A solution of a substance containing 1.05 g | vedclass

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(D) Two solutions are isotonic if they have the same osmotic pressure,which implies they have the same molar concentration $(C_1 = C_2)$.For the gluco

Vedclass · Accepted Answer

$63$

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(D) Two solutions are isotonic if they have the same osmotic pressure,which implies they have the same molar concentration $(C_1 = C_2)$.For the glucose solution,$3 \% (w/v)$ means $3 \ g$ of glucose in $100 \ mL$ of solution.The molar mass of glucose $(C_6H_{12}O_6)$ is $180 \ g/mol$.The molarity of the glucose solution is $C_2 = \frac{3 \ g / 180 \ g/mol}{0.1 \ L} = \frac{1}{6} \ M$.For the substance solution,$1.05 \ g$ is present in $100 \ mL$,so $C_1 = \frac{1.05 / M_{sub}}{0.1} = \frac{10.5}{M_{sub}}$.Equating the concentrations: $\frac{10.5}{M_{sub}} = \frac{1}{6}$.Therefore,$M_{sub} = 10.5 	imes 6 = 63 \ g/mol$.

$A$ solution of a substance containing $1.05 \ g$ per $100 \ mL$ was found to be isotonic with $3 \% (w/v)$ glucose solution. The molecular mass of the substance is :

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