20 mL of the pure formic acid HCOOH liquid is mixed with 40 mL of pure water to prepare a solution- If the density of pure formic acid liquid- pure water and the solution are 1-15 g - mL - 1 g - mL and 9-8 g - mL- then-A- The volume expansion upon mixing the two pure liquids is 3-33 -B- Mole fraction of formic acid in the solution is 9-50C- The volume contraction upon mixing the two pure liquids is 6-66 -D- - w - v of the resulting solution will be 38 1-3-

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20 mL of the ...

Question

$20 mL$ of the pure formic acid $(H C O O H)$ liquid is mixed with $40 mL$ of pure water to prepare a solution. If the density of pure formic acid liquid, pure water and the solution are $1.15 g/mL, 1 g / m L$ and $\frac{9}{8} g/mL$ , then:

The volume expansion upon mixing the two pure liquids is

3.33 %

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The volume contraction upon mixing the two pure liquids is

6.66 %

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Mole fraction of formic acid in the solution is

\frac{9}{50}

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% w / v

of the resulting solution will be

38 \frac{1}{3} %

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Solution

The correct option is B The volume contraction upon mixing the two pure liquids is $6.66 %$
$(V_{m i x})_{a c t u a l} = \frac{m_{m i x}}{d_{m i x}} = \frac{m_{H C O O H} + m_{H_{2} O}}{d_{m i x}}$
$= \frac{1.15 \times 20 + 1 \times 40}{9 / 8} = 56 mL$
$(V_{m i x})_{e x p e c t e d} = 20 + 40 = 60 mL$
$∴$ % contraction in volume
$= \frac{60 - 56}{60} \times 100 = 6.66 % X_{H C O O H} = \frac{\frac{1.15 \times 20}{46}}{\frac{1.15 \times 20}{46} + \frac{40}{18}} = \frac{9}{49}$
$\Rightarrow % w / v = \frac{w t_{H C O O H}}{V_{s o l}} \times 100 = \frac{23}{56} \times 100 = 41.07 %$

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20 mL of the pure formic acid (HCOOH) liquid is mixed with 40 mL of pure water to prepare a solution. If the density of pure formic acid liquid, pure water and the solution are 1.15 g/mL ,1 g/mL and 98 g/mL, then:

$20 mL$ of the pure formic acid $(H C O O H)$ liquid is mixed with $40 mL$ of pure water to prepare a solution. If the density of pure formic acid liquid, pure water and the solution are $1.15 g/mL, 1 g / m L$ and $\frac{9}{8} g/mL$ , then: