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Question: The mass of glucose to be dissolved in \(50g\) of water to get \(0.3\) \(Molal\) solution is : A....

The mass of glucose to be dissolved in 50g50g of water to get 0.30.3 MolalMolal solution is :
A.27g27g
B.0.27g0.27g
C.2.7g2.7g
D.5.4g5.4g

Explanation

Solution

To solve this question we must remember the formula of molality . Molality is a quantity to measure concentration of a solution and it is calculated by number of moles of solute present in 1kg1kg of solvent and its unit is mol/kgmol/kg and 1mol/kg1mol/kg is also termed as 1Molal1Molal solution .

Complete step by step answer:
If nn is the number of moles of solute and wwis the mass of solvent in kilograms present in the solution then molality =nw = \dfrac{n}{w}.
In the given question water is solvent and glucose is solute and mass of water present in solution is 50g=501000kg=0.05kg50g = \dfrac{{50}}{{1000}}kg = 0.05kg and molality of the solution is 0.3Molal0.3Molal .
Now with the given data we can easily calculate the required number of moles of glucose.
n=0.3×0.05n = 0.3 \times 0.05 == 0.0150.015
So 0.0150.015 moles of glucose is required to make the solution of 0.3Molal0.3Molal . But in the given question we have to calculate mass of glucose so we need to convert it into grams .
We know that the number of moles is calculated by dividing the mass of the compound in grams divided by its molar mass .
The molar mass of glucose =180g = 180g
Therefore required mass of glucose ==molar mass of glucose ×\times number of moles of glucose
Required mass of glucose =(0.015×180)g = \left( {0.015 \times 180} \right)g =2.7g = 2.7g
Hence option (C) is the correct answer .

Note:
While calculating concentration terms like molarity or molality we should take care of units like in when we calculate molality we should put the value of mass of solvent in kilograms . Molarity is defined as the number of moles of solute present per litre of the solution so in the case of molarity we have to divide the number of moles to the volume of the whole solution .