40% (w/v) of naoh solution (specific gravity = 1.12) is equivalent to : ( of naoh = 40 g/mol)
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40% (w/v) of NaCl solution, means 40g of NaCl is present in 100ml of solution.
number of mole of NaCl = 40/58.5
given, specific gravity = 1.12
we know, specific gravity is nothing, just relative density of a substance.
so, relative density of solution = 1.12
or, density of solution/density of water at 4°C = 1.12
or, density of solution / 1g/ml = 1.12
or, density of solution = 1.12 g/ml
mass of solution = volume of solution × density = 100ml × 1.12 g/ml = 112g
now, mass of solvent = 112g - 40g = 72g
molality = number of mole of solute/mass of solvent in kg
= (40/58.5) × 1000/72
= 40000/(58.5 × 72)
≈ 9.5m
hence, 40% (w/v) NaCl solution is equivalent to 9.5m molality of solution.
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