How much KOH is needed to make 2 molality solution if water used was 200 gm.
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Answer: 22.4g
Explanation:
molality = moles of solute / kilograms of solvent
ATQ,
molality = 2
amount of solvent = 200gm = 0.2kg
∴ 2 = no. of moles of KOH/ 0.2
=> no. of moles of KOH = 2 x 0.2 = 0.4 mol
we know that amount of substance(in gm) = no. of moles x molar mass of substance
molar mass of KOH= 39+16+1 = 56 amu
∴, amount of KOH= 0.4 x 56 = 22.4g
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