The weight of sodium carbonates required to prepare $700 m l$ of deci-molar solution is__

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Solution

A decimolar molar means one tenth of molar solution( $\frac{1}{10} M$ )
Given V $= 700$ ml= $0.7$ L
From the formula

Molarity $= \frac{M o l e s o f s o l u t e}{V o l u m e i n L i t r e}$

$\frac{1}{10} = \frac{M o l e s}{0.7}$

Moles of solute $= 0.07$

Mole number $= \frac{m a s s}{m o l a r m a s s}$

Molar mass of sodium

carbonate $N a_{2} C O_{3} = 23 \times 2 + 12 + 16 \times 3 = 106$ g

Mass required $= 106 \times 0.07 = 7.42$ g

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