The freezing point of a 4 % by weight aqueous | vedclass

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(D) For the same freezing point,the molality $(m)$ of both aqueous solutions must be equal,assuming both are non-electrolytes.$m_A = m_B$The formula f

Vedclass · Accepted Answer

$160$

Vedclass · Answer

(D) For the same freezing point,the molality $(m)$ of both aqueous solutions must be equal,assuming both are non-electrolytes.$m_A = m_B$The formula for molality is $m = \frac{	ext{mass of solute (g)} 	imes 1000}{	ext{molecular weight of solute} 	imes 	ext{mass of solvent (g)}}$.For a $4 \%$ solution of $A$,mass of $A = 4 \ g$ and mass of water $= 96 \ g$.For a $10 \%$ solution of $B$,mass of $B = 10 \ g$ and mass of water $= 90 \ g$.Substituting the values: $\frac{4 	imes 1000}{60 	imes 96} = \frac{10 	imes 1000}{M_B 	imes 90}$.Simplifying the equation: $\frac{4}{60 	imes 96} = \frac{10}{M_B 	imes 90}$.$M_B = \frac{10 	imes 60 	imes 96}{4 	imes 90} = \frac{57600}{360} = 160$.

The freezing point of a $4 \%$ by weight aqueous solution of $A$ is equal to the freezing point of a $10 \%$ by weight aqueous solution of $B$. If the molecular weight of $A$ is $60$,then the molecular weight of $B$ is:

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