Assertion: One molal aqueous solution of gluc | vedclass

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Question

(A) $Molality = \frac{	ext{Number of moles of solute}}{	ext{Weight of solvent in kg}}$For glucose $(C_6H_{12}O_6)$,the molar mass is $180 \, g/mol$.

Vedclass · Accepted Answer

If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

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(A) $Molality = \frac{	ext{Number of moles of solute}}{	ext{Weight of solvent in kg}}$For glucose $(C_6H_{12}O_6)$,the molar mass is $180 \, g/mol$.Number of moles of glucose = $\frac{180 \, g}{180 \, g/mol} = 1 \, mol$.Since the solvent weight is $1 \, kg$,the molality is $\frac{1 \, mol}{1 \, kg} = 1 \, m$.Thus,the assertion is correct.The definition of a one molal solution is a solution containing $1 \, mole$ of solute in $1000 \, g$ $(1 \, kg)$ of solvent,which makes the reason correct and the correct explanation for the assertion.

Assertion: One molal aqueous solution of glucose contains $180 \, g$ of glucose in $1 \, kg$ water.
Reason: Solution containing one mole of solute in $1000 \, g$ of solvent is called one molal solution.

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Assertion: One molal aqueous solution of glucose contains $180 \, g$ of glucose in $1 \, kg$ water.Reason: Solution containing one mole of solute in $1000 \, g$ of solvent is called one molal solution.