The evaporation of water from a solution in a single effect evaporator requires a certain amount of steam. In this case, we are given that 1kg of water needs to be evaporated, and we need to determine the amount of steam required.

The amount of steam required for evaporation can be calculated using the concept of latent heat of vaporization. The latent heat of vaporization is the amount of heat required to convert a liquid into a vapor at a constant temperature and pressure. In this case, we assume that the evaporation takes place at a constant temperature and pressure.

The latent heat of vaporization of water is approximately 2260 kJ/kg. This means that for every kilogram of water that needs to be evaporated, 2260 kJ of heat is required.

To convert the heat into steam, we need to consider the specific enthalpy of steam. The specific enthalpy of steam is the amount of heat required to convert a unit mass of liquid water into steam at a specific temperature and pressure.

The specific enthalpy of steam at atmospheric pressure (1 bar) and 100°C is approximately 2676 kJ/kg. This means that for every kilogram of steam produced, 2676 kJ of heat is required.

To calculate the amount of steam required, we can use the following equation:

Amount of steam required = Heat required / Specific enthalpy of steam

Substituting the values, we get:

Amount of steam required = 2260 kJ / 2676 kJ/kg

Amount of steam required = 0.844 kg

Therefore, the evaporation of 1kg of water from a solution in a single effect evaporator requires approximately 0.844 kg of steam.

Since option B, 1-1.3 kg, includes the calculated value of 0.844 kg, it is the correct answer.