Question

80 gm of NaoH is present in 500ml of water having density 1.2gm/ cm find molarity and molality​

Answer 1

Answer:

Mass of 1 mole of NaOH = 40g

There are 2 moles of NaOH in 80g

Molarity = $\frac {moles \: of \: solute} {volume \: of \: solvent in L}$

= $\frac {2} {0.5}$

= 4 mol/L

Density of solution = 1.2g/ml

Mass of solution = 500 * 1.2 = 600g

Mass of Solvent = 600-80 = 540g

Molality = $\frac {moles \: of \: solute} {mass \: of \: solvent \: in \: Kg}$

= $\frac {2} {0.54}$

= 3.7 molal

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