = 1.08g/cc×500ml=540g Wt. of NaOH + = No. of moles x molecular Wt. = 500×21000×40g=40g Wt. of solvent i.e water in original solution = 540-40 = 500 g if V2 be the volume of H3PO3 required for neutralizing the solution then- M1V1=M2V2 ⇒V2=M1V1M2=2×5001×2ml=500ml Note-Basicity of H3PO3=2 Wt. of water added = Density of water × Volume = 1gcc×500ml=500g Total Wt. of solvent i.e water = 500 + 500 = 1000g Molality of solution = MolesofsoluteWt.ofsolventinkg=40/401000/1000=1mole/kg Hence Optin-(b) is the correct Answer.
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