Question

if 60g of NaoH mixed in 340g of water find its molality

Answer 1

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Formula to be used
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$= > \bf \: molality = \frac{w1}{m} \times \frac{10 00}{w2} \\ \\ where \: \: \: w1 = \: mass \: \: of \: \: solute \: \: in \: \: gm \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: w2 = mass \: \: of \: \: solvent \: \: in \: \: gm \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: m = molecular \: \: mass$
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=> Weight of solute (w1) =60 gm

=> molecular mass (m) = 23+16+1 = 40 gm

=> weight of solvent (w2) = 340 gm

So,

$= > molilty \: = \bf \: \frac{60}{40} \times \frac{1000}{340} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \bf \: 4.41 \: \: moles \: \: {kg}^{ - 1}$
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