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Molality

A 300 g, 30% ...

Question

A $300 g$ , $30 %$ (w/w) NaOH solution is mixed with $500 g$ $40 %$ (w/w) $N a O H$ solution. What is the molality of final solution?

A

1.422 m

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B

14.22 m

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C

15.22 m

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D

N o n e

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Solution

The correct option is B $14.22 m$
Mole = $\frac{T o t a l M a s s}{M o l a r m a s s}$
Mole of solute $(N a O H)$ $= \frac{290}{40} = 7.25$
Mass of solvent = Total mass - Mass of solute $= 800 - 290 = 510 g$
Molality = $\frac{M o l e s o f s o l u t e}{M a s s o f s o l v e n t i n k g}$
$= \frac{7.25}{510} \times 1000 = 14.22 m$

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