Question

Calculate the molality of an aqueous solution in which mass percentage of glucose is 30%​

Answer 1

30% mass percentage of glucose means that:

• In 100 gram of water, there is 30 grams of glucose.

Now, moles of glucose will be :

$moles \: = \dfrac{given \: mass}{molar \: mass}$

$\implies moles \: = \dfrac{30}{180}$

$\implies moles \: = \dfrac{1}{6} \: mol$

Now, molality will be :

$molality = \dfrac{moles}{weight \: of \: solvent(in \: kg)}$

$\implies molality = \dfrac{( \frac{1}{6} )}{( \frac{100}{1000} )}$

$\implies molality = \dfrac{10}{6}$

$\implies molality = 1.67 \: m$

So, solution is 1.67 molal.

Answer 2

Explanation:

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