The amount of anhydrous Na_2CO_3 present in 2 | vedclass

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(D) The molarity formula is given by $M = \frac{w}{M_m 	imes V(L)}$,where $w$ is the mass in grams,$M_m$ is the molar mass,and $V$ is the volume in l

Vedclass · Accepted Answer

$6.63$

Vedclass · Answer

(D) The molarity formula is given by $M = \frac{w}{M_m 	imes V(L)}$,where $w$ is the mass in grams,$M_m$ is the molar mass,and $V$ is the volume in liters.For $Na_2CO_3$,the molar mass $M_m = (2 	imes 23) + 12 + (3 	imes 16) = 106 \ g/mol$.Given $M = 0.25 \ M$ and $V = 250 \ mL = 0.25 \ L$.Substituting the values: $0.25 = \frac{w}{106 	imes 0.25}$.$w = 0.25 	imes 106 	imes 0.25 = 6.625 \ g$.Rounding to two decimal places,we get $6.63 \ g$.

The amount of anhydrous $Na_2CO_3$ present in $250 \ mL$ of $0.25 \ M$ solution is ........ $g$.

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