If a 6.84% (w/v) solution of sucrose (molar | vedclass

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(B) For isotonic solutions,the osmotic pressure $(\pi)$ is equal,which implies their molar concentrations $(C)$ are equal at the same temperature.$C_1

Vedclass · Accepted Answer

$76$

Vedclass · Answer

(B) For isotonic solutions,the osmotic pressure $(\pi)$ is equal,which implies their molar concentrations $(C)$ are equal at the same temperature.$C_1 = C_2$$\frac{w_1}{M_1 	imes V_1} = \frac{w_2}{M_2 	imes V_2}$Given:For sucrose: $w_1 = 6.84 \ g$,$V_1 = 100 \ mL$,$M_1 = 342 \ g/mol$For thiocarbamide: $w_2 = 1.52 \ g$,$V_2 = 100 \ mL$,$M_2 = ?$Substituting the values:$\frac{6.84}{342 	imes 0.1} = \frac{1.52}{M_2 	imes 0.1}$$\frac{6.84}{342} = \frac{1.52}{M_2}$$0.02 = \frac{1.52}{M_2}$$M_2 = \frac{1.52}{0.02} = 76 \ g/mol$Thus,the molar mass of thiocarbamide is $76 \ g/mol$.

If a $6.84\% \ (w/v)$ solution of sucrose (molar mass $342 \ g/mol$) is isotonic with a $1.52\% \ (w/v)$ solution of thiocarbamide,what is the molar mass of thiocarbamide?

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