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Chapter 3: Problem 35

Calculate the following quantities: (a) mass, in grams, of 0.105 mol sucrose $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)$ (b) moles of \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) in $143.50 \mathrm{~g}$ of this substance (c) number of molecules in $1.0 \times 10^{-6} \mathrm{~mol} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$ (d) number of \(\mathrm{N}\) atoms in \(0.410 \mathrm{~mol} \mathrm{NH}_{3}\)

Short Answer

Expert verified
(a) The mass of 0.105 mol of sucrose (C12H22O11) is 35.95 g. (b) There are 0.7574 moles of Zn(NO3)2 in 143.50 g of the substance. (c) There are 6.022 x 10^17 molecules in 1.0 x 10^-6 mol of CH3CH2OH. (d) There are 2.47 x 10^23 N atoms in 0.410 mol of NH3.

Step by step solution

01

a) Mass of 0.105 mol of sucrose in grams

First, we need to find the molar mass of sucrose (C12H22O11). The molar mass of each element can be found on the periodic table: C = 12.01 g/mol, H = 1.01 g/mol, O = 16.00 g/mol. Molar mass of sucrose: C12H22O11 = 12 * 12.01 g/mol + 22 * 1.01 g/mol + 11 * 16.00 g/mol = 144.12 g/mol + 22.22 g/mol + 176.00 g/mol = 342.34 g/mol Now, we will use the relationship between moles, mass, and molar mass: Mass = Moles × Molar mass Mass of 0.105 mol of sucrose = 0.105 mol × 342.34 g/mol = 35.95 g So, the mass of 0.105 mol of sucrose is 35.95 g.
02

b) Moles of Zn(NO3)2 in 143.50 g of this substance

First, we need to find the molar mass of Zn(NO3)2: Molar mass of Zn(NO3)2 = (Zn) + 2 * (N + 3 * O) Molar masses: Zn = 65.38 g/mol, N = 14.01 g/mol, O = 16.00 g/mol Molar mass of Zn(NO3)2 = 65.38 g/mol + 2 * (14.01 g/mol + 3 * 16.00 g/mol) = 65.38 g/mol + 2 * (62.01 g/mol) = 189.40 g/mol Now, we will use the relationship between moles, mass, and molar mass, with the mass of Zn(NO3)2 given: Moles = Mass ÷ Molar mass Moles of Zn(NO3)2 = 143.50 g ÷ 189.40 g/mol = 0.7574 mol So, there are 0.7574 moles of Zn(NO3)2 in 143.50 g of the substance.
03

c) Number of molecules in 1.0 x 10^-6 mol of CH3CH2OH

To find the number of molecules, we will use Avogadro's number, where 1 mole of any substance contains 6.022 x 10^23 particles (atoms, ions, or molecules). Number of molecules = Moles × Avogadro's number Number of molecules in 1.0 x 10^-6 mol of CH3CH2OH = 1.0 x 10^-6 mol × 6.022 x 10^23 molecules/mol = 6.022 x 10^17 molecules So, there are 6.022 x 10^17 molecules in 1.0 x 10^-6 mol of CH3CH2OH.
04

d) Number of N atoms in 0.410 mol of NH3

First, note that there is 1 N atom in each NH3 molecule. So, to find the number of N atoms, we can directly use Avogadro's number. Number of N atoms = Moles of NH3 × Avogadro's number Number of N atoms in 0.410 mol of NH3 = 0.410 mol × 6.022 x 10^23 atoms/mol = 2.47 x 10^23 N atoms So, there are 2.47 x 10^23 N atoms in 0.410 mol of NH3.

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Most popular questions from this chapter

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