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GREYUIT [131]
3 years ago
8

A 2.567-g sample of a monoprotic acid was dissolved in water. It took 15.24 mL of a 0.1578 M NaOH solution to neutralize the aci

d. Calculate the molar mass of the acid.
Chemistry
1 answer:
olchik [2.2K]3 years ago
8 0

Answer:

The molar mass of the acid is 1067.42 g/mol  

Explanation:

A monotropic acid can donate one proton only in a acd base reaction thus

Monotropic acid + NaOH → NaSalt + H₂O

From the above reaction, one mole of NaOH reacts with one mole of monotropc acid

however there are 0.1578 M NaOH from 15.24 mL hence we have

Number of moles of NaOH = 0.1578 M/L × (15.24/1000) L = 0.002404872 M of NaOH

The number of moles of the monotropic acid = number of moles of NaOH = 0.0024 M. However Number of moles = mass/(molar mass)

Therefore molar mass = mass/number of moles = 2.567(2.4×10⁻³) = 1067.42 g/mol  

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What volume (in liters) of a solution contains 0.14 mol of KCl?
oksano4ka [1.4K]

Answer:

\boxed {\boxed {\sf 0.078 \ L }}

Explanation:

We are asked to find the volume of a solution given the moles of solute and molarity.

Molarity is a measure of concentration in moles per liter. It is calculated using the following formula:

molarity= \frac{moles \ of \ solute}{liters \ of \ solution}

We know there are 0.14 moles of potassium chloride (KCl), which is the solute. The molarity of the solution is 1.8 molar or 1.8 moles of potassium chloride per liter.

  • moles of solute = 0.14 mol KCl
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Substitute these values/variables into the formula.

1.8 \ mol \ KCl/ L = \frac { 0.14 \ mol \ KCl}{x}

We are solving for x, so we must isolate the variable. First, cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.

\frac {1.8 \ mol \ KCl/L}{1} = \frac{0.14 \ mol \ KCl}{x}

1.8 \ mol \ KCl/ L *x = 1*0.14 \ mol \ KCl

1.8 \ mol \ KCl/ L *x = 0.14 \ mol \ KCl

Now x is being multiplied by 1.8 moles of potassium chloride per liter. The inverse operation of multiplication is division, so we divide both sides by 1.8 mol KCl/L.

\frac {1.8 \ mol \ KCl/ L *x}{1.8 \ mol \ KCl/L} = \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}

x= \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}

The units of moles of potassium chloride cancel.

x= \frac{0.14 }{1.8 L}

x=0.07777777778 \ L

The original measurements of moles and molarity have 2 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 7 in the ten-thousandth place tells us to round the 7 up to a 8.

x \approx 0.078 \ L

There are approximately <u>0.078 liters of solution.</u>

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