Question

The density of ethylene glycol (antifreeze, HOCH2CH2OH) is 1.09 g/mL. How
many grams of ethylene glycol should be mixed with 375 mL of water to make a
7.50% (v/v) mixture?

Answer 1

Answer :-

33.136 grams of ethylene glycol should be mixed .

Explanation :-

We have :-

→ Denisty of ethylene glycol = 1.09 g/mL

→ Volume of water = 375 mL

→ Concentration (v/v %) = 7.50 %

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Let the volume of ethylene glycol (solute) be 'x' mL .

So, volume of the solution :-

= Volume of solute + Volume of solvent

= (x + 375) mL

Concentration (v/v %) :-

= Vol. of solute/Vol. of solution × 100

⇒ x/(x + 375) × 100 = 7.5

⇒ 100x = 7.5(x + 375)

⇒ 100x = 7.5x + 2812.5

⇒ 100x - 7.5x = 2812.5

⇒ 92.5x = 2812.5

⇒ x = 2812.5/92.5

⇒ x = 30.4 mL

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Mass of ethylene glycol :-

= Volume × Density

= 30.4 × 1.09

= 33.136 g

Answer 2

$\large\underline{\sf{\red{Given}}}$

✭ Density of ethylene glycol = 1.09 g/mL

✭ Volume of water = 375 mL

✭ Concentration (v/v) = 7.5 %

$\large\underline{\sf{\blue{To\:Find}}}$

✭ How many grams of ethylene glycol should be mixed with 375 mL of water to make a 7.50% (v/v) mixture?

$\large\underline{\sf{\gray{Solution}}}$

◈ Let volume of solute (ethylene glycol) be 'v' mL

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Finding volume of the solution,

$\small\sf\Rightarrow\: Volume_{(solution)} = Volume_{(solute)} + Volume_{(solvent)}$

$\small\Rightarrow\:\sf Volume_{(solution)} = v + 375$

$\small\Rightarrow\:\sf \pink {Volume_{(solution)} = (v + 375)\:mL}$

Now,

➳ $\small\sf Concentration\:(v/v) = \dfrac{Volume_{(solute)}}{Volume_{(solution)}}\:\times\:100$

➳ $\small\sf 7.5 = \dfrac{v}{v + 375}\:\times\:100$

➳ $\small\sf 7.5 = \dfrac{100v}{v + 375}$

➳ $\small\sf 7.5(v + 375) = 100v$

➳ $\small\sf 7.5v + 2812.5 = 100v$

➳ $\small\sf 2812.5 = 100v - 7.5v$

➳ $\small\sf 2812.5 = 92.5v$

➳ $\small\sf v = \dfrac{2812.5}{92.5}$

➳ $\small\sf v = {\cancel{\dfrac{2812.5}{92.5}}}$

➳ $\small\sf\pink{ v = 30.4\:mL}$

We know that,

⪼ $\underline{\boxed{\sf{Mass = Volume\:\times\:Density}}}$

Finding mass of ethylene solution to be mixed,

$\small\sf \rightarrow\: Mass_{(ethylene\:glycol)} = Volume_{(ethylene\:glycol)} \:\times\: Density_{(ethylene\:glycol)}$

$\small\rightarrow\:\sf Mass_{(ethylene\:glycol)} = 30.4 \:\times\:1.09$

$\small\rightarrow\:\sf\pink{Mass_{(ethylene\:glycol)} = 33.136\:g}$

$\small\therefore\:{\underline{\sf{Mass\:of\:ethylene\:glycol\:to\:be\:mixed\:=\:{\textsf{\textbf{33.136\:g}}}}}}$

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