Question

2. (a) Calculate the number of molecules in 50 g of CaCO3.
(b) Calculate the mass of 0.5 moles of nitrogen gas.
(c) Calculate the number of moles in 50 g of NaCl. [Atomic mass of Ca = 40 u, C = 12 u,
O = 16 u, N = 14 u, Na = 23 u, Cl = 35.5 u, Na = 6.022 x 1023 mol-']​

Answer 1

Answer:

(a) The number of molecules in 50 g of $CaCO_{3}$ is $3.011*10^{23}$ molecules.

(b) The mass of 0.5 moles of nitrogen gas is 14 g.

(c) The number of moles in 50 g of NaCl is 0.85 moles.

Explanation:

(a) Number of moles = Given mass ÷ molar mass

Given mass = 50 g

Molar mass of $CaCO_{3}$ = (1×40) + (1×12) +(3×16)

                                    = 100 g/mol

Number of moles = $\frac{50}{100}$

                             = 0.5 moles

1 mole contains $6.022*10^{23}$ molecules

0.5 mole contains $0.5*6.022*10^{23}$ molecules

                             = $3.011*10^{23}$ molecules

(b) Number of moles = Given mass ÷ molar mass

Molar mass of $N_{2}$ = 28 g/mol

Given number of moles = 0.5 moles

Mass of $N_{2}$ = 0.5 × 28

                   = 14 g

(c) Number of moles = Given mass ÷ molar mass

Molar mass of NaCl = (1×23) +(1×35.5)

                                 = 58.5 g/mol

Given mass = 50 g

Number of moles = Given mass ÷ molar mass

                             =$\frac{50}{58.5}$

                             = 0.85 moles