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3 years ago

an aqueous solution is 65% (v/v) rubbing alcohol. How many millilitres of water are in a 95-ml sample of this solution?

Chemistry

1 answer:

KengaRu [80]3 years ago

4 0

Answer:

The amount of water in a 95 mL sample of the solution is 33.25 mL

Explanation:

65% (v/v) rubbing alcohol is a solution containing 65 mL rubbing alcohol in 100 mL solution of the rubbing alcohol and water

Therefore, 95 mL sample of the solution will contain x mL of rubbing alcohol;

x = 65% × 95 = 0.65 × 95 = 61.75 mL rubbing alcohol and therefore, the amount of water contained = 95 - 61.75 = 33.25 mL

The amount of water in a 95 mL sample of the solution = 33.25 mL.

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Nitella [24]

Answer:

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Explanation:

The expression for Clausius-Clapeyron Equation is shown below as:

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Where,

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R is the gas constant (8.314×10⁻³ kJ /mol K)

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For two situations and phases, the equation becomes:

$\ln \left( \dfrac{P_1}{P_2} \right) = \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right)$

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T(K) = T( °C) + 273.15

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$T_2$ = 75°C = 348.15 K

So,

$\ln \:\left(\:\frac{13.95}{144.78}\right)\:=\:\frac{\Delta \:H_{vap}}{8.314}\:\left(\:\frac{1}{348.15}-\:\frac{1}{298.15}\:\right)$

$\Delta \:H_{vap}=\ln \left(\frac{13.95}{144.78}\right)\frac{8.314}{\left(\frac{1}{348.15}-\frac{1}{298.15}\right)}$

$\Delta \:H_{vap}=\frac{8.314}{\frac{1}{348.15}-\frac{1}{298.15}}\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)$

$\Delta \:H_{vap}=\left(-\frac{863000.86966\dots }{50}\right)\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)$

$\Delta \:H_{vap}=40383.88\ J/mol$

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