Answer to Question #112036 in Chemistry for Joy

Question

1.How many moles of molecules are there in
a)35.5g of chlorine molecules
b)160gof bromine molecules
2.Find the relative atomic mass of Y (L=6.0×10²³) particles per mole.An atom of a certain element Y has mass of 6.5×10²³g
how many molecules are there in
a)0.5moles of oxygen
b)0.15moles of chlorine molecules
c)2moles of nitrogen molecules
4.calc the no of molecules in a)16g of oxygen molecules
b)240g of bromine molecules
c)0.1g of hydrogen molecules
d)106g of chlorine molecules

Answer 1

1. According to the equation:

n = m / Mr

where n - number of moles, m - mass, Mr - molecular weight.

a). n (Cl₂) = 35.5 g / 70.0 g/mol = 0.5 mol

b). n (Br2) = 160 g / 160.0 g/mol = 1 mol

2. The total mass of Y particles per mole equals:

m (Y) = 6.0 × 10²³ × 6.5 × 10^-23 g = 39 g

As 6.0 × 1023 particles are present in 1 mol, the atomic mass of Y is 39 g/mol.

3. According to Avogadro's number, 1 mol of a substance contains 6.0 × 10²³ particles.

a). N(O₂) = 0.5 mol × 6.0 × 10²³ = 3.0 × 10²³ molecules

b). N(Cl₂) = 0.15 mol × 6.0 × 10²³ = 9.0 × 10²² molecules

c). N(N2) = 2 mol × 6.0 × 10²³ = 1.2 × 10²⁴ molecules

4. According to the equation:

N = m × N_A / Mr

where N - number of particles, m - mass, N_A - Avogadro's number, Mr - molecular weight

a). N(O₂) = 16 g × 6.0 × 10²³ / 32 g/mol = 3.0 × 10²³ molecules

b). N(Br₂) = 240 g × 6.0 × 10²³ / 160 g/mol = 9 × 10²³ molecules

c). N(H₂) = 0.1 g × 6.0 × 10²³ / 2 g/mol = 3.0 × 10²² molecules

d). N(Cl₂) = 106 g × 6.0 × 10²³ / 70 g/mol = 9.1 × 10²³ molecules