63 g of a compound (Mol. Wt. = 126) was diss | vedclass

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(B) Step $1$: Calculate the number of moles of the solute. Moles $= \frac{	ext{Mass}}{	ext{Molar Mass}} = \frac{63 \ g}{126 \ g/mol} = 0.5 \ mol$. S

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(B) Step $1$: Calculate the number of moles of the solute. Moles $= \frac{	ext{Mass}}{	ext{Molar Mass}} = \frac{63 \ g}{126 \ g/mol} = 0.5 \ mol$. Step $2$: Calculate the total mass of the solution. Mass of solution $= 	ext{Mass of solute} + 	ext{Mass of solvent} = 63 \ g + 500 \ g = 563 \ g$. Step $3$: Calculate the volume of the solution using density. Volume $= \frac{	ext{Mass}}{	ext{Density}} = \frac{563 \ g}{1.126 \ g/mL} = 500 \ mL = 0.5 \ L$. Step $4$: Calculate the molarity. Molarity $(M) = \frac{	ext{Moles of solute}}{	ext{Volume of solution in } L} = \frac{0.5 \ mol}{0.5 \ L} = 1.0 \ M$.

List-$I$ (Example)	List-$II$ (Type of Solution)
$A$. Humidity	$I$. Solid in solid
$B$. Alloys	$II$. Liquid in gas
$C$. Amalgams	$III$. Solid in gas
$D$. Smoke	$IV$. Liquid in solid

$63 \ g$ of a compound (Mol. Wt. $= 126$) was dissolved in $500 \ g$ of distilled water. The density of the resultant solution is $1.126 \ g / mL$. The molarity of the solution is: (in $M$)

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