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Q.

What is the mass ration of ethylene glycol $(C_{2} H_{6} O_{2}$ molar mass = 62 g/mol) required for making 500 g for 0.25 molal aqueous solution and 250 ml of 0.25 molal aqueous solution?

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a

3:1

b

1:2

c

1:1

d

2:1

answer is B.

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Detailed Solution

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$\begin{array}{l} \frac{m_{1}}{m_{2}} = \frac{n_{1} / w_{1}}{n_{2} / w_{2}} \Rightarrow \frac{0.25}{0.25} = \frac{n_{1}}{n_{2}} \times \frac{250}{500} \\ \frac{n_{1}}{n_{2}} = \frac{2}{1} = \frac{(mass of solute)_{1}}{(mass of solute)_{2}} \end{array}$

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