Solution:

Molarity of B in the solution:

Molarity is the number of moles of solute present in one liter of solution. In this case, we have solution A and B mixed together. We are given that the mole fraction of B is 0.6, which means that 60% of the total moles in the solution are due to B. We can assume that the total volume of the solution is 1 liter for simplicity.

Let nB be the number of moles of B present in the solution. Then, we know that:

nB = 0.6 * 1 liter * (density of solution) / (molecular weight of B)

We are given that the density of the solution is 0.7 g/ml. We also know the molecular weight of B is 10 g/mol. Substituting these values, we get:

nB = 0.6 * 1 liter * 0.7 g/ml / 10 g/mol = 0.042 moles

Therefore, the molarity of B in the solution is:

Molarity of B = nB / volume of solution = 0.042 moles / 1 liter = 0.042 M

Molality of B in the solution:

Molality is the number of moles of solute present in 1 kg of solvent. In this case, we are not given the mass of the solvent, but we can calculate it using the density of the solution and the mass of the solution.

Let mB be the number of moles of B present in 1 kg of solvent. Then, we know that:

mB = nB / mass of solvent in kg

We can calculate the mass of the solution as:

mass of solution = volume of solution * density of solution = 1 liter * 0.7 g/ml = 0.7 kg

The mass of solvent can be calculated as:

mass of solvent = mass of solution - mass of solute = 0.7 kg - (0.6 * 10 g) = 0.04 kg

Substituting these values, we get:

mB = 0.042 moles / 0.04 kg = 1.05 mol/kg

Therefore, the molality of B in the solution is:

Molality of B = 1.05 mol/kg