Question

The density of 3 M solution of NaCl is 1.25g mL^-1.
Calculate the molality of the solution.​

Answer 1

Q. The density of 3 M solution of NaCl is 1.25g mL^-1.Calculate the molality of the solution.

Ans:- 2.79 m

Explanation:-

$3 \: m \: solution \: means \: that \: 3 \: moles \: of \: nacl \: are \: present \: in \: 1l \: solution$

$mass \: of \: nacl \: in \: 1l \: solution \: = 3 \times 58.5$

$= 175.5g$

$molecular \: mass \: of \: nacl \: = 58.5$

$density \: of \: solution \: = 1.25g {ml}^{ - 1}$

$mass \: of \: 1l \: solution \: = 1000 \times 1.25$

$= 1250g$

$mass \: of \: water \: in \: solution \: = 1250 - 175.5$

$= 1074.5g$

$molality = \frac{mass \: of \: solute \: }{mass \: of \: solvent \: (in \: g)} \times 1000$

$= \frac{3}{1074.5} \times 1000$

$2.79m$

Answer 2

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Is in the pic

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