Assertion: One molal aqueous solution of gluc | vedclass

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(A) $	ext{Molality} = \frac{	ext{Number of moles of solute}}{	ext{Mass of solvent in kg}}$$	ext{Number of moles of glucose} = \frac{	ext{Given ma

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

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(A) $	ext{Molality} = \frac{	ext{Number of moles of solute}}{	ext{Mass of solvent in kg}}$$	ext{Number of moles of glucose} = \frac{	ext{Given mass}}{	ext{Molar mass}} = \frac{180 \ g}{180 \ g/mol} = 1 \ mol$.$	ext{Molality} = \frac{1 \ mol}{1 \ kg} = 1 \ m$.Since $1 \ mol$ of glucose $(180 \ g)$ is dissolved in $1 \ kg$ $(1000 \ g)$ of solvent,the solution is $1 \ molal$.Thus,both Assertion and Reason are correct,and the Reason is the correct explanation of the Assertion.

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

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