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  • What is the mass ratio of ethylene glycol <img alt="\left(\mathrm{C}_2 \mathrm{H}_6 \mathrm{O}_2\right., \text{molar mass }\left.=62 \mathrm{~g} / \mathrm{mol}\right)" src="https://entrancecorner.o

What is the mass ratio of ethylene glycol \left(\mathrm{C}_2 \mathrm{H}_6 \mathrm{O}_2\right., \text{molar mass }\left.=62 \mathrm{~g} / \mathrm{mol}\right) required for making 500 \mathrm{~g} of 0.25 molal aqueous solution and 250 \mathrm{~mL} of 0.25 molal aqueous solution?

Option: 1

1: 1


Option: 2

2: 1


Option: 3

1: 2


Option: 4

3: 1


Answers (1)

best_answer

Case I

\mathrm{x\; gm} \: \mathrm{C}_2 \mathrm{H}_6 \mathrm{O}_2\; \text{present}
0.25=\frac{\mathrm{x} / 62}{500-\mathrm{x}} \times 1000
125=\left(\frac{1000}{62}+0.25\right) \mathrm{x}\quad \dots(1)


Case II

\mathrm{y\; gm} \: \mathrm{C}_2 \mathrm{H}_6 \mathrm{O}_2\; \text{present}.
0.25=\frac{y / 62}{250-y} \times 1000
62.5-0.25 \mathrm{y}=\frac{1000}{62} y
62.5=\left(\frac{1000}{62}+0.25\right) \mathrm{y}\quad \dots(2)

equation (1) \div equation (2)
\mathrm{\frac{x}{y}= \frac{125}{62.5}= \frac{2}{1}}

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