Calculate the molality of this aquaeous sol'n = 48.2% by mass KBr sol'n??

how do you answer this? pls help me. thank u sooo much...

Answer:
48.2% means 48.2 g KBr per 100 g solution
so...
482g KBr per 1000g solution (1 liter)
482 g KBr(1mol KBr/119 g KBr per mole)=
482/119=4.05 mole KBr in 1 liter thus 4.05 Molar solution

kz
You are given 48.2 gm of KBr in 100 gm of solution.
48.2 gm Kbr + 51.8 gm H2O = 100 gm solution
Molality = moles per 1000 gm water
48.2g x 1000 / 51.8 = 930.5 gm KBr per 1000 gm water
930.5 / 119 = 7.82 moles of KBr
7.82 m (molality)
Molality = moles/litre.
% by mass = grams of solute in 1,000g of Solution
48.2% of 1,000g = 482g
KBr Mol.mass = 39 + 80 = 119g/mole.
Moles = Mass ÷ Mol.mass = 482 ÷ 119 = 4.050moles.
Molality = 4.05m.

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