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1 year ago

What is the molarity of a 1.44 (m/v)% solution of an HCOOOH (Formic acid)?

1 answer:

8 0

This problem is providing the mass-volume percent of a formic acid solution so its molarity is required and found to be 0.313 M after the following calculations.

<h3>Molarity</h3>

In chemistry, units of concentration provide a measurable understanding of the relationship between the relative amounts of both solute and solvent. In the case of molarity, one must relate moles of solute and liters of solution as follows:

$M=\frac{mol\ solute}{Volume\ solution \ in \ L}$

In such a way, when given this mass-volume percent of 1.44% for the formic acid in the solution, one can assume there is 100 mL of solution and 1.44 g of solute (formic acid), which means one must convert the volume to liters and the mass to moles with:

$mol\ solute=\frac{1.44g}{46.03g/mol} =0.0313mol\\\\Volume\ in \liters: 100mL*\frac{1L}{1000mL}=0.100L$

Finally, we plug in these numbers in the equation for the calculation of molarity:

$M=\frac{0.0313mol}{0.100L}\\\\M=0.313M$

Learn more about molarity: brainly.com/question/10053901

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3 years ago

What is the mass of 1.7 × 1023 atoms of zinc (Zn)?

Rudiy27

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The mathematical formula is:

Number of moles = $\frac{given mass in g}{molar mass}$ (1)

Number of zinc atoms is equal to $1.7\times10^{23}$ , by Avogadro number, number of moles can be calculated.

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$1.7\times10^{23}\times\frac{1 mol}{6.022\times10^{23} }$

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0.2822 mol = $\frac{mass in g}{65.4 g/mol}$

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